Closed form for ${largeint}_0^1frac{ln^4(1+x)ln x}x , dx$
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Can someone compute
$$ int_0^1frac{ln^4(1+x)ln x}x ,dx$$
in closed form?
I conjecture that the answer can be expressed as a polynomial function with rational coefficients in constants of the form $operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $mathrm{Li}_n$ is the $n$th polylogarithm.
The reason for my conjecture is that
$$ int_0^1frac{ln^2(1+x)ln x}x ; dx =frac{pi^4}{24}-frac16ln^42+frac{pi^2}6ln^22-frac72zeta(3)ln2-4operatorname{Li}_4!left(frac12right) $$
and as shown here on Math StackExchange
$$ int_0^1frac{ln^3(1+x)ln x}x ; dx = frac{pi^2}3ln^32-frac25ln^52+frac{pi^2}2zeta(3)+frac{99}{16}zeta(5)-frac{21}4zeta(3)ln^22\-12operatorname{Li}_4left(frac12right)ln2-12operatorname{Li}_5left(frac12right).$$
The Riemann zeta function obeys
$$ zeta_n = operatorname{Li}_n(1) $$
and $pi^{2n}$ is a rational number times $zeta_{2n}$. Also,
$$ operatorname{Li}_1(x) = -ln(1 - x) .$$
So, the two integrals above are polynomials with rational coefficients in constants of the form $operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $mathrm{Li}_n$ is the $n$th polylogarithm. Maybe this pattern continues!
If my conjecture is true, next I'll ask about
$$ int_0^1frac{ln^k(1+x)ln x}x ,dx$$
for $k = 4, 5, 6, dots $
sequences-and-series definite-integrals closed-form harmonic-numbers polylogarithm
edited Aug 31 '17 at 4:43
Lucian
40.9k159130
asked Jun 27 '17 at 12:54
John Baez
26718
|
show 2 more comments
up vote
3
down vote
favorite
Can someone compute
$$ int_0^1frac{ln^4(1+x)ln x}x ,dx$$
in closed form?
I conjecture that the answer can be expressed as a polynomial function with rational coefficients in constants of the form $operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $mathrm{Li}_n$ is the $n$th polylogarithm.
The reason for my conjecture is that
$$ int_0^1frac{ln^2(1+x)ln x}x ; dx =frac{pi^4}{24}-frac16ln^42+frac{pi^2}6ln^22-frac72zeta(3)ln2-4operatorname{Li}_4!left(frac12right) $$
and as shown here on Math StackExchange
$$ int_0^1frac{ln^3(1+x)ln x}x ; dx = frac{pi^2}3ln^32-frac25ln^52+frac{pi^2}2zeta(3)+frac{99}{16}zeta(5)-frac{21}4zeta(3)ln^22\-12operatorname{Li}_4left(frac12right)ln2-12operatorname{Li}_5left(frac12right).$$
The Riemann zeta function obeys
$$ zeta_n = operatorname{Li}_n(1) $$
and $pi^{2n}$ is a rational number times $zeta_{2n}$. Also,
$$ operatorname{Li}_1(x) = -ln(1 - x) .$$
So, the two integrals above are polynomials with rational coefficients in constants of the form $operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $mathrm{Li}_n$ is the $n$th polylogarithm. Maybe this pattern continues!
If my conjecture is true, next I'll ask about
$$ int_0^1frac{ln^k(1+x)ln x}x ,dx$$
for $k = 4, 5, 6, dots $
sequences-and-series definite-integrals closed-form harmonic-numbers polylogarithm
edited Aug 31 '17 at 4:43
Lucian
40.9k159130
asked Jun 27 '17 at 12:54
John Baez
26718
Your conjecture is based on those two examples? Or do you have more clues to guess the closed form of the integrals you are considering?
– user37238
Jun 27 '17 at 13:05
We have $$ log^4(1-x)=sum_{ngeq 1}frac{4x^n}{n}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right] tag{1}$$ hence the given integral equals $$ I=sum_{ngeq 1}frac{4(-1)^{n+1}}{n^3}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right]tag{2} $$
– Jack D'Aurizio
Jun 27 '17 at 13:15
Plenty of Euler sums are involved. What is the purpose of reaching a closed form depending on values of $zeta$ at the integers and $text{Li}_n$ at $frac{1}{2}$?
– Jack D'Aurizio
Jun 27 '17 at 13:16
Might be a fun problem to apply the PSQL algorithm to: crd-legacy.lbl.gov/~dhbailey/dhbpapers/ppslq.pdf
– user14717
Jun 27 '17 at 17:24
5
$$ int_0^1frac{ln^4(1+x)ln x}x ; dx = frac{pi^2}3ln^42-frac{17}{30}ln^62-frac{pi^4}{60}ln^22+frac{26}{315}{pi^6}-2{pi^2}zeta(3)ln2+12(zeta(3))^2+frac{3}{4}zeta(5)ln2-3zeta(3)ln^32-24operatorname{Li}_4left(frac12right)ln^22-72operatorname{Li}_5left(frac12right)ln2-96operatorname{Li}_6left(frac12right)+12S.$$ $$ S=sum^infty_{n=1}frac{H_n}{(n+1)^52^n}=0.017446006115193776854329$$
– user178256
Jun 28 '17 at 6:04
|
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Can someone compute
$$ int_0^1frac{ln^4(1+x)ln x}x ,dx$$
in closed form?
I conjecture that the answer can be expressed as a polynomial function with rational coefficients in constants of the form $operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $mathrm{Li}_n$ is the $n$th polylogarithm.
The reason for my conjecture is that
$$ int_0^1frac{ln^2(1+x)ln x}x ; dx =frac{pi^4}{24}-frac16ln^42+frac{pi^2}6ln^22-frac72zeta(3)ln2-4operatorname{Li}_4!left(frac12right) $$
and as shown here on Math StackExchange
$$ int_0^1frac{ln^3(1+x)ln x}x ; dx = frac{pi^2}3ln^32-frac25ln^52+frac{pi^2}2zeta(3)+frac{99}{16}zeta(5)-frac{21}4zeta(3)ln^22\-12operatorname{Li}_4left(frac12right)ln2-12operatorname{Li}_5left(frac12right).$$
The Riemann zeta function obeys
$$ zeta_n = operatorname{Li}_n(1) $$
and $pi^{2n}$ is a rational number times $zeta_{2n}$. Also,
$$ operatorname{Li}_1(x) = -ln(1 - x) .$$
So, the two integrals above are polynomials with rational coefficients in constants of the form $operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $mathrm{Li}_n$ is the $n$th polylogarithm. Maybe this pattern continues!
If my conjecture is true, next I'll ask about
$$ int_0^1frac{ln^k(1+x)ln x}x ,dx$$
for $k = 4, 5, 6, dots $
sequences-and-series definite-integrals closed-form harmonic-numbers polylogarithm
edited Aug 31 '17 at 4:43
Lucian
40.9k159130
asked Jun 27 '17 at 12:54
John Baez
26718
Can someone compute
$$ int_0^1frac{ln^4(1+x)ln x}x ,dx$$
in closed form?
I conjecture that the answer can be expressed as a polynomial function with rational coefficients in constants of the form $operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $mathrm{Li}_n$ is the $n$th polylogarithm.
The reason for my conjecture is that
$$ int_0^1frac{ln^2(1+x)ln x}x ; dx =frac{pi^4}{24}-frac16ln^42+frac{pi^2}6ln^22-frac72zeta(3)ln2-4operatorname{Li}_4!left(frac12right) $$
and as shown here on Math StackExchange
$$ int_0^1frac{ln^3(1+x)ln x}x ; dx = frac{pi^2}3ln^32-frac25ln^52+frac{pi^2}2zeta(3)+frac{99}{16}zeta(5)-frac{21}4zeta(3)ln^22\-12operatorname{Li}_4left(frac12right)ln2-12operatorname{Li}_5left(frac12right).$$
The Riemann zeta function obeys
$$ zeta_n = operatorname{Li}_n(1) $$
and $pi^{2n}$ is a rational number times $zeta_{2n}$. Also,
$$ operatorname{Li}_1(x) = -ln(1 - x) .$$
So, the two integrals above are polynomials with rational coefficients in constants of the form $operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $mathrm{Li}_n$ is the $n$th polylogarithm. Maybe this pattern continues!
If my conjecture is true, next I'll ask about
$$ int_0^1frac{ln^k(1+x)ln x}x ,dx$$
for $k = 4, 5, 6, dots $
sequences-and-series definite-integrals closed-form harmonic-numbers polylogarithm
sequences-and-series definite-integrals closed-form harmonic-numbers polylogarithm
edited Aug 31 '17 at 4:43
Lucian
40.9k159130
asked Jun 27 '17 at 12:54
John Baez
26718
edited Aug 31 '17 at 4:43
Lucian
40.9k159130
asked Jun 27 '17 at 12:54
John Baez
26718
edited Aug 31 '17 at 4:43
Lucian
40.9k159130
edited Aug 31 '17 at 4:43
Lucian
40.9k159130
edited Aug 31 '17 at 4:43
Lucian
40.9k159130
40.9k159130
asked Jun 27 '17 at 12:54
John Baez
26718
asked Jun 27 '17 at 12:54
John Baez
26718
asked Jun 27 '17 at 12:54
John Baez
26718
26718
Your conjecture is based on those two examples? Or do you have more clues to guess the closed form of the integrals you are considering?
– user37238
Jun 27 '17 at 13:05
We have $$ log^4(1-x)=sum_{ngeq 1}frac{4x^n}{n}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right] tag{1}$$ hence the given integral equals $$ I=sum_{ngeq 1}frac{4(-1)^{n+1}}{n^3}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right]tag{2} $$
– Jack D'Aurizio
Jun 27 '17 at 13:15
Plenty of Euler sums are involved. What is the purpose of reaching a closed form depending on values of $zeta$ at the integers and $text{Li}_n$ at $frac{1}{2}$?
– Jack D'Aurizio
Jun 27 '17 at 13:16
Might be a fun problem to apply the PSQL algorithm to: crd-legacy.lbl.gov/~dhbailey/dhbpapers/ppslq.pdf
– user14717
Jun 27 '17 at 17:24
5
$$ int_0^1frac{ln^4(1+x)ln x}x ; dx = frac{pi^2}3ln^42-frac{17}{30}ln^62-frac{pi^4}{60}ln^22+frac{26}{315}{pi^6}-2{pi^2}zeta(3)ln2+12(zeta(3))^2+frac{3}{4}zeta(5)ln2-3zeta(3)ln^32-24operatorname{Li}_4left(frac12right)ln^22-72operatorname{Li}_5left(frac12right)ln2-96operatorname{Li}_6left(frac12right)+12S.$$ $$ S=sum^infty_{n=1}frac{H_n}{(n+1)^52^n}=0.017446006115193776854329$$
– user178256
Jun 28 '17 at 6:04
|
show 2 more comments
Your conjecture is based on those two examples? Or do you have more clues to guess the closed form of the integrals you are considering?
– user37238
Jun 27 '17 at 13:05
We have $$ log^4(1-x)=sum_{ngeq 1}frac{4x^n}{n}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right] tag{1}$$ hence the given integral equals $$ I=sum_{ngeq 1}frac{4(-1)^{n+1}}{n^3}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right]tag{2} $$
– Jack D'Aurizio
Jun 27 '17 at 13:15
Plenty of Euler sums are involved. What is the purpose of reaching a closed form depending on values of $zeta$ at the integers and $text{Li}_n$ at $frac{1}{2}$?
– Jack D'Aurizio
Jun 27 '17 at 13:16
Might be a fun problem to apply the PSQL algorithm to: crd-legacy.lbl.gov/~dhbailey/dhbpapers/ppslq.pdf
– user14717
Jun 27 '17 at 17:24
5
$$ int_0^1frac{ln^4(1+x)ln x}x ; dx = frac{pi^2}3ln^42-frac{17}{30}ln^62-frac{pi^4}{60}ln^22+frac{26}{315}{pi^6}-2{pi^2}zeta(3)ln2+12(zeta(3))^2+frac{3}{4}zeta(5)ln2-3zeta(3)ln^32-24operatorname{Li}_4left(frac12right)ln^22-72operatorname{Li}_5left(frac12right)ln2-96operatorname{Li}_6left(frac12right)+12S.$$ $$ S=sum^infty_{n=1}frac{H_n}{(n+1)^52^n}=0.017446006115193776854329$$
– user178256
Jun 28 '17 at 6:04
Your conjecture is based on those two examples? Or do you have more clues to guess the closed form of the integrals you are considering?
– user37238
Jun 27 '17 at 13:05
Your conjecture is based on those two examples? Or do you have more clues to guess the closed form of the integrals you are considering?
– user37238
Jun 27 '17 at 13:05
We have $$ log^4(1-x)=sum_{ngeq 1}frac{4x^n}{n}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right] tag{1}$$ hence the given integral equals $$ I=sum_{ngeq 1}frac{4(-1)^{n+1}}{n^3}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right]tag{2} $$
– Jack D'Aurizio
Jun 27 '17 at 13:15
We have $$ log^4(1-x)=sum_{ngeq 1}frac{4x^n}{n}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right] tag{1}$$ hence the given integral equals $$ I=sum_{ngeq 1}frac{4(-1)^{n+1}}{n^3}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right]tag{2} $$
– Jack D'Aurizio
Jun 27 '17 at 13:15
Plenty of Euler sums are involved. What is the purpose of reaching a closed form depending on values of $zeta$ at the integers and $text{Li}_n$ at $frac{1}{2}$?
– Jack D'Aurizio
Jun 27 '17 at 13:16
Plenty of Euler sums are involved. What is the purpose of reaching a closed form depending on values of $zeta$ at the integers and $text{Li}_n$ at $frac{1}{2}$?
– Jack D'Aurizio
Jun 27 '17 at 13:16
Might be a fun problem to apply the PSQL algorithm to: crd-legacy.lbl.gov/~dhbailey/dhbpapers/ppslq.pdf
– user14717
Jun 27 '17 at 17:24
Might be a fun problem to apply the PSQL algorithm to: crd-legacy.lbl.gov/~dhbailey/dhbpapers/ppslq.pdf
– user14717
Jun 27 '17 at 17:24
5
5
$$ int_0^1frac{ln^4(1+x)ln x}x ; dx = frac{pi^2}3ln^42-frac{17}{30}ln^62-frac{pi^4}{60}ln^22+frac{26}{315}{pi^6}-2{pi^2}zeta(3)ln2+12(zeta(3))^2+frac{3}{4}zeta(5)ln2-3zeta(3)ln^32-24operatorname{Li}_4left(frac12right)ln^22-72operatorname{Li}_5left(frac12right)ln2-96operatorname{Li}_6left(frac12right)+12S.$$ $$ S=sum^infty_{n=1}frac{H_n}{(n+1)^52^n}=0.017446006115193776854329$$
– user178256
Jun 28 '17 at 6:04
$$ int_0^1frac{ln^4(1+x)ln x}x ; dx = frac{pi^2}3ln^42-frac{17}{30}ln^62-frac{pi^4}{60}ln^22+frac{26}{315}{pi^6}-2{pi^2}zeta(3)ln2+12(zeta(3))^2+frac{3}{4}zeta(5)ln2-3zeta(3)ln^32-24operatorname{Li}_4left(frac12right)ln^22-72operatorname{Li}_5left(frac12right)ln2-96operatorname{Li}_6left(frac12right)+12S.$$ $$ S=sum^infty_{n=1}frac{H_n}{(n+1)^52^n}=0.017446006115193776854329$$
– user178256
Jun 28 '17 at 6:04
|
show 2 more comments
4 Answers
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The Stirling numbers of the first kind $left[ begin{array}{c} n \ k end{array} right]$ are usually defined by: $$ sumlimits_{k=0}^n left[ begin{array}{c} n \ k end{array} right] x^k := x(x+1)…(x+n-1) $$
The definition of $enspaceeta_n(m)enspace$ in $enspace$ Evaluate $int_{0}^{pi }theta ^{3}log^{3}left ( 2sinfrac{theta }{2} right )mathrm{d}theta $ is
$$eta_n(m):=sumlimits_{k=1}^infty frac{(-1)^{k-1}}{k^m}left(frac{n!}{(k-1)!}left[begin{array}{c} k \ n+1 end{array} right]right)$$
with $enspace m>0$, $enspace ninmathbb{N}_0$ , $enspaceeta_0(m)=eta(m)$
and $enspacedisplaystyle frac{1}{(k-1)!} left[ begin{array}{c} k \ {n+1} end{array} right]= sumlimits_{i_1=1}^{k-1}sumlimits_{i_2=i_1+1}^{k-1}…sumlimits_{i_n=i_{n-1}+1}^{k-1}frac{1}{i_1 i_2 … i_n}$ .
To combine this series with the polylogarithm is a separate problem.
Now we can write:
$displaystyle sumlimits_{k=1}^infty frac{z^k}{k!}intlimits_0^1 frac{ln^k(1+x)ln x}{x} dx = intlimits_0^1 frac{((1+x)^z-1)ln x}{x} dx = - sumlimits_{k=1}^infty binom z k frac{1}{k^2}$
$hspace{5.3cm}displaystyle = - sumlimits_{k=1}^infty z^k sumlimits_{v=k}^infty frac{(-1)^{k-v}}{v^2 v!} left[ begin{array}{c} v \ k end{array} right]= sumlimits_{k=1}^infty frac{(-z)^k}{(k-1)!} mu_{k-1}(3)$
It follows
$$intlimits_0^1 frac{ln^k(1+x)ln x}{x} dx = (-1)^k,k,mu_{k-1}(3)$$
for $kinmathbb{N}$ .
answered Jun 28 '17 at 9:26
user90369
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It's not a complete answer but it is too lengthy for a comment.
@user14717:
Program GP PARI contains a routine to perform something like PSQL stuff.
Here a script
p 200
A1=Pi^6
A2=Pi^4*log(2)^2
A3=Pi^2*log(2)^4
A4=polylog(6,1/2)
A5=polylog(5,1/2)*log(2)
A6=polylog(4,1/2)*log(2)^2
A7=polylog(3,1/2)*log(2)^3
A8=polylog(2,1/2)*log(2)^4
A9=log(2)^6
A10=zeta(3)*log(2)^3
A11=zeta(5)*log(2)
A12=zeta(3)^2
A13=Pi^2*log(2)*zeta(3)
A14=polylog(3,1/2)*Pi^2*log(2)
A15=polylog(4,1/2)*Pi^2
A16=polylog(3,1/2)^2
A17=polylog(2,1/2)^2*log(2)^2
A18=polylog(2,1/2)^2*Pi^2
A19=polylog(2,1/2)^3
A20=polylog(2,1/2)*Pi^2*log(2)^2
A21=polylog(2,1/2)*Pi^4
A22=polylog(3,1/2)*zeta(3)
J=intnum(x=0,1,log(1+x)^4*log(x)/x)
lindep([J,A1,A2,A4,A5,A6,A7,A8,A9,A11,A12,A13,A15,A22])
Last command returns an integer relation of Ai's equals to 0.
Notice that some Ai's are linearly dependant on integers.
Anyway i wasn't able to find out such integer relation using these constants.
answered Jun 29 '17 at 19:11
FDP
4,47411221
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1
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I am very skeptical as to whether the conjecture is true (except of course when $k=2,3$). Let me explain why.
Note that the following identity holds:
begin{eqnarray}
&&-2intlimits_0^1 frac{log(1+x)^k}{x} log(x) dx=\
&&imath pi sumlimits_{l=1}^{k+1} (-1)^l k_{(l-1)} (Li_l(2)-Li_{k+1}(1)1_{l=k+1}) log[2]^{k+1-l}+\
&&sumlimits_{l=2}^k (-1)^{l-1} k_{(l-1)} intlimits_1^2 frac{Li_{l}(x)}{1-x} [log(x)]^{k+1-l} dx+\
&&(-1)^{k+1} k! intlimits_0^1 frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx
end{eqnarray}
The identity above comes from the knowledge of the anti-derivative of the fraction in the integrand and from integrating by parts once.
Now, if we take a look at the last integral above we can actualy write down the antiderivative. We have:
begin{eqnarray}
&&int frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx=\
&&left{
begin{array}{rr}
sumlimits_{l=1}^{k/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{k/2+1} frac{1}{2} [Li_{k/2+1}(1+x)]^2 - Li_{k+1}(1) log(x)& mbox{if $k$ is even}\
sumlimits_{l=1}^{(k+1)/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{(k+1)/2+1} int frac{[Li_{(k+1)/2}(1+x)]^2}{1+x} dx - Li_{k+1}(1) log(x)& mbox{if $k$ is odd}
end{array}
right.
end{eqnarray}
As we can see it is only when $k$ is even that the result reduces to poly-logarithms only otherwise a new unknown antiderivative involving a square of a polylogarithm remains. As a matter of fact it is not difficult to see that by integrating by parts and changing and swapping order of integration appropriately all the remaining integrals in the third line from the top in the first formula above reduce to poly-logarithms and following quantities:
begin{equation}
S^{(2,p)}_q := intlimits_1^2 frac{[Li_q(x)]^2}{x}cdot [log(x)]^p dx
end{equation}
where $pge1$ and $qge 1$.
Let us now take $k=4$ as an example. Here the integrals in question read:
begin{eqnarray}
intlimits_1^2 frac{Li_2(x)}{1-x} cdot [log(x)]^3 dx &=& -frac{3}{2} text{Li}_2(2){}^2 log ^2(2)-i pi text{Li}_2(2) log ^3(2) + 3 S^{(2,1)}_2 - S^{(2,3)}_1\
intlimits_1^2 frac{Li_3(x)}{1-x} cdot [log(x)]^2 dx &=& text{Li}_3(2){}^2-2 text{Li}_2(2) text{Li}_3(2) log (2)-zeta (3)^2-frac{1}{32} left(pi ^4+28 i pi zeta (3)right) log ^2(2) + 3 S^{(2,1)}_2\
intlimits_1^2 frac{Li_4(x)}{1-x} cdot [log(x)]^1 dx &=& text{Li}_3(2){}^2-frac{pi ^2 text{Li}_4(2)}{4}-text{Li}_2(2) text{Li}_3(2) log (2)-zeta (3)^2+frac{pi ^6}{540} + S^{(2,1)}_2\
intlimits_0^1 frac{Li_5(1+x)-Li_5(1)}{x} &=& frac{1}{2} left(-text{Li}_3(2){}^2+2 text{Li}_2(2) text{Li}_4(2)+zeta (3)^2right)+i pi (text{Li}_5(2)-zeta (5))-frac{pi ^6}{540}
end{eqnarray}
Now, when we bring everything together the quantities $S^{(2,1)}_2$ miraculously cancel out and then by substituting $xleftarrow 1/x$ in the quantity $S^{(2,3)}_1$ we obtain the following result:
begin{eqnarray}
&&intlimits_0^1 frac{[log(1+x)]^4}{x} cdot log(x) dx =\
&&-log (2) left(96 text{Li}_5left(frac{1}{2}right)+48 text{Li}_4left(frac{1}{2}right) log (2)+14 zeta (3) log ^2(2)+log ^5(2)right)+\
&&-96 text{Li}_6left(frac{1}{2}right)+frac{32 pi ^6}{315}+pi ^2 log ^4(2)+2 intlimits_{1/2}^1 frac{[log(1-x)]^2}{x} cdot [log(x)]^3 dx=\
&&12 left(zeta (3)^2-10 text{Li}_6left(frac{1}{2}right)right)+frac{3}{4} log (2) left(zeta (5)-96 text{Li}_5left(frac{1}{2}right)right)-24 text{Li}_4left(frac{1}{2}right) log ^2(2)+frac{1}{3} pi ^2
left(log ^4(2)-6 zeta (3) log (2)right)-3 zeta (3) log ^3(2)+frac{26 pi ^6}{315}-frac{17 log ^6(2)}{30}-frac{1}{60} pi ^4 log ^2(2) + 24 {bf H}^{(1)}_5(1/2)
end{eqnarray}
where in the last line we expressed the remaining integral through Euler sums and used the following results:
begin{eqnarray}
{bf H}^{(1)}_2(1/2) &=& zeta (3)-frac{1}{12} pi ^2 log (2)\
{bf H}^{(1)}_3(1/2) &=& text{Li}_4left(frac{1}{2}right)-frac{1}{8} zeta (3) log (2)+frac{pi ^4}{720}+frac{log ^4(2)}{24}\
{bf H}^{(1)}_4(1/2) &=& 2 text{Li}_5left(frac{1}{2}right)+text{Li}_4left(frac{1}{2}right) log (2)+frac{zeta (5)}{32}-frac{1}{36} pi ^2 left(3 zeta (3)+log ^3(2)right)+frac{1}{2} zeta (3) log ^2(2)+frac{log
^5(2)}{40}-frac{1}{720} pi ^4 log (2)
end{eqnarray}
Now, the question whether ${bf H}^{(1)}_5(1/2)$ is expressible in terms of poly-logarithms is most likely answered negatively. Indeed I used the web interface http://wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi to find possible linear dependencies. The code
lindep([ zp(2,6)+zp(2,5,1), z(6), z(3)*z(3), z(3)*z(2)*log(2), z(3)*log(2)^3, z(2)*log(2)^4, z(2)^2*log(2)^2, z(5)*log(2), log(2)^6, zp(2,4)*log(2)^2, zp(2,5)*log(2), zp(2,6)])
being run for two different number of digits of precision gives completely different results which suggests that such linear dependency most likely does not exist.
answered Dec 13 '17 at 13:24
Przemo
4,1321928
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[#ffd,10px]{ds{%
int_{0}^{1}{ln^{4}pars{1 + x}lnpars{x} over x},dd x}}
,,,stackrel{1 + x mapsto x}{=},,,
int_{1}^{2}{ln^{4}pars{x}lnpars{x - 1} over x - 1},dd x
\[5mm] stackrel{x mapsto 1/x}{=},,,&
int_{1}^{1/2}{ln^{4}pars{1/x}lnpars{1/x - 1} over 1/x - 1}
,pars{-,{dd x over x^{2}}} =
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} - ln^{5}pars{x} over
xpars{1 - x}},dd x
\[1cm] = &
underbrace{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x}
_{ds{mc{I}_{1}}} -
underbrace{int_{1/2}^{1}{ln^{5}pars{x} over x},dd x}
_{ds{= -,{ln^{6}pars{2} over 6}}} +
underbrace{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over 1 - x},dd x}
_{ds{mc{I}_{2}}}
\[1mm] &
-int_{1/2}^{1}{ln^{5}pars{x} over 1 - x},dd x
label{1}tag{1}
end{align}
Note that
begin{align}
int_{1/2}^{1}{ln^{5}pars{x} over 1 - x},dd x & =
-ln^{6}pars{2} +
5int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x =
-ln^{6}pars{2} + 5,mc{I}_{1}
end{align}
such that eqref{1} becomes
begin{equation}
bbx{bbox[#ffd,10px]{ds{%
int_{0}^{1}{ln^{4}pars{1 + x}lnpars{x} over x},dd x}} =
{7ln^{6}pars{2} over 6} - 4,mc{I}_{1} + mc{I}_{2}}label{2}tag{2}
end{equation}
$ds{Hugemc{I}_{1}: ?}$.
begin{align}
mc{I}_{1} & equiv
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x =
-int_{1/2}^{1}mrm{Li}_{2}'pars{x}ln^{4}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4int_{1/2}^{1}mrm{Li}_{3}'pars{x}ln^{3}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} -
12int_{1/2}^{1}mrm{Li}_{4}'pars{x}ln^{2}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24int_{1/2}^{1}mrm{Li}_{5}'pars{x}lnpars{x},dd x
\[1cm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24,mrm{Li}_{5}pars{1 over 2}lnpars{2}
\[1mm] & - 24int_{1/2}^{1}mrm{Li}_{6}'pars{x},dd x
\[1cm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24,mrm{Li}_{5}pars{1 over 2}lnpars{2}
\[1mm] & -24,mrm{Li}_{6}pars{1} + 24,mrm{Li}_{6}pars{1 over 2}
end{align}
Note that
$ds{mrm{Li}_{2}pars{1 over 2} = {pi^{2} over 12} -
{ln^{2}pars{2} over 2}}$ and
$ds{mrm{Li}_{3}pars{1 over 2} = {ln^{3}pars{2} over 6} -
{pi^{2}lnpars{2} over 12} + {7zetapars{3} over 8}}$. Moreover,
$ds{vphantom{Huge A}mrm{Li}_{6}pars{1} = zetapars{6} =
{pi^{6} over 945}}$.
Then,
$$
begin{array}{|rcl|}hline mbox{}&&\
ds{quadmc{I}_{1}} & ds{equiv} &
ds{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x}
\[5mm] & ds{=} &
ds{-,{8pi^{6} over 315} - {pi^{2}ln^{4}pars{2} over 4} +
{ln^{6}pars{2} over 6} + 12ln^{2}pars{2},mrm{Li}_{4}pars{1 over 2} +
24lnpars{2},mrm{Li}_{5}pars{1 over 2}quad}
\[1mm] &&
ds{%
+ 24,mrm{Li}_{6}pars{1 over 2} + {7ln^{3}pars{2},zetapars{3} over 2}
approx -0.0269}
\ mbox{}&&
\ hline
end{array}
$$
$ds{Hugemc{I}_{2}: ?}$. This one isn't trivial at all. An attempt is given by
begin{align}
mc{I}_{2} & equiv
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over 1 - x},dd x
,,,stackrel{x mapsto 1 - x}{=},,,
int_{0}^{1/2}{ln^{4}pars{1 - x}lnpars{x} over x},dd x
\[5mm] & =
left.partiald[4]{}{mu}partiald{}{nu}
int_{0}^{1/2}bracks{pars{1 - x}^{mu} - 1}x^{nu - 1}
,dd x,rightvert_{ mu = 0,, nu = 0^{+}}
\[5mm] & =
partiald[4]{}{mu}partiald{}{nu}
bracks{int_{0}^{1/2}pars{1 - x}^{mu},x^{nu - 1}
,dd x - {1 over 2^{nu}nu}}_{ mu = 0,, nu = 0^{+}}
\[5mm] & =
partiald[4]{}{mu}partiald{}{nu}
bracks{mrm{B}pars{{1 over 2},nu,1 + mu} - {1 over 2^{nu}nu}}_{ mu = 0,, nu = 0^{+}}
end{align}
$ds{mrm{B}}$ is the Incomplete Beta Function. It'll continue$ldots$
answered Feb 1 at 23:17
Felix Marin
66.4k7107139
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4 Answers
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4 Answers
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up vote
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The Stirling numbers of the first kind $left[ begin{array}{c} n \ k end{array} right]$ are usually defined by: $$ sumlimits_{k=0}^n left[ begin{array}{c} n \ k end{array} right] x^k := x(x+1)…(x+n-1) $$
The definition of $enspaceeta_n(m)enspace$ in $enspace$ Evaluate $int_{0}^{pi }theta ^{3}log^{3}left ( 2sinfrac{theta }{2} right )mathrm{d}theta $ is
$$eta_n(m):=sumlimits_{k=1}^infty frac{(-1)^{k-1}}{k^m}left(frac{n!}{(k-1)!}left[begin{array}{c} k \ n+1 end{array} right]right)$$
with $enspace m>0$, $enspace ninmathbb{N}_0$ , $enspaceeta_0(m)=eta(m)$
and $enspacedisplaystyle frac{1}{(k-1)!} left[ begin{array}{c} k \ {n+1} end{array} right]= sumlimits_{i_1=1}^{k-1}sumlimits_{i_2=i_1+1}^{k-1}…sumlimits_{i_n=i_{n-1}+1}^{k-1}frac{1}{i_1 i_2 … i_n}$ .
To combine this series with the polylogarithm is a separate problem.
Now we can write:
$displaystyle sumlimits_{k=1}^infty frac{z^k}{k!}intlimits_0^1 frac{ln^k(1+x)ln x}{x} dx = intlimits_0^1 frac{((1+x)^z-1)ln x}{x} dx = - sumlimits_{k=1}^infty binom z k frac{1}{k^2}$
$hspace{5.3cm}displaystyle = - sumlimits_{k=1}^infty z^k sumlimits_{v=k}^infty frac{(-1)^{k-v}}{v^2 v!} left[ begin{array}{c} v \ k end{array} right]= sumlimits_{k=1}^infty frac{(-z)^k}{(k-1)!} mu_{k-1}(3)$
It follows
$$intlimits_0^1 frac{ln^k(1+x)ln x}{x} dx = (-1)^k,k,mu_{k-1}(3)$$
for $kinmathbb{N}$ .
answered Jun 28 '17 at 9:26
user90369
8,173925
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up vote
3
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The Stirling numbers of the first kind $left[ begin{array}{c} n \ k end{array} right]$ are usually defined by: $$ sumlimits_{k=0}^n left[ begin{array}{c} n \ k end{array} right] x^k := x(x+1)…(x+n-1) $$
The definition of $enspaceeta_n(m)enspace$ in $enspace$ Evaluate $int_{0}^{pi }theta ^{3}log^{3}left ( 2sinfrac{theta }{2} right )mathrm{d}theta $ is
$$eta_n(m):=sumlimits_{k=1}^infty frac{(-1)^{k-1}}{k^m}left(frac{n!}{(k-1)!}left[begin{array}{c} k \ n+1 end{array} right]right)$$
with $enspace m>0$, $enspace ninmathbb{N}_0$ , $enspaceeta_0(m)=eta(m)$
and $enspacedisplaystyle frac{1}{(k-1)!} left[ begin{array}{c} k \ {n+1} end{array} right]= sumlimits_{i_1=1}^{k-1}sumlimits_{i_2=i_1+1}^{k-1}…sumlimits_{i_n=i_{n-1}+1}^{k-1}frac{1}{i_1 i_2 … i_n}$ .
To combine this series with the polylogarithm is a separate problem.
Now we can write:
$displaystyle sumlimits_{k=1}^infty frac{z^k}{k!}intlimits_0^1 frac{ln^k(1+x)ln x}{x} dx = intlimits_0^1 frac{((1+x)^z-1)ln x}{x} dx = - sumlimits_{k=1}^infty binom z k frac{1}{k^2}$
$hspace{5.3cm}displaystyle = - sumlimits_{k=1}^infty z^k sumlimits_{v=k}^infty frac{(-1)^{k-v}}{v^2 v!} left[ begin{array}{c} v \ k end{array} right]= sumlimits_{k=1}^infty frac{(-z)^k}{(k-1)!} mu_{k-1}(3)$
It follows
$$intlimits_0^1 frac{ln^k(1+x)ln x}{x} dx = (-1)^k,k,mu_{k-1}(3)$$
for $kinmathbb{N}$ .
answered Jun 28 '17 at 9:26
user90369
8,173925
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up vote
3
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up vote
3
down vote
The Stirling numbers of the first kind $left[ begin{array}{c} n \ k end{array} right]$ are usually defined by: $$ sumlimits_{k=0}^n left[ begin{array}{c} n \ k end{array} right] x^k := x(x+1)…(x+n-1) $$
The definition of $enspaceeta_n(m)enspace$ in $enspace$ Evaluate $int_{0}^{pi }theta ^{3}log^{3}left ( 2sinfrac{theta }{2} right )mathrm{d}theta $ is
$$eta_n(m):=sumlimits_{k=1}^infty frac{(-1)^{k-1}}{k^m}left(frac{n!}{(k-1)!}left[begin{array}{c} k \ n+1 end{array} right]right)$$
with $enspace m>0$, $enspace ninmathbb{N}_0$ , $enspaceeta_0(m)=eta(m)$
and $enspacedisplaystyle frac{1}{(k-1)!} left[ begin{array}{c} k \ {n+1} end{array} right]= sumlimits_{i_1=1}^{k-1}sumlimits_{i_2=i_1+1}^{k-1}…sumlimits_{i_n=i_{n-1}+1}^{k-1}frac{1}{i_1 i_2 … i_n}$ .
To combine this series with the polylogarithm is a separate problem.
Now we can write:
$displaystyle sumlimits_{k=1}^infty frac{z^k}{k!}intlimits_0^1 frac{ln^k(1+x)ln x}{x} dx = intlimits_0^1 frac{((1+x)^z-1)ln x}{x} dx = - sumlimits_{k=1}^infty binom z k frac{1}{k^2}$
$hspace{5.3cm}displaystyle = - sumlimits_{k=1}^infty z^k sumlimits_{v=k}^infty frac{(-1)^{k-v}}{v^2 v!} left[ begin{array}{c} v \ k end{array} right]= sumlimits_{k=1}^infty frac{(-z)^k}{(k-1)!} mu_{k-1}(3)$
It follows
$$intlimits_0^1 frac{ln^k(1+x)ln x}{x} dx = (-1)^k,k,mu_{k-1}(3)$$
for $kinmathbb{N}$ .
answered Jun 28 '17 at 9:26
user90369
8,173925
The Stirling numbers of the first kind $left[ begin{array}{c} n \ k end{array} right]$ are usually defined by: $$ sumlimits_{k=0}^n left[ begin{array}{c} n \ k end{array} right] x^k := x(x+1)…(x+n-1) $$
The definition of $enspaceeta_n(m)enspace$ in $enspace$ Evaluate $int_{0}^{pi }theta ^{3}log^{3}left ( 2sinfrac{theta }{2} right )mathrm{d}theta $ is
$$eta_n(m):=sumlimits_{k=1}^infty frac{(-1)^{k-1}}{k^m}left(frac{n!}{(k-1)!}left[begin{array}{c} k \ n+1 end{array} right]right)$$
with $enspace m>0$, $enspace ninmathbb{N}_0$ , $enspaceeta_0(m)=eta(m)$
and $enspacedisplaystyle frac{1}{(k-1)!} left[ begin{array}{c} k \ {n+1} end{array} right]= sumlimits_{i_1=1}^{k-1}sumlimits_{i_2=i_1+1}^{k-1}…sumlimits_{i_n=i_{n-1}+1}^{k-1}frac{1}{i_1 i_2 … i_n}$ .
To combine this series with the polylogarithm is a separate problem.
Now we can write:
$displaystyle sumlimits_{k=1}^infty frac{z^k}{k!}intlimits_0^1 frac{ln^k(1+x)ln x}{x} dx = intlimits_0^1 frac{((1+x)^z-1)ln x}{x} dx = - sumlimits_{k=1}^infty binom z k frac{1}{k^2}$
$hspace{5.3cm}displaystyle = - sumlimits_{k=1}^infty z^k sumlimits_{v=k}^infty frac{(-1)^{k-v}}{v^2 v!} left[ begin{array}{c} v \ k end{array} right]= sumlimits_{k=1}^infty frac{(-z)^k}{(k-1)!} mu_{k-1}(3)$
It follows
$$intlimits_0^1 frac{ln^k(1+x)ln x}{x} dx = (-1)^k,k,mu_{k-1}(3)$$
for $kinmathbb{N}$ .
answered Jun 28 '17 at 9:26
user90369
8,173925
edited Nov 21 at 12:08
answered Jun 28 '17 at 9:26
user90369
8,173925
answered Jun 28 '17 at 9:26
user90369
8,173925
answered Jun 28 '17 at 9:26
user90369
8,173925
8,173925
add a comment |
add a comment |
up vote
2
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It's not a complete answer but it is too lengthy for a comment.
@user14717:
Program GP PARI contains a routine to perform something like PSQL stuff.
Here a script
p 200
A1=Pi^6
A2=Pi^4*log(2)^2
A3=Pi^2*log(2)^4
A4=polylog(6,1/2)
A5=polylog(5,1/2)*log(2)
A6=polylog(4,1/2)*log(2)^2
A7=polylog(3,1/2)*log(2)^3
A8=polylog(2,1/2)*log(2)^4
A9=log(2)^6
A10=zeta(3)*log(2)^3
A11=zeta(5)*log(2)
A12=zeta(3)^2
A13=Pi^2*log(2)*zeta(3)
A14=polylog(3,1/2)*Pi^2*log(2)
A15=polylog(4,1/2)*Pi^2
A16=polylog(3,1/2)^2
A17=polylog(2,1/2)^2*log(2)^2
A18=polylog(2,1/2)^2*Pi^2
A19=polylog(2,1/2)^3
A20=polylog(2,1/2)*Pi^2*log(2)^2
A21=polylog(2,1/2)*Pi^4
A22=polylog(3,1/2)*zeta(3)
J=intnum(x=0,1,log(1+x)^4*log(x)/x)
lindep([J,A1,A2,A4,A5,A6,A7,A8,A9,A11,A12,A13,A15,A22])
Last command returns an integer relation of Ai's equals to 0.
Notice that some Ai's are linearly dependant on integers.
Anyway i wasn't able to find out such integer relation using these constants.
answered Jun 29 '17 at 19:11
FDP
4,47411221
add a comment |
up vote
2
down vote
It's not a complete answer but it is too lengthy for a comment.
@user14717:
Program GP PARI contains a routine to perform something like PSQL stuff.
Here a script
p 200
A1=Pi^6
A2=Pi^4*log(2)^2
A3=Pi^2*log(2)^4
A4=polylog(6,1/2)
A5=polylog(5,1/2)*log(2)
A6=polylog(4,1/2)*log(2)^2
A7=polylog(3,1/2)*log(2)^3
A8=polylog(2,1/2)*log(2)^4
A9=log(2)^6
A10=zeta(3)*log(2)^3
A11=zeta(5)*log(2)
A12=zeta(3)^2
A13=Pi^2*log(2)*zeta(3)
A14=polylog(3,1/2)*Pi^2*log(2)
A15=polylog(4,1/2)*Pi^2
A16=polylog(3,1/2)^2
A17=polylog(2,1/2)^2*log(2)^2
A18=polylog(2,1/2)^2*Pi^2
A19=polylog(2,1/2)^3
A20=polylog(2,1/2)*Pi^2*log(2)^2
A21=polylog(2,1/2)*Pi^4
A22=polylog(3,1/2)*zeta(3)
J=intnum(x=0,1,log(1+x)^4*log(x)/x)
lindep([J,A1,A2,A4,A5,A6,A7,A8,A9,A11,A12,A13,A15,A22])
Last command returns an integer relation of Ai's equals to 0.
Notice that some Ai's are linearly dependant on integers.
Anyway i wasn't able to find out such integer relation using these constants.
answered Jun 29 '17 at 19:11
FDP
4,47411221
add a comment |
up vote
2
down vote
up vote
2
down vote
It's not a complete answer but it is too lengthy for a comment.
@user14717:
Program GP PARI contains a routine to perform something like PSQL stuff.
Here a script
p 200
A1=Pi^6
A2=Pi^4*log(2)^2
A3=Pi^2*log(2)^4
A4=polylog(6,1/2)
A5=polylog(5,1/2)*log(2)
A6=polylog(4,1/2)*log(2)^2
A7=polylog(3,1/2)*log(2)^3
A8=polylog(2,1/2)*log(2)^4
A9=log(2)^6
A10=zeta(3)*log(2)^3
A11=zeta(5)*log(2)
A12=zeta(3)^2
A13=Pi^2*log(2)*zeta(3)
A14=polylog(3,1/2)*Pi^2*log(2)
A15=polylog(4,1/2)*Pi^2
A16=polylog(3,1/2)^2
A17=polylog(2,1/2)^2*log(2)^2
A18=polylog(2,1/2)^2*Pi^2
A19=polylog(2,1/2)^3
A20=polylog(2,1/2)*Pi^2*log(2)^2
A21=polylog(2,1/2)*Pi^4
A22=polylog(3,1/2)*zeta(3)
J=intnum(x=0,1,log(1+x)^4*log(x)/x)
lindep([J,A1,A2,A4,A5,A6,A7,A8,A9,A11,A12,A13,A15,A22])
Last command returns an integer relation of Ai's equals to 0.
Notice that some Ai's are linearly dependant on integers.
Anyway i wasn't able to find out such integer relation using these constants.
answered Jun 29 '17 at 19:11
FDP
4,47411221
It's not a complete answer but it is too lengthy for a comment.
@user14717:
Program GP PARI contains a routine to perform something like PSQL stuff.
Here a script
p 200
A1=Pi^6
A2=Pi^4*log(2)^2
A3=Pi^2*log(2)^4
A4=polylog(6,1/2)
A5=polylog(5,1/2)*log(2)
A6=polylog(4,1/2)*log(2)^2
A7=polylog(3,1/2)*log(2)^3
A8=polylog(2,1/2)*log(2)^4
A9=log(2)^6
A10=zeta(3)*log(2)^3
A11=zeta(5)*log(2)
A12=zeta(3)^2
A13=Pi^2*log(2)*zeta(3)
A14=polylog(3,1/2)*Pi^2*log(2)
A15=polylog(4,1/2)*Pi^2
A16=polylog(3,1/2)^2
A17=polylog(2,1/2)^2*log(2)^2
A18=polylog(2,1/2)^2*Pi^2
A19=polylog(2,1/2)^3
A20=polylog(2,1/2)*Pi^2*log(2)^2
A21=polylog(2,1/2)*Pi^4
A22=polylog(3,1/2)*zeta(3)
J=intnum(x=0,1,log(1+x)^4*log(x)/x)
lindep([J,A1,A2,A4,A5,A6,A7,A8,A9,A11,A12,A13,A15,A22])
Last command returns an integer relation of Ai's equals to 0.
Notice that some Ai's are linearly dependant on integers.
Anyway i wasn't able to find out such integer relation using these constants.
answered Jun 29 '17 at 19:11
FDP
4,47411221
answered Jun 29 '17 at 19:11
FDP
4,47411221
answered Jun 29 '17 at 19:11
FDP
4,47411221
answered Jun 29 '17 at 19:11
FDP
4,47411221
4,47411221
add a comment |
add a comment |
up vote
1
down vote
I am very skeptical as to whether the conjecture is true (except of course when $k=2,3$). Let me explain why.
Note that the following identity holds:
begin{eqnarray}
&&-2intlimits_0^1 frac{log(1+x)^k}{x} log(x) dx=\
&&imath pi sumlimits_{l=1}^{k+1} (-1)^l k_{(l-1)} (Li_l(2)-Li_{k+1}(1)1_{l=k+1}) log[2]^{k+1-l}+\
&&sumlimits_{l=2}^k (-1)^{l-1} k_{(l-1)} intlimits_1^2 frac{Li_{l}(x)}{1-x} [log(x)]^{k+1-l} dx+\
&&(-1)^{k+1} k! intlimits_0^1 frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx
end{eqnarray}
The identity above comes from the knowledge of the anti-derivative of the fraction in the integrand and from integrating by parts once.
Now, if we take a look at the last integral above we can actualy write down the antiderivative. We have:
begin{eqnarray}
&&int frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx=\
&&left{
begin{array}{rr}
sumlimits_{l=1}^{k/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{k/2+1} frac{1}{2} [Li_{k/2+1}(1+x)]^2 - Li_{k+1}(1) log(x)& mbox{if $k$ is even}\
sumlimits_{l=1}^{(k+1)/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{(k+1)/2+1} int frac{[Li_{(k+1)/2}(1+x)]^2}{1+x} dx - Li_{k+1}(1) log(x)& mbox{if $k$ is odd}
end{array}
right.
end{eqnarray}
As we can see it is only when $k$ is even that the result reduces to poly-logarithms only otherwise a new unknown antiderivative involving a square of a polylogarithm remains. As a matter of fact it is not difficult to see that by integrating by parts and changing and swapping order of integration appropriately all the remaining integrals in the third line from the top in the first formula above reduce to poly-logarithms and following quantities:
begin{equation}
S^{(2,p)}_q := intlimits_1^2 frac{[Li_q(x)]^2}{x}cdot [log(x)]^p dx
end{equation}
where $pge1$ and $qge 1$.
Let us now take $k=4$ as an example. Here the integrals in question read:
begin{eqnarray}
intlimits_1^2 frac{Li_2(x)}{1-x} cdot [log(x)]^3 dx &=& -frac{3}{2} text{Li}_2(2){}^2 log ^2(2)-i pi text{Li}_2(2) log ^3(2) + 3 S^{(2,1)}_2 - S^{(2,3)}_1\
intlimits_1^2 frac{Li_3(x)}{1-x} cdot [log(x)]^2 dx &=& text{Li}_3(2){}^2-2 text{Li}_2(2) text{Li}_3(2) log (2)-zeta (3)^2-frac{1}{32} left(pi ^4+28 i pi zeta (3)right) log ^2(2) + 3 S^{(2,1)}_2\
intlimits_1^2 frac{Li_4(x)}{1-x} cdot [log(x)]^1 dx &=& text{Li}_3(2){}^2-frac{pi ^2 text{Li}_4(2)}{4}-text{Li}_2(2) text{Li}_3(2) log (2)-zeta (3)^2+frac{pi ^6}{540} + S^{(2,1)}_2\
intlimits_0^1 frac{Li_5(1+x)-Li_5(1)}{x} &=& frac{1}{2} left(-text{Li}_3(2){}^2+2 text{Li}_2(2) text{Li}_4(2)+zeta (3)^2right)+i pi (text{Li}_5(2)-zeta (5))-frac{pi ^6}{540}
end{eqnarray}
Now, when we bring everything together the quantities $S^{(2,1)}_2$ miraculously cancel out and then by substituting $xleftarrow 1/x$ in the quantity $S^{(2,3)}_1$ we obtain the following result:
begin{eqnarray}
&&intlimits_0^1 frac{[log(1+x)]^4}{x} cdot log(x) dx =\
&&-log (2) left(96 text{Li}_5left(frac{1}{2}right)+48 text{Li}_4left(frac{1}{2}right) log (2)+14 zeta (3) log ^2(2)+log ^5(2)right)+\
&&-96 text{Li}_6left(frac{1}{2}right)+frac{32 pi ^6}{315}+pi ^2 log ^4(2)+2 intlimits_{1/2}^1 frac{[log(1-x)]^2}{x} cdot [log(x)]^3 dx=\
&&12 left(zeta (3)^2-10 text{Li}_6left(frac{1}{2}right)right)+frac{3}{4} log (2) left(zeta (5)-96 text{Li}_5left(frac{1}{2}right)right)-24 text{Li}_4left(frac{1}{2}right) log ^2(2)+frac{1}{3} pi ^2
left(log ^4(2)-6 zeta (3) log (2)right)-3 zeta (3) log ^3(2)+frac{26 pi ^6}{315}-frac{17 log ^6(2)}{30}-frac{1}{60} pi ^4 log ^2(2) + 24 {bf H}^{(1)}_5(1/2)
end{eqnarray}
where in the last line we expressed the remaining integral through Euler sums and used the following results:
begin{eqnarray}
{bf H}^{(1)}_2(1/2) &=& zeta (3)-frac{1}{12} pi ^2 log (2)\
{bf H}^{(1)}_3(1/2) &=& text{Li}_4left(frac{1}{2}right)-frac{1}{8} zeta (3) log (2)+frac{pi ^4}{720}+frac{log ^4(2)}{24}\
{bf H}^{(1)}_4(1/2) &=& 2 text{Li}_5left(frac{1}{2}right)+text{Li}_4left(frac{1}{2}right) log (2)+frac{zeta (5)}{32}-frac{1}{36} pi ^2 left(3 zeta (3)+log ^3(2)right)+frac{1}{2} zeta (3) log ^2(2)+frac{log
^5(2)}{40}-frac{1}{720} pi ^4 log (2)
end{eqnarray}
Now, the question whether ${bf H}^{(1)}_5(1/2)$ is expressible in terms of poly-logarithms is most likely answered negatively. Indeed I used the web interface http://wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi to find possible linear dependencies. The code
lindep([ zp(2,6)+zp(2,5,1), z(6), z(3)*z(3), z(3)*z(2)*log(2), z(3)*log(2)^3, z(2)*log(2)^4, z(2)^2*log(2)^2, z(5)*log(2), log(2)^6, zp(2,4)*log(2)^2, zp(2,5)*log(2), zp(2,6)])
being run for two different number of digits of precision gives completely different results which suggests that such linear dependency most likely does not exist.
answered Dec 13 '17 at 13:24
Przemo
4,1321928
add a comment |
up vote
1
down vote
I am very skeptical as to whether the conjecture is true (except of course when $k=2,3$). Let me explain why.
Note that the following identity holds:
begin{eqnarray}
&&-2intlimits_0^1 frac{log(1+x)^k}{x} log(x) dx=\
&&imath pi sumlimits_{l=1}^{k+1} (-1)^l k_{(l-1)} (Li_l(2)-Li_{k+1}(1)1_{l=k+1}) log[2]^{k+1-l}+\
&&sumlimits_{l=2}^k (-1)^{l-1} k_{(l-1)} intlimits_1^2 frac{Li_{l}(x)}{1-x} [log(x)]^{k+1-l} dx+\
&&(-1)^{k+1} k! intlimits_0^1 frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx
end{eqnarray}
The identity above comes from the knowledge of the anti-derivative of the fraction in the integrand and from integrating by parts once.
Now, if we take a look at the last integral above we can actualy write down the antiderivative. We have:
begin{eqnarray}
&&int frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx=\
&&left{
begin{array}{rr}
sumlimits_{l=1}^{k/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{k/2+1} frac{1}{2} [Li_{k/2+1}(1+x)]^2 - Li_{k+1}(1) log(x)& mbox{if $k$ is even}\
sumlimits_{l=1}^{(k+1)/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{(k+1)/2+1} int frac{[Li_{(k+1)/2}(1+x)]^2}{1+x} dx - Li_{k+1}(1) log(x)& mbox{if $k$ is odd}
end{array}
right.
end{eqnarray}
As we can see it is only when $k$ is even that the result reduces to poly-logarithms only otherwise a new unknown antiderivative involving a square of a polylogarithm remains. As a matter of fact it is not difficult to see that by integrating by parts and changing and swapping order of integration appropriately all the remaining integrals in the third line from the top in the first formula above reduce to poly-logarithms and following quantities:
begin{equation}
S^{(2,p)}_q := intlimits_1^2 frac{[Li_q(x)]^2}{x}cdot [log(x)]^p dx
end{equation}
where $pge1$ and $qge 1$.
Let us now take $k=4$ as an example. Here the integrals in question read:
begin{eqnarray}
intlimits_1^2 frac{Li_2(x)}{1-x} cdot [log(x)]^3 dx &=& -frac{3}{2} text{Li}_2(2){}^2 log ^2(2)-i pi text{Li}_2(2) log ^3(2) + 3 S^{(2,1)}_2 - S^{(2,3)}_1\
intlimits_1^2 frac{Li_3(x)}{1-x} cdot [log(x)]^2 dx &=& text{Li}_3(2){}^2-2 text{Li}_2(2) text{Li}_3(2) log (2)-zeta (3)^2-frac{1}{32} left(pi ^4+28 i pi zeta (3)right) log ^2(2) + 3 S^{(2,1)}_2\
intlimits_1^2 frac{Li_4(x)}{1-x} cdot [log(x)]^1 dx &=& text{Li}_3(2){}^2-frac{pi ^2 text{Li}_4(2)}{4}-text{Li}_2(2) text{Li}_3(2) log (2)-zeta (3)^2+frac{pi ^6}{540} + S^{(2,1)}_2\
intlimits_0^1 frac{Li_5(1+x)-Li_5(1)}{x} &=& frac{1}{2} left(-text{Li}_3(2){}^2+2 text{Li}_2(2) text{Li}_4(2)+zeta (3)^2right)+i pi (text{Li}_5(2)-zeta (5))-frac{pi ^6}{540}
end{eqnarray}
Now, when we bring everything together the quantities $S^{(2,1)}_2$ miraculously cancel out and then by substituting $xleftarrow 1/x$ in the quantity $S^{(2,3)}_1$ we obtain the following result:
begin{eqnarray}
&&intlimits_0^1 frac{[log(1+x)]^4}{x} cdot log(x) dx =\
&&-log (2) left(96 text{Li}_5left(frac{1}{2}right)+48 text{Li}_4left(frac{1}{2}right) log (2)+14 zeta (3) log ^2(2)+log ^5(2)right)+\
&&-96 text{Li}_6left(frac{1}{2}right)+frac{32 pi ^6}{315}+pi ^2 log ^4(2)+2 intlimits_{1/2}^1 frac{[log(1-x)]^2}{x} cdot [log(x)]^3 dx=\
&&12 left(zeta (3)^2-10 text{Li}_6left(frac{1}{2}right)right)+frac{3}{4} log (2) left(zeta (5)-96 text{Li}_5left(frac{1}{2}right)right)-24 text{Li}_4left(frac{1}{2}right) log ^2(2)+frac{1}{3} pi ^2
left(log ^4(2)-6 zeta (3) log (2)right)-3 zeta (3) log ^3(2)+frac{26 pi ^6}{315}-frac{17 log ^6(2)}{30}-frac{1}{60} pi ^4 log ^2(2) + 24 {bf H}^{(1)}_5(1/2)
end{eqnarray}
where in the last line we expressed the remaining integral through Euler sums and used the following results:
begin{eqnarray}
{bf H}^{(1)}_2(1/2) &=& zeta (3)-frac{1}{12} pi ^2 log (2)\
{bf H}^{(1)}_3(1/2) &=& text{Li}_4left(frac{1}{2}right)-frac{1}{8} zeta (3) log (2)+frac{pi ^4}{720}+frac{log ^4(2)}{24}\
{bf H}^{(1)}_4(1/2) &=& 2 text{Li}_5left(frac{1}{2}right)+text{Li}_4left(frac{1}{2}right) log (2)+frac{zeta (5)}{32}-frac{1}{36} pi ^2 left(3 zeta (3)+log ^3(2)right)+frac{1}{2} zeta (3) log ^2(2)+frac{log
^5(2)}{40}-frac{1}{720} pi ^4 log (2)
end{eqnarray}
Now, the question whether ${bf H}^{(1)}_5(1/2)$ is expressible in terms of poly-logarithms is most likely answered negatively. Indeed I used the web interface http://wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi to find possible linear dependencies. The code
lindep([ zp(2,6)+zp(2,5,1), z(6), z(3)*z(3), z(3)*z(2)*log(2), z(3)*log(2)^3, z(2)*log(2)^4, z(2)^2*log(2)^2, z(5)*log(2), log(2)^6, zp(2,4)*log(2)^2, zp(2,5)*log(2), zp(2,6)])
being run for two different number of digits of precision gives completely different results which suggests that such linear dependency most likely does not exist.
answered Dec 13 '17 at 13:24
Przemo
4,1321928
add a comment |
up vote
1
down vote
up vote
1
down vote
I am very skeptical as to whether the conjecture is true (except of course when $k=2,3$). Let me explain why.
Note that the following identity holds:
begin{eqnarray}
&&-2intlimits_0^1 frac{log(1+x)^k}{x} log(x) dx=\
&&imath pi sumlimits_{l=1}^{k+1} (-1)^l k_{(l-1)} (Li_l(2)-Li_{k+1}(1)1_{l=k+1}) log[2]^{k+1-l}+\
&&sumlimits_{l=2}^k (-1)^{l-1} k_{(l-1)} intlimits_1^2 frac{Li_{l}(x)}{1-x} [log(x)]^{k+1-l} dx+\
&&(-1)^{k+1} k! intlimits_0^1 frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx
end{eqnarray}
The identity above comes from the knowledge of the anti-derivative of the fraction in the integrand and from integrating by parts once.
Now, if we take a look at the last integral above we can actualy write down the antiderivative. We have:
begin{eqnarray}
&&int frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx=\
&&left{
begin{array}{rr}
sumlimits_{l=1}^{k/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{k/2+1} frac{1}{2} [Li_{k/2+1}(1+x)]^2 - Li_{k+1}(1) log(x)& mbox{if $k$ is even}\
sumlimits_{l=1}^{(k+1)/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{(k+1)/2+1} int frac{[Li_{(k+1)/2}(1+x)]^2}{1+x} dx - Li_{k+1}(1) log(x)& mbox{if $k$ is odd}
end{array}
right.
end{eqnarray}
As we can see it is only when $k$ is even that the result reduces to poly-logarithms only otherwise a new unknown antiderivative involving a square of a polylogarithm remains. As a matter of fact it is not difficult to see that by integrating by parts and changing and swapping order of integration appropriately all the remaining integrals in the third line from the top in the first formula above reduce to poly-logarithms and following quantities:
begin{equation}
S^{(2,p)}_q := intlimits_1^2 frac{[Li_q(x)]^2}{x}cdot [log(x)]^p dx
end{equation}
where $pge1$ and $qge 1$.
Let us now take $k=4$ as an example. Here the integrals in question read:
begin{eqnarray}
intlimits_1^2 frac{Li_2(x)}{1-x} cdot [log(x)]^3 dx &=& -frac{3}{2} text{Li}_2(2){}^2 log ^2(2)-i pi text{Li}_2(2) log ^3(2) + 3 S^{(2,1)}_2 - S^{(2,3)}_1\
intlimits_1^2 frac{Li_3(x)}{1-x} cdot [log(x)]^2 dx &=& text{Li}_3(2){}^2-2 text{Li}_2(2) text{Li}_3(2) log (2)-zeta (3)^2-frac{1}{32} left(pi ^4+28 i pi zeta (3)right) log ^2(2) + 3 S^{(2,1)}_2\
intlimits_1^2 frac{Li_4(x)}{1-x} cdot [log(x)]^1 dx &=& text{Li}_3(2){}^2-frac{pi ^2 text{Li}_4(2)}{4}-text{Li}_2(2) text{Li}_3(2) log (2)-zeta (3)^2+frac{pi ^6}{540} + S^{(2,1)}_2\
intlimits_0^1 frac{Li_5(1+x)-Li_5(1)}{x} &=& frac{1}{2} left(-text{Li}_3(2){}^2+2 text{Li}_2(2) text{Li}_4(2)+zeta (3)^2right)+i pi (text{Li}_5(2)-zeta (5))-frac{pi ^6}{540}
end{eqnarray}
Now, when we bring everything together the quantities $S^{(2,1)}_2$ miraculously cancel out and then by substituting $xleftarrow 1/x$ in the quantity $S^{(2,3)}_1$ we obtain the following result:
begin{eqnarray}
&&intlimits_0^1 frac{[log(1+x)]^4}{x} cdot log(x) dx =\
&&-log (2) left(96 text{Li}_5left(frac{1}{2}right)+48 text{Li}_4left(frac{1}{2}right) log (2)+14 zeta (3) log ^2(2)+log ^5(2)right)+\
&&-96 text{Li}_6left(frac{1}{2}right)+frac{32 pi ^6}{315}+pi ^2 log ^4(2)+2 intlimits_{1/2}^1 frac{[log(1-x)]^2}{x} cdot [log(x)]^3 dx=\
&&12 left(zeta (3)^2-10 text{Li}_6left(frac{1}{2}right)right)+frac{3}{4} log (2) left(zeta (5)-96 text{Li}_5left(frac{1}{2}right)right)-24 text{Li}_4left(frac{1}{2}right) log ^2(2)+frac{1}{3} pi ^2
left(log ^4(2)-6 zeta (3) log (2)right)-3 zeta (3) log ^3(2)+frac{26 pi ^6}{315}-frac{17 log ^6(2)}{30}-frac{1}{60} pi ^4 log ^2(2) + 24 {bf H}^{(1)}_5(1/2)
end{eqnarray}
where in the last line we expressed the remaining integral through Euler sums and used the following results:
begin{eqnarray}
{bf H}^{(1)}_2(1/2) &=& zeta (3)-frac{1}{12} pi ^2 log (2)\
{bf H}^{(1)}_3(1/2) &=& text{Li}_4left(frac{1}{2}right)-frac{1}{8} zeta (3) log (2)+frac{pi ^4}{720}+frac{log ^4(2)}{24}\
{bf H}^{(1)}_4(1/2) &=& 2 text{Li}_5left(frac{1}{2}right)+text{Li}_4left(frac{1}{2}right) log (2)+frac{zeta (5)}{32}-frac{1}{36} pi ^2 left(3 zeta (3)+log ^3(2)right)+frac{1}{2} zeta (3) log ^2(2)+frac{log
^5(2)}{40}-frac{1}{720} pi ^4 log (2)
end{eqnarray}
Now, the question whether ${bf H}^{(1)}_5(1/2)$ is expressible in terms of poly-logarithms is most likely answered negatively. Indeed I used the web interface http://wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi to find possible linear dependencies. The code
lindep([ zp(2,6)+zp(2,5,1), z(6), z(3)*z(3), z(3)*z(2)*log(2), z(3)*log(2)^3, z(2)*log(2)^4, z(2)^2*log(2)^2, z(5)*log(2), log(2)^6, zp(2,4)*log(2)^2, zp(2,5)*log(2), zp(2,6)])
being run for two different number of digits of precision gives completely different results which suggests that such linear dependency most likely does not exist.
answered Dec 13 '17 at 13:24
Przemo
4,1321928
I am very skeptical as to whether the conjecture is true (except of course when $k=2,3$). Let me explain why.
Note that the following identity holds:
begin{eqnarray}
&&-2intlimits_0^1 frac{log(1+x)^k}{x} log(x) dx=\
&&imath pi sumlimits_{l=1}^{k+1} (-1)^l k_{(l-1)} (Li_l(2)-Li_{k+1}(1)1_{l=k+1}) log[2]^{k+1-l}+\
&&sumlimits_{l=2}^k (-1)^{l-1} k_{(l-1)} intlimits_1^2 frac{Li_{l}(x)}{1-x} [log(x)]^{k+1-l} dx+\
&&(-1)^{k+1} k! intlimits_0^1 frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx
end{eqnarray}
The identity above comes from the knowledge of the anti-derivative of the fraction in the integrand and from integrating by parts once.
Now, if we take a look at the last integral above we can actualy write down the antiderivative. We have:
begin{eqnarray}
&&int frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx=\
&&left{
begin{array}{rr}
sumlimits_{l=1}^{k/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{k/2+1} frac{1}{2} [Li_{k/2+1}(1+x)]^2 - Li_{k+1}(1) log(x)& mbox{if $k$ is even}\
sumlimits_{l=1}^{(k+1)/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{(k+1)/2+1} int frac{[Li_{(k+1)/2}(1+x)]^2}{1+x} dx - Li_{k+1}(1) log(x)& mbox{if $k$ is odd}
end{array}
right.
end{eqnarray}
As we can see it is only when $k$ is even that the result reduces to poly-logarithms only otherwise a new unknown antiderivative involving a square of a polylogarithm remains. As a matter of fact it is not difficult to see that by integrating by parts and changing and swapping order of integration appropriately all the remaining integrals in the third line from the top in the first formula above reduce to poly-logarithms and following quantities:
begin{equation}
S^{(2,p)}_q := intlimits_1^2 frac{[Li_q(x)]^2}{x}cdot [log(x)]^p dx
end{equation}
where $pge1$ and $qge 1$.
Let us now take $k=4$ as an example. Here the integrals in question read:
begin{eqnarray}
intlimits_1^2 frac{Li_2(x)}{1-x} cdot [log(x)]^3 dx &=& -frac{3}{2} text{Li}_2(2){}^2 log ^2(2)-i pi text{Li}_2(2) log ^3(2) + 3 S^{(2,1)}_2 - S^{(2,3)}_1\
intlimits_1^2 frac{Li_3(x)}{1-x} cdot [log(x)]^2 dx &=& text{Li}_3(2){}^2-2 text{Li}_2(2) text{Li}_3(2) log (2)-zeta (3)^2-frac{1}{32} left(pi ^4+28 i pi zeta (3)right) log ^2(2) + 3 S^{(2,1)}_2\
intlimits_1^2 frac{Li_4(x)}{1-x} cdot [log(x)]^1 dx &=& text{Li}_3(2){}^2-frac{pi ^2 text{Li}_4(2)}{4}-text{Li}_2(2) text{Li}_3(2) log (2)-zeta (3)^2+frac{pi ^6}{540} + S^{(2,1)}_2\
intlimits_0^1 frac{Li_5(1+x)-Li_5(1)}{x} &=& frac{1}{2} left(-text{Li}_3(2){}^2+2 text{Li}_2(2) text{Li}_4(2)+zeta (3)^2right)+i pi (text{Li}_5(2)-zeta (5))-frac{pi ^6}{540}
end{eqnarray}
Now, when we bring everything together the quantities $S^{(2,1)}_2$ miraculously cancel out and then by substituting $xleftarrow 1/x$ in the quantity $S^{(2,3)}_1$ we obtain the following result:
begin{eqnarray}
&&intlimits_0^1 frac{[log(1+x)]^4}{x} cdot log(x) dx =\
&&-log (2) left(96 text{Li}_5left(frac{1}{2}right)+48 text{Li}_4left(frac{1}{2}right) log (2)+14 zeta (3) log ^2(2)+log ^5(2)right)+\
&&-96 text{Li}_6left(frac{1}{2}right)+frac{32 pi ^6}{315}+pi ^2 log ^4(2)+2 intlimits_{1/2}^1 frac{[log(1-x)]^2}{x} cdot [log(x)]^3 dx=\
&&12 left(zeta (3)^2-10 text{Li}_6left(frac{1}{2}right)right)+frac{3}{4} log (2) left(zeta (5)-96 text{Li}_5left(frac{1}{2}right)right)-24 text{Li}_4left(frac{1}{2}right) log ^2(2)+frac{1}{3} pi ^2
left(log ^4(2)-6 zeta (3) log (2)right)-3 zeta (3) log ^3(2)+frac{26 pi ^6}{315}-frac{17 log ^6(2)}{30}-frac{1}{60} pi ^4 log ^2(2) + 24 {bf H}^{(1)}_5(1/2)
end{eqnarray}
where in the last line we expressed the remaining integral through Euler sums and used the following results:
begin{eqnarray}
{bf H}^{(1)}_2(1/2) &=& zeta (3)-frac{1}{12} pi ^2 log (2)\
{bf H}^{(1)}_3(1/2) &=& text{Li}_4left(frac{1}{2}right)-frac{1}{8} zeta (3) log (2)+frac{pi ^4}{720}+frac{log ^4(2)}{24}\
{bf H}^{(1)}_4(1/2) &=& 2 text{Li}_5left(frac{1}{2}right)+text{Li}_4left(frac{1}{2}right) log (2)+frac{zeta (5)}{32}-frac{1}{36} pi ^2 left(3 zeta (3)+log ^3(2)right)+frac{1}{2} zeta (3) log ^2(2)+frac{log
^5(2)}{40}-frac{1}{720} pi ^4 log (2)
end{eqnarray}
Now, the question whether ${bf H}^{(1)}_5(1/2)$ is expressible in terms of poly-logarithms is most likely answered negatively. Indeed I used the web interface http://wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi to find possible linear dependencies. The code
lindep([ zp(2,6)+zp(2,5,1), z(6), z(3)*z(3), z(3)*z(2)*log(2), z(3)*log(2)^3, z(2)*log(2)^4, z(2)^2*log(2)^2, z(5)*log(2), log(2)^6, zp(2,4)*log(2)^2, zp(2,5)*log(2), zp(2,6)])
being run for two different number of digits of precision gives completely different results which suggests that such linear dependency most likely does not exist.
answered Dec 13 '17 at 13:24
Przemo
4,1321928
edited Dec 14 '17 at 11:06
answered Dec 13 '17 at 13:24
Przemo
4,1321928
answered Dec 13 '17 at 13:24
Przemo
4,1321928
answered Dec 13 '17 at 13:24
Przemo
4,1321928
4,1321928
add a comment |
add a comment |
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0
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[#ffd,10px]{ds{%
int_{0}^{1}{ln^{4}pars{1 + x}lnpars{x} over x},dd x}}
,,,stackrel{1 + x mapsto x}{=},,,
int_{1}^{2}{ln^{4}pars{x}lnpars{x - 1} over x - 1},dd x
\[5mm] stackrel{x mapsto 1/x}{=},,,&
int_{1}^{1/2}{ln^{4}pars{1/x}lnpars{1/x - 1} over 1/x - 1}
,pars{-,{dd x over x^{2}}} =
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} - ln^{5}pars{x} over
xpars{1 - x}},dd x
\[1cm] = &
underbrace{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x}
_{ds{mc{I}_{1}}} -
underbrace{int_{1/2}^{1}{ln^{5}pars{x} over x},dd x}
_{ds{= -,{ln^{6}pars{2} over 6}}} +
underbrace{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over 1 - x},dd x}
_{ds{mc{I}_{2}}}
\[1mm] &
-int_{1/2}^{1}{ln^{5}pars{x} over 1 - x},dd x
label{1}tag{1}
end{align}
Note that
begin{align}
int_{1/2}^{1}{ln^{5}pars{x} over 1 - x},dd x & =
-ln^{6}pars{2} +
5int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x =
-ln^{6}pars{2} + 5,mc{I}_{1}
end{align}
such that eqref{1} becomes
begin{equation}
bbx{bbox[#ffd,10px]{ds{%
int_{0}^{1}{ln^{4}pars{1 + x}lnpars{x} over x},dd x}} =
{7ln^{6}pars{2} over 6} - 4,mc{I}_{1} + mc{I}_{2}}label{2}tag{2}
end{equation}
$ds{Hugemc{I}_{1}: ?}$.
begin{align}
mc{I}_{1} & equiv
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x =
-int_{1/2}^{1}mrm{Li}_{2}'pars{x}ln^{4}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4int_{1/2}^{1}mrm{Li}_{3}'pars{x}ln^{3}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} -
12int_{1/2}^{1}mrm{Li}_{4}'pars{x}ln^{2}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24int_{1/2}^{1}mrm{Li}_{5}'pars{x}lnpars{x},dd x
\[1cm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24,mrm{Li}_{5}pars{1 over 2}lnpars{2}
\[1mm] & - 24int_{1/2}^{1}mrm{Li}_{6}'pars{x},dd x
\[1cm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24,mrm{Li}_{5}pars{1 over 2}lnpars{2}
\[1mm] & -24,mrm{Li}_{6}pars{1} + 24,mrm{Li}_{6}pars{1 over 2}
end{align}
Note that
$ds{mrm{Li}_{2}pars{1 over 2} = {pi^{2} over 12} -
{ln^{2}pars{2} over 2}}$ and
$ds{mrm{Li}_{3}pars{1 over 2} = {ln^{3}pars{2} over 6} -
{pi^{2}lnpars{2} over 12} + {7zetapars{3} over 8}}$. Moreover,
$ds{vphantom{Huge A}mrm{Li}_{6}pars{1} = zetapars{6} =
{pi^{6} over 945}}$.
Then,
$$
begin{array}{|rcl|}hline mbox{}&&\
ds{quadmc{I}_{1}} & ds{equiv} &
ds{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x}
\[5mm] & ds{=} &
ds{-,{8pi^{6} over 315} - {pi^{2}ln^{4}pars{2} over 4} +
{ln^{6}pars{2} over 6} + 12ln^{2}pars{2},mrm{Li}_{4}pars{1 over 2} +
24lnpars{2},mrm{Li}_{5}pars{1 over 2}quad}
\[1mm] &&
ds{%
+ 24,mrm{Li}_{6}pars{1 over 2} + {7ln^{3}pars{2},zetapars{3} over 2}
approx -0.0269}
\ mbox{}&&
\ hline
end{array}
$$
$ds{Hugemc{I}_{2}: ?}$. This one isn't trivial at all. An attempt is given by
begin{align}
mc{I}_{2} & equiv
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over 1 - x},dd x
,,,stackrel{x mapsto 1 - x}{=},,,
int_{0}^{1/2}{ln^{4}pars{1 - x}lnpars{x} over x},dd x
\[5mm] & =
left.partiald[4]{}{mu}partiald{}{nu}
int_{0}^{1/2}bracks{pars{1 - x}^{mu} - 1}x^{nu - 1}
,dd x,rightvert_{ mu = 0,, nu = 0^{+}}
\[5mm] & =
partiald[4]{}{mu}partiald{}{nu}
bracks{int_{0}^{1/2}pars{1 - x}^{mu},x^{nu - 1}
,dd x - {1 over 2^{nu}nu}}_{ mu = 0,, nu = 0^{+}}
\[5mm] & =
partiald[4]{}{mu}partiald{}{nu}
bracks{mrm{B}pars{{1 over 2},nu,1 + mu} - {1 over 2^{nu}nu}}_{ mu = 0,, nu = 0^{+}}
end{align}
$ds{mrm{B}}$ is the Incomplete Beta Function. It'll continue$ldots$
answered Feb 1 at 23:17
Felix Marin
66.4k7107139
add a comment |
up vote
0
down vote
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[#ffd,10px]{ds{%
int_{0}^{1}{ln^{4}pars{1 + x}lnpars{x} over x},dd x}}
,,,stackrel{1 + x mapsto x}{=},,,
int_{1}^{2}{ln^{4}pars{x}lnpars{x - 1} over x - 1},dd x
\[5mm] stackrel{x mapsto 1/x}{=},,,&
int_{1}^{1/2}{ln^{4}pars{1/x}lnpars{1/x - 1} over 1/x - 1}
,pars{-,{dd x over x^{2}}} =
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} - ln^{5}pars{x} over
xpars{1 - x}},dd x
\[1cm] = &
underbrace{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x}
_{ds{mc{I}_{1}}} -
underbrace{int_{1/2}^{1}{ln^{5}pars{x} over x},dd x}
_{ds{= -,{ln^{6}pars{2} over 6}}} +
underbrace{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over 1 - x},dd x}
_{ds{mc{I}_{2}}}
\[1mm] &
-int_{1/2}^{1}{ln^{5}pars{x} over 1 - x},dd x
label{1}tag{1}
end{align}
Note that
begin{align}
int_{1/2}^{1}{ln^{5}pars{x} over 1 - x},dd x & =
-ln^{6}pars{2} +
5int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x =
-ln^{6}pars{2} + 5,mc{I}_{1}
end{align}
such that eqref{1} becomes
begin{equation}
bbx{bbox[#ffd,10px]{ds{%
int_{0}^{1}{ln^{4}pars{1 + x}lnpars{x} over x},dd x}} =
{7ln^{6}pars{2} over 6} - 4,mc{I}_{1} + mc{I}_{2}}label{2}tag{2}
end{equation}
$ds{Hugemc{I}_{1}: ?}$.
begin{align}
mc{I}_{1} & equiv
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x =
-int_{1/2}^{1}mrm{Li}_{2}'pars{x}ln^{4}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4int_{1/2}^{1}mrm{Li}_{3}'pars{x}ln^{3}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} -
12int_{1/2}^{1}mrm{Li}_{4}'pars{x}ln^{2}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24int_{1/2}^{1}mrm{Li}_{5}'pars{x}lnpars{x},dd x
\[1cm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24,mrm{Li}_{5}pars{1 over 2}lnpars{2}
\[1mm] & - 24int_{1/2}^{1}mrm{Li}_{6}'pars{x},dd x
\[1cm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24,mrm{Li}_{5}pars{1 over 2}lnpars{2}
\[1mm] & -24,mrm{Li}_{6}pars{1} + 24,mrm{Li}_{6}pars{1 over 2}
end{align}
Note that
$ds{mrm{Li}_{2}pars{1 over 2} = {pi^{2} over 12} -
{ln^{2}pars{2} over 2}}$ and
$ds{mrm{Li}_{3}pars{1 over 2} = {ln^{3}pars{2} over 6} -
{pi^{2}lnpars{2} over 12} + {7zetapars{3} over 8}}$. Moreover,
$ds{vphantom{Huge A}mrm{Li}_{6}pars{1} = zetapars{6} =
{pi^{6} over 945}}$.
Then,
$$
begin{array}{|rcl|}hline mbox{}&&\
ds{quadmc{I}_{1}} & ds{equiv} &
ds{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x}
\[5mm] & ds{=} &
ds{-,{8pi^{6} over 315} - {pi^{2}ln^{4}pars{2} over 4} +
{ln^{6}pars{2} over 6} + 12ln^{2}pars{2},mrm{Li}_{4}pars{1 over 2} +
24lnpars{2},mrm{Li}_{5}pars{1 over 2}quad}
\[1mm] &&
ds{%
+ 24,mrm{Li}_{6}pars{1 over 2} + {7ln^{3}pars{2},zetapars{3} over 2}
approx -0.0269}
\ mbox{}&&
\ hline
end{array}
$$
$ds{Hugemc{I}_{2}: ?}$. This one isn't trivial at all. An attempt is given by
begin{align}
mc{I}_{2} & equiv
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over 1 - x},dd x
,,,stackrel{x mapsto 1 - x}{=},,,
int_{0}^{1/2}{ln^{4}pars{1 - x}lnpars{x} over x},dd x
\[5mm] & =
left.partiald[4]{}{mu}partiald{}{nu}
int_{0}^{1/2}bracks{pars{1 - x}^{mu} - 1}x^{nu - 1}
,dd x,rightvert_{ mu = 0,, nu = 0^{+}}
\[5mm] & =
partiald[4]{}{mu}partiald{}{nu}
bracks{int_{0}^{1/2}pars{1 - x}^{mu},x^{nu - 1}
,dd x - {1 over 2^{nu}nu}}_{ mu = 0,, nu = 0^{+}}
\[5mm] & =
partiald[4]{}{mu}partiald{}{nu}
bracks{mrm{B}pars{{1 over 2},nu,1 + mu} - {1 over 2^{nu}nu}}_{ mu = 0,, nu = 0^{+}}
end{align}
$ds{mrm{B}}$ is the Incomplete Beta Function. It'll continue$ldots$
answered Feb 1 at 23:17
Felix Marin
66.4k7107139
add a comment |
up vote
0
down vote
up vote
0
down vote
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[#ffd,10px]{ds{%
int_{0}^{1}{ln^{4}pars{1 + x}lnpars{x} over x},dd x}}
,,,stackrel{1 + x mapsto x}{=},,,
int_{1}^{2}{ln^{4}pars{x}lnpars{x - 1} over x - 1},dd x
\[5mm] stackrel{x mapsto 1/x}{=},,,&
int_{1}^{1/2}{ln^{4}pars{1/x}lnpars{1/x - 1} over 1/x - 1}
,pars{-,{dd x over x^{2}}} =
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} - ln^{5}pars{x} over
xpars{1 - x}},dd x
\[1cm] = &
underbrace{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x}
_{ds{mc{I}_{1}}} -
underbrace{int_{1/2}^{1}{ln^{5}pars{x} over x},dd x}
_{ds{= -,{ln^{6}pars{2} over 6}}} +
underbrace{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over 1 - x},dd x}
_{ds{mc{I}_{2}}}
\[1mm] &
-int_{1/2}^{1}{ln^{5}pars{x} over 1 - x},dd x
label{1}tag{1}
end{align}
Note that
begin{align}
int_{1/2}^{1}{ln^{5}pars{x} over 1 - x},dd x & =
-ln^{6}pars{2} +
5int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x =
-ln^{6}pars{2} + 5,mc{I}_{1}
end{align}
such that eqref{1} becomes
begin{equation}
bbx{bbox[#ffd,10px]{ds{%
int_{0}^{1}{ln^{4}pars{1 + x}lnpars{x} over x},dd x}} =
{7ln^{6}pars{2} over 6} - 4,mc{I}_{1} + mc{I}_{2}}label{2}tag{2}
end{equation}
$ds{Hugemc{I}_{1}: ?}$.
begin{align}
mc{I}_{1} & equiv
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x =
-int_{1/2}^{1}mrm{Li}_{2}'pars{x}ln^{4}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4int_{1/2}^{1}mrm{Li}_{3}'pars{x}ln^{3}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} -
12int_{1/2}^{1}mrm{Li}_{4}'pars{x}ln^{2}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24int_{1/2}^{1}mrm{Li}_{5}'pars{x}lnpars{x},dd x
\[1cm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24,mrm{Li}_{5}pars{1 over 2}lnpars{2}
\[1mm] & - 24int_{1/2}^{1}mrm{Li}_{6}'pars{x},dd x
\[1cm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24,mrm{Li}_{5}pars{1 over 2}lnpars{2}
\[1mm] & -24,mrm{Li}_{6}pars{1} + 24,mrm{Li}_{6}pars{1 over 2}
end{align}
Note that
$ds{mrm{Li}_{2}pars{1 over 2} = {pi^{2} over 12} -
{ln^{2}pars{2} over 2}}$ and
$ds{mrm{Li}_{3}pars{1 over 2} = {ln^{3}pars{2} over 6} -
{pi^{2}lnpars{2} over 12} + {7zetapars{3} over 8}}$. Moreover,
$ds{vphantom{Huge A}mrm{Li}_{6}pars{1} = zetapars{6} =
{pi^{6} over 945}}$.
Then,
$$
begin{array}{|rcl|}hline mbox{}&&\
ds{quadmc{I}_{1}} & ds{equiv} &
ds{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x}
\[5mm] & ds{=} &
ds{-,{8pi^{6} over 315} - {pi^{2}ln^{4}pars{2} over 4} +
{ln^{6}pars{2} over 6} + 12ln^{2}pars{2},mrm{Li}_{4}pars{1 over 2} +
24lnpars{2},mrm{Li}_{5}pars{1 over 2}quad}
\[1mm] &&
ds{%
+ 24,mrm{Li}_{6}pars{1 over 2} + {7ln^{3}pars{2},zetapars{3} over 2}
approx -0.0269}
\ mbox{}&&
\ hline
end{array}
$$
$ds{Hugemc{I}_{2}: ?}$. This one isn't trivial at all. An attempt is given by
begin{align}
mc{I}_{2} & equiv
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over 1 - x},dd x
,,,stackrel{x mapsto 1 - x}{=},,,
int_{0}^{1/2}{ln^{4}pars{1 - x}lnpars{x} over x},dd x
\[5mm] & =
left.partiald[4]{}{mu}partiald{}{nu}
int_{0}^{1/2}bracks{pars{1 - x}^{mu} - 1}x^{nu - 1}
,dd x,rightvert_{ mu = 0,, nu = 0^{+}}
\[5mm] & =
partiald[4]{}{mu}partiald{}{nu}
bracks{int_{0}^{1/2}pars{1 - x}^{mu},x^{nu - 1}
,dd x - {1 over 2^{nu}nu}}_{ mu = 0,, nu = 0^{+}}
\[5mm] & =
partiald[4]{}{mu}partiald{}{nu}
bracks{mrm{B}pars{{1 over 2},nu,1 + mu} - {1 over 2^{nu}nu}}_{ mu = 0,, nu = 0^{+}}
end{align}
$ds{mrm{B}}$ is the Incomplete Beta Function. It'll continue$ldots$
answered Feb 1 at 23:17
Felix Marin
66.4k7107139
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[#ffd,10px]{ds{%
int_{0}^{1}{ln^{4}pars{1 + x}lnpars{x} over x},dd x}}
,,,stackrel{1 + x mapsto x}{=},,,
int_{1}^{2}{ln^{4}pars{x}lnpars{x - 1} over x - 1},dd x
\[5mm] stackrel{x mapsto 1/x}{=},,,&
int_{1}^{1/2}{ln^{4}pars{1/x}lnpars{1/x - 1} over 1/x - 1}
,pars{-,{dd x over x^{2}}} =
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} - ln^{5}pars{x} over
xpars{1 - x}},dd x
\[1cm] = &
underbrace{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x}
_{ds{mc{I}_{1}}} -
underbrace{int_{1/2}^{1}{ln^{5}pars{x} over x},dd x}
_{ds{= -,{ln^{6}pars{2} over 6}}} +
underbrace{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over 1 - x},dd x}
_{ds{mc{I}_{2}}}
\[1mm] &
-int_{1/2}^{1}{ln^{5}pars{x} over 1 - x},dd x
label{1}tag{1}
end{align}
Note that
begin{align}
int_{1/2}^{1}{ln^{5}pars{x} over 1 - x},dd x & =
-ln^{6}pars{2} +
5int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x =
-ln^{6}pars{2} + 5,mc{I}_{1}
end{align}
such that eqref{1} becomes
begin{equation}
bbx{bbox[#ffd,10px]{ds{%
int_{0}^{1}{ln^{4}pars{1 + x}lnpars{x} over x},dd x}} =
{7ln^{6}pars{2} over 6} - 4,mc{I}_{1} + mc{I}_{2}}label{2}tag{2}
end{equation}
$ds{Hugemc{I}_{1}: ?}$.
begin{align}
mc{I}_{1} & equiv
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x =
-int_{1/2}^{1}mrm{Li}_{2}'pars{x}ln^{4}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4int_{1/2}^{1}mrm{Li}_{3}'pars{x}ln^{3}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} -
12int_{1/2}^{1}mrm{Li}_{4}'pars{x}ln^{2}pars{x},dd x
\[5mm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24int_{1/2}^{1}mrm{Li}_{5}'pars{x}lnpars{x},dd x
\[1cm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24,mrm{Li}_{5}pars{1 over 2}lnpars{2}
\[1mm] & - 24int_{1/2}^{1}mrm{Li}_{6}'pars{x},dd x
\[1cm] & =
mrm{Li}_{2}pars{1 over 2}ln^{4}pars{2} +
4,mrm{Li}_{3}pars{1 over 2}ln^{3}pars{2} +
12,mrm{Li}_{4}pars{1 over 2}ln^{2}pars{2} +
24,mrm{Li}_{5}pars{1 over 2}lnpars{2}
\[1mm] & -24,mrm{Li}_{6}pars{1} + 24,mrm{Li}_{6}pars{1 over 2}
end{align}
Note that
$ds{mrm{Li}_{2}pars{1 over 2} = {pi^{2} over 12} -
{ln^{2}pars{2} over 2}}$ and
$ds{mrm{Li}_{3}pars{1 over 2} = {ln^{3}pars{2} over 6} -
{pi^{2}lnpars{2} over 12} + {7zetapars{3} over 8}}$. Moreover,
$ds{vphantom{Huge A}mrm{Li}_{6}pars{1} = zetapars{6} =
{pi^{6} over 945}}$.
Then,
$$
begin{array}{|rcl|}hline mbox{}&&\
ds{quadmc{I}_{1}} & ds{equiv} &
ds{int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over x},dd x}
\[5mm] & ds{=} &
ds{-,{8pi^{6} over 315} - {pi^{2}ln^{4}pars{2} over 4} +
{ln^{6}pars{2} over 6} + 12ln^{2}pars{2},mrm{Li}_{4}pars{1 over 2} +
24lnpars{2},mrm{Li}_{5}pars{1 over 2}quad}
\[1mm] &&
ds{%
+ 24,mrm{Li}_{6}pars{1 over 2} + {7ln^{3}pars{2},zetapars{3} over 2}
approx -0.0269}
\ mbox{}&&
\ hline
end{array}
$$
$ds{Hugemc{I}_{2}: ?}$. This one isn't trivial at all. An attempt is given by
begin{align}
mc{I}_{2} & equiv
int_{1/2}^{1}{ln^{4}pars{x}lnpars{1 - x} over 1 - x},dd x
,,,stackrel{x mapsto 1 - x}{=},,,
int_{0}^{1/2}{ln^{4}pars{1 - x}lnpars{x} over x},dd x
\[5mm] & =
left.partiald[4]{}{mu}partiald{}{nu}
int_{0}^{1/2}bracks{pars{1 - x}^{mu} - 1}x^{nu - 1}
,dd x,rightvert_{ mu = 0,, nu = 0^{+}}
\[5mm] & =
partiald[4]{}{mu}partiald{}{nu}
bracks{int_{0}^{1/2}pars{1 - x}^{mu},x^{nu - 1}
,dd x - {1 over 2^{nu}nu}}_{ mu = 0,, nu = 0^{+}}
\[5mm] & =
partiald[4]{}{mu}partiald{}{nu}
bracks{mrm{B}pars{{1 over 2},nu,1 + mu} - {1 over 2^{nu}nu}}_{ mu = 0,, nu = 0^{+}}
end{align}
$ds{mrm{B}}$ is the Incomplete Beta Function. It'll continue$ldots$
answered Feb 1 at 23:17
Felix Marin
66.4k7107139
answered Feb 1 at 23:17
Felix Marin
66.4k7107139
answered Feb 1 at 23:17
Felix Marin
66.4k7107139
answered Feb 1 at 23:17
Felix Marin
66.4k7107139
66.4k7107139
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Your conjecture is based on those two examples? Or do you have more clues to guess the closed form of the integrals you are considering?
– user37238
Jun 27 '17 at 13:05
We have $$ log^4(1-x)=sum_{ngeq 1}frac{4x^n}{n}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right] tag{1}$$ hence the given integral equals $$ I=sum_{ngeq 1}frac{4(-1)^{n+1}}{n^3}left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}right]tag{2} $$
– Jack D'Aurizio
Jun 27 '17 at 13:15
Plenty of Euler sums are involved. What is the purpose of reaching a closed form depending on values of $zeta$ at the integers and $text{Li}_n$ at $frac{1}{2}$?
– Jack D'Aurizio
Jun 27 '17 at 13:16
Might be a fun problem to apply the PSQL algorithm to: crd-legacy.lbl.gov/~dhbailey/dhbpapers/ppslq.pdf
– user14717
Jun 27 '17 at 17:24
5
$$ int_0^1frac{ln^4(1+x)ln x}x ; dx = frac{pi^2}3ln^42-frac{17}{30}ln^62-frac{pi^4}{60}ln^22+frac{26}{315}{pi^6}-2{pi^2}zeta(3)ln2+12(zeta(3))^2+frac{3}{4}zeta(5)ln2-3zeta(3)ln^32-24operatorname{Li}_4left(frac12right)ln^22-72operatorname{Li}_5left(frac12right)ln2-96operatorname{Li}_6left(frac12right)+12S.$$ $$ S=sum^infty_{n=1}frac{H_n}{(n+1)^52^n}=0.017446006115193776854329$$
– user178256
Jun 28 '17 at 6:04