Prove $int_0^1frac{log(t^2-t+1)}{t^2-t}mathrm dt=frac{pi^2}9$












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I am in the middle of proving that
$$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac{pi^2}{18}$$
And I have reduced the series to
$$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac12int_0^1frac{log(t^2-t+1)}{t^2-t}mathrm dt$$
But this integral is giving me issues. I broke up the integral
$$int_0^1frac{log(t^2-t+1)}{t^2-t}mathrm dt=int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt-int_0^1frac{log(t^2-t+1)}tmathrm dt$$
I preformed the substitution $t-1=u$ on the first integral, then split it up:
$$int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt=int_{-1}^0frac{log(2u+isqrt3+1)}umathrm du+int_{-1}^0frac{log(2u-isqrt3+1)}umathrm du-2log2int_{-1}^0frac{mathrm du}u$$
But the last term diverges, but I don't know what I did wrong. In any case, I would be surprised if there wasn't an easier way to go about this. Any suggestions? Thanks.






integration sequences-and-series alternative-proof







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  • $begingroup$
    The other two integrals, with the $isqrt3$'s, diverge as well. You're essentially winding up with an expression of the form $infty+infty-2infty$.
    $endgroup$
    – Barry Cipra
    Dec 3 '18 at 18:09








  • 1




    $begingroup$
    The correct substitution is $t=1-u$. Then the two integrals become the same
    $endgroup$
    – Federico
    Dec 3 '18 at 18:12










  • $begingroup$
    Sorry to bump this, but can you tell me how did you arrive at that integral from the sum?
    $endgroup$
    – Zacky
    Dec 5 '18 at 1:37








  • 1




    $begingroup$
    @Zacky first we note that $$frac1{k^2{2kchoose k}}=frac{Gamma^2(k)}{2kGamma(2k)}$$ We then use the definition of the beta function to find $$frac{Gamma^2(k)}{2kGamma(2k)}=frac1{2k}int_0^1 t^{k-1}(1-t)^{k-1}mathrm dt$$ Finally, using the Taylor series $$frac{log(t+1)}t=sum_{kgeq1}frac{(-1)^{k-1}t^{k-1}}k$$ we interchange the sum and integral signs to arrive at $$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac12int_0^1 frac{log(t^2-t+1)}{t^2-t}mathrm dt$$
    $endgroup$
    – clathratus
    Dec 5 '18 at 4:09












  • $begingroup$
    An equivalent problem is discussed at page 31 of my notes.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 0:10
















6












$begingroup$


I am in the middle of proving that
$$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac{pi^2}{18}$$
And I have reduced the series to
$$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac12int_0^1frac{log(t^2-t+1)}{t^2-t}mathrm dt$$
But this integral is giving me issues. I broke up the integral
$$int_0^1frac{log(t^2-t+1)}{t^2-t}mathrm dt=int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt-int_0^1frac{log(t^2-t+1)}tmathrm dt$$
I preformed the substitution $t-1=u$ on the first integral, then split it up:
$$int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt=int_{-1}^0frac{log(2u+isqrt3+1)}umathrm du+int_{-1}^0frac{log(2u-isqrt3+1)}umathrm du-2log2int_{-1}^0frac{mathrm du}u$$
But the last term diverges, but I don't know what I did wrong. In any case, I would be surprised if there wasn't an easier way to go about this. Any suggestions? Thanks.






integration sequences-and-series alternative-proof







share|cite|improve this question











$endgroup$












  • $begingroup$
    The other two integrals, with the $isqrt3$'s, diverge as well. You're essentially winding up with an expression of the form $infty+infty-2infty$.
    $endgroup$
    – Barry Cipra
    Dec 3 '18 at 18:09








  • 1




    $begingroup$
    The correct substitution is $t=1-u$. Then the two integrals become the same
    $endgroup$
    – Federico
    Dec 3 '18 at 18:12










  • $begingroup$
    Sorry to bump this, but can you tell me how did you arrive at that integral from the sum?
    $endgroup$
    – Zacky
    Dec 5 '18 at 1:37








  • 1




    $begingroup$
    @Zacky first we note that $$frac1{k^2{2kchoose k}}=frac{Gamma^2(k)}{2kGamma(2k)}$$ We then use the definition of the beta function to find $$frac{Gamma^2(k)}{2kGamma(2k)}=frac1{2k}int_0^1 t^{k-1}(1-t)^{k-1}mathrm dt$$ Finally, using the Taylor series $$frac{log(t+1)}t=sum_{kgeq1}frac{(-1)^{k-1}t^{k-1}}k$$ we interchange the sum and integral signs to arrive at $$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac12int_0^1 frac{log(t^2-t+1)}{t^2-t}mathrm dt$$
    $endgroup$
    – clathratus
    Dec 5 '18 at 4:09












  • $begingroup$
    An equivalent problem is discussed at page 31 of my notes.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 0:10














6












6








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4



$begingroup$


I am in the middle of proving that
$$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac{pi^2}{18}$$
And I have reduced the series to
$$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac12int_0^1frac{log(t^2-t+1)}{t^2-t}mathrm dt$$
But this integral is giving me issues. I broke up the integral
$$int_0^1frac{log(t^2-t+1)}{t^2-t}mathrm dt=int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt-int_0^1frac{log(t^2-t+1)}tmathrm dt$$
I preformed the substitution $t-1=u$ on the first integral, then split it up:
$$int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt=int_{-1}^0frac{log(2u+isqrt3+1)}umathrm du+int_{-1}^0frac{log(2u-isqrt3+1)}umathrm du-2log2int_{-1}^0frac{mathrm du}u$$
But the last term diverges, but I don't know what I did wrong. In any case, I would be surprised if there wasn't an easier way to go about this. Any suggestions? Thanks.






integration sequences-and-series alternative-proof







share|cite|improve this question











$endgroup$




I am in the middle of proving that
$$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac{pi^2}{18}$$
And I have reduced the series to
$$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac12int_0^1frac{log(t^2-t+1)}{t^2-t}mathrm dt$$
But this integral is giving me issues. I broke up the integral
$$int_0^1frac{log(t^2-t+1)}{t^2-t}mathrm dt=int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt-int_0^1frac{log(t^2-t+1)}tmathrm dt$$
I preformed the substitution $t-1=u$ on the first integral, then split it up:
$$int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt=int_{-1}^0frac{log(2u+isqrt3+1)}umathrm du+int_{-1}^0frac{log(2u-isqrt3+1)}umathrm du-2log2int_{-1}^0frac{mathrm du}u$$
But the last term diverges, but I don't know what I did wrong. In any case, I would be surprised if there wasn't an easier way to go about this. Any suggestions? Thanks.






integration sequences-and-series alternative-proof




integration sequences-and-series alternative-proof






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edited Dec 3 '18 at 18:01







clathratus

















asked Dec 3 '18 at 17:29









clathratusclathratus

3,740333




3,740333












  • $begingroup$
    The other two integrals, with the $isqrt3$'s, diverge as well. You're essentially winding up with an expression of the form $infty+infty-2infty$.
    $endgroup$
    – Barry Cipra
    Dec 3 '18 at 18:09








  • 1




    $begingroup$
    The correct substitution is $t=1-u$. Then the two integrals become the same
    $endgroup$
    – Federico
    Dec 3 '18 at 18:12










  • $begingroup$
    Sorry to bump this, but can you tell me how did you arrive at that integral from the sum?
    $endgroup$
    – Zacky
    Dec 5 '18 at 1:37








  • 1




    $begingroup$
    @Zacky first we note that $$frac1{k^2{2kchoose k}}=frac{Gamma^2(k)}{2kGamma(2k)}$$ We then use the definition of the beta function to find $$frac{Gamma^2(k)}{2kGamma(2k)}=frac1{2k}int_0^1 t^{k-1}(1-t)^{k-1}mathrm dt$$ Finally, using the Taylor series $$frac{log(t+1)}t=sum_{kgeq1}frac{(-1)^{k-1}t^{k-1}}k$$ we interchange the sum and integral signs to arrive at $$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac12int_0^1 frac{log(t^2-t+1)}{t^2-t}mathrm dt$$
    $endgroup$
    – clathratus
    Dec 5 '18 at 4:09












  • $begingroup$
    An equivalent problem is discussed at page 31 of my notes.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 0:10


















  • $begingroup$
    The other two integrals, with the $isqrt3$'s, diverge as well. You're essentially winding up with an expression of the form $infty+infty-2infty$.
    $endgroup$
    – Barry Cipra
    Dec 3 '18 at 18:09








  • 1




    $begingroup$
    The correct substitution is $t=1-u$. Then the two integrals become the same
    $endgroup$
    – Federico
    Dec 3 '18 at 18:12










  • $begingroup$
    Sorry to bump this, but can you tell me how did you arrive at that integral from the sum?
    $endgroup$
    – Zacky
    Dec 5 '18 at 1:37








  • 1




    $begingroup$
    @Zacky first we note that $$frac1{k^2{2kchoose k}}=frac{Gamma^2(k)}{2kGamma(2k)}$$ We then use the definition of the beta function to find $$frac{Gamma^2(k)}{2kGamma(2k)}=frac1{2k}int_0^1 t^{k-1}(1-t)^{k-1}mathrm dt$$ Finally, using the Taylor series $$frac{log(t+1)}t=sum_{kgeq1}frac{(-1)^{k-1}t^{k-1}}k$$ we interchange the sum and integral signs to arrive at $$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac12int_0^1 frac{log(t^2-t+1)}{t^2-t}mathrm dt$$
    $endgroup$
    – clathratus
    Dec 5 '18 at 4:09












  • $begingroup$
    An equivalent problem is discussed at page 31 of my notes.
    $endgroup$
    – Jack D'Aurizio
    Dec 6 '18 at 0:10
















$begingroup$
The other two integrals, with the $isqrt3$'s, diverge as well. You're essentially winding up with an expression of the form $infty+infty-2infty$.
$endgroup$
– Barry Cipra
Dec 3 '18 at 18:09






$begingroup$
The other two integrals, with the $isqrt3$'s, diverge as well. You're essentially winding up with an expression of the form $infty+infty-2infty$.
$endgroup$
– Barry Cipra
Dec 3 '18 at 18:09






1




1




$begingroup$
The correct substitution is $t=1-u$. Then the two integrals become the same
$endgroup$
– Federico
Dec 3 '18 at 18:12




$begingroup$
The correct substitution is $t=1-u$. Then the two integrals become the same
$endgroup$
– Federico
Dec 3 '18 at 18:12












$begingroup$
Sorry to bump this, but can you tell me how did you arrive at that integral from the sum?
$endgroup$
– Zacky
Dec 5 '18 at 1:37






$begingroup$
Sorry to bump this, but can you tell me how did you arrive at that integral from the sum?
$endgroup$
– Zacky
Dec 5 '18 at 1:37






1




1




$begingroup$
@Zacky first we note that $$frac1{k^2{2kchoose k}}=frac{Gamma^2(k)}{2kGamma(2k)}$$ We then use the definition of the beta function to find $$frac{Gamma^2(k)}{2kGamma(2k)}=frac1{2k}int_0^1 t^{k-1}(1-t)^{k-1}mathrm dt$$ Finally, using the Taylor series $$frac{log(t+1)}t=sum_{kgeq1}frac{(-1)^{k-1}t^{k-1}}k$$ we interchange the sum and integral signs to arrive at $$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac12int_0^1 frac{log(t^2-t+1)}{t^2-t}mathrm dt$$
$endgroup$
– clathratus
Dec 5 '18 at 4:09






$begingroup$
@Zacky first we note that $$frac1{k^2{2kchoose k}}=frac{Gamma^2(k)}{2kGamma(2k)}$$ We then use the definition of the beta function to find $$frac{Gamma^2(k)}{2kGamma(2k)}=frac1{2k}int_0^1 t^{k-1}(1-t)^{k-1}mathrm dt$$ Finally, using the Taylor series $$frac{log(t+1)}t=sum_{kgeq1}frac{(-1)^{k-1}t^{k-1}}k$$ we interchange the sum and integral signs to arrive at $$sum_{kgeq1}frac1{k^2{2kchoose k}}=frac12int_0^1 frac{log(t^2-t+1)}{t^2-t}mathrm dt$$
$endgroup$
– clathratus
Dec 5 '18 at 4:09














$begingroup$
An equivalent problem is discussed at page 31 of my notes.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 0:10




$begingroup$
An equivalent problem is discussed at page 31 of my notes.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 0:10










4 Answers
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Let's multiply by $1$ the integral found in @Federico's answer.
$$int_0^1frac{log(1-x+x^2)}x dx =int_0^1frac{ln(1+x^3)-ln(1+x)}{x}dx$$
$$int_0^1frac{ln(1+x^3)}{x}dxoverset{x=t^{1/3}}=frac13int_0^1 frac{ln(1+t)}{t^{1/3}},t^{1/3-1}dtoverset{t=x}=frac13int_0^1frac{ln(1+x)}{x}dx$$
$$sum_{n=1}^infty frac1{n^2{2nchoose n}}
=frac23 int_0^1 frac{ln(1+x)}{x}dx=frac23sum_{n=1}^infty int_0^1frac{(-1)^{n-1}x^{n-1}}{n}dx$$$$=frac23sum_{n=1}^infty frac{(-1)^{n-1}}{n^2}=frac23cdotfrac{pi^2}{12}=frac{pi^2}{18}$$
Above I used: $ displaystyle{ln(1+x)=sum_{n=1}^infty frac{(-1)^{n-1}x^n}{n}}$






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  • $begingroup$
    What was the $1$ you multiplied by in the first step?
    $endgroup$
    – clathratus
    Dec 4 '18 at 0:02






  • 1




    $begingroup$
    $frac{1+x}{1+x}$
    $endgroup$
    – Zacky
    Dec 4 '18 at 0:03








  • 1




    $begingroup$
    Perfect! thank you!
    $endgroup$
    – clathratus
    Dec 4 '18 at 0:04



















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I start from
$$
int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt-int_0^1frac{log(t^2-t+1)}tmathrm dt .
$$
In the first integral, substitute $t=1-u$. Then
$$
int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt
=-int_0^1frac{log(u^2-u+1)}{u}mathrm du .
$$



So you get
$$
sum_{kgeq1}frac1{k^2{2kchoose k}}
= -int_0^1frac{log(t^2-t+1)}tmathrm dt .
$$



ADDENDUM



After some sleep, I managed to compute the integral with the help of polylogarithm.
For $ninmathbb R$, define
$$
mathrm{Li}_n(z) = sum_{k=1}^infty frac{z^k}{k^n}.
$$



Some simple facts.





  • $mathrm{Li}_1(z)=-log(1-z)$.


  • $mathrm{Li}_2(-1)=sum_{k=1}^infty frac{(-1)^k}{k^2} = -frac{pi^2}{12}$.


  • $zfrac{d}{dz} mathrm{Li}_n(z) = mathrm{Li}_{n-1}(z)$.


  • $mathrm{Li}_n(z) = int_0^z frac{mathrm{Li}_{n-1}(s)}{s},ds$.


Then
$$
begin{split}
-int_0^1frac{log(t^2-t+1)}tmathrm dt
&= -int_0^1frac{log(1+t^3)-log(1+t)}tmathrm dt \
&= frac13int_0^1frac{mathrm{Li}_1(-t^3)}{t^3}3t^2mathrm dt
- int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= frac13 int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt
- int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= -frac23 int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= -frac23 int_0^{-1}frac{mathrm{Li}_1(t)}{t}mathrm dt \
&= -frac23 mathrm{Li}_2(-1)
= -frac23 left(-frac{pi^2}{12}right)
= frac{pi^2}{18}.
end{split}
$$






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  • $begingroup$
    can I have a teeny tiny hint on how to start with this remaining integral?
    $endgroup$
    – clathratus
    Dec 3 '18 at 18:25






  • 2




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    Oh well, I don't have the slightest idea right now :D
    $endgroup$
    – Federico
    Dec 3 '18 at 18:32






  • 2




    $begingroup$
    But at least you have only one integral and not two!
    $endgroup$
    – Federico
    Dec 3 '18 at 18:32










  • $begingroup$
    Thanks for this addendum, it's pretty neat.
    $endgroup$
    – clathratus
    Dec 4 '18 at 17:44



















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Another Approach is to employ Feynman's Trick:



Let



$$I(x) = int_{0}^{1} frac{lnleft| x^2left(t^2 - tright) + 1right|}{t^2 - t}:dt$$



Note $I = I(1)$ and $I(0) = 0$



Thus



begin{align}
I'(x) &= int_{0}^{1} frac{2xleft(t^2 - tright)}{left(x^2left(t^2 - tright) + 1right)left( t^2 - tright)}:dt = frac{2}{x}int_{0}^{1} frac{1}{left(t - frac{1}{2}right)^2 + frac{4 - x^2}{4x^2}}:dt\
&= frac{4}{x}int_{0}^{frac{1}{2}} frac{1}{t^2 + frac{4 - x^2}{4x^2}}:dt = frac{8}{sqrt{4 - x^2}}arctanleft(frac{x}{sqrt{4 -x^2}} right)
end{align}



We now integrate to solve $I(x)$



$$I(x) = intfrac{8}{sqrt{4 - x^2}}arctanleft(frac{x}{sqrt{4 -x^2}} right) :dx = 4left[arctanleft( frac{x}{sqrt{4 - x^2}}right) right]^2 + C $$



Where $C$ is a constant of integration. As $I(0) = 0$ we find $C = 0$ and so:



$$I(x) = 4left[arctanleft( frac{x}{sqrt{4 - x^2}}right) right]^2$$



And finally



$$ I = I(1) = 4left[arctanleft( frac{1}{sqrt{3}}right) right]^2 = frac{pi^2}{9}$$






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  • $begingroup$
    This is a really cool approach! In retrospect, seeing the $$frac{log(t^2-t+1)}{t^2-t}$$ really begs for the parameterization which you introduced. Great job.
    $endgroup$
    – clathratus
    Dec 5 '18 at 3:57






  • 1




    $begingroup$
    I'm glad you enjoyed it. I've been on a Feynman 'kick' lately and have started to develop pattern matching skills. One element that may seem odd on the surface is the use of $x^2$ rather than just $x$. If you are interested, try both. You will find the $x$ creates an incredibly nasty indefinite integral to solve at the end and yet $x^2$ makes the process incredibly easy.
    $endgroup$
    – DavidG
    Dec 5 '18 at 4:33










  • $begingroup$
    I've done quite a bit with series, but not much with the 'Feynman' trick. Any resources that you've found especially helpful in learning about when and how to use it?
    $endgroup$
    – clathratus
    Dec 5 '18 at 4:37






  • 1




    $begingroup$
    @clathratus - Please refer to my page. I've almost exclusively being posting questions/answers in this area. In some parts I combine both Feynman's Trick and Laplace Transforms. I've also used multiple introduced parameters to solve systems. I would love to have a 'Feynman Trick' mate (/buddy for the US Folk) so please contact me at any time. I want to put together a simple LaTeX document with examples of this method and would love contributors.
    $endgroup$
    – DavidG
    Dec 5 '18 at 4:58










  • $begingroup$
    How can I contact you? I'm afraid that I may not have much to offer with the Feynman trick, but I do have more experience with generalized series, factorials/products, and I have been recently investigating special functions and finding closed forms, which may be useful. We could combine forces and do some cool stuff. By the way, you may contact me via the email address joverton2020@gmail.com
    $endgroup$
    – clathratus
    Dec 5 '18 at 5:22



















1












$begingroup$

Firstly observe that, for $x$ real,



begin{align}(1-x)^2-(1-x)+1&=(1-2x+x^2)-1+x+1\
&=x^2-x+1
end{align}



begin{align}J&=int_0^1 frac{ln(x^2-x+1)}{x(x-1)},dx\
&=-int_0^1 frac{ln(x^2-x+1)}{1-x},dx-int_0^1 frac{ln(x^2-x+1)}{x},dx
end{align}



In the first integral perform the change of variable $y=1-x$,



begin{align}J&=-2int_0^1 frac{ln(x^2-x+1)}{x},dx\
&=-2int_0^1 frac{lnleft(frac{x^3+1}{x+1}right)}{x},dx\
&=2int_0^1 frac{lnleft(x+1right)}{x},dx-2int_0^1 frac{lnleft(x^3+1right)}{x},dx\
&=2int_0^1 frac{lnleft(x+1right)}{x},dx-2int_0^1 frac{x^2lnleft(x^3+1right)}{x^3},dx\
end{align}



In the latter integral perform the change of variable $y=x^3$,



begin{align}J&=2int_0^1 frac{lnleft(x+1right)}{x},dx-frac{2}{3}int_0^1
frac{lnleft(x+1right)}{x},dx\
&=frac{4}{3}int_0^1 frac{lnleft(x+1right)}{x},dx\
&=frac{4}{3}int_0^1 frac{lnleft(1-x^2right)}{x},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=frac{4}{3}int_0^1 frac{xlnleft(1-x^2right)}{x^2},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx
end{align}



In the first integral perform the change of variable $y=x^2$,



begin{align}J&=frac{2}{3}int_0^1 frac{lnleft(1-xright)}{x},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=-frac{2}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=-frac{2}{3}Big[ln xln(1-x)Big]_0^1 -frac{2}{3}int_0^1 frac{ln x}{1-x},dx\
&=-frac{2}{3}int_0^1 frac{ln x}{1-x},dx\
&=frac{2}{3}zeta(2)\
&=frac{2}{3}times frac{pi^2}{6}\
&=boxed{frac{pi^2}{9}}
end{align}






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  • $begingroup$
    I really appreciate the detailed and aesthetically pleasing nature of this response, not to mention the elegance. Thanks for this great answer.
    $endgroup$
    – clathratus
    Dec 5 '18 at 17:47










  • $begingroup$
    Thanks, but after reading all the anwers i'm afraid i have only summed up computations already written by other people (with some details ok)
    $endgroup$
    – FDP
    Dec 5 '18 at 18:53











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4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Let's multiply by $1$ the integral found in @Federico's answer.
$$int_0^1frac{log(1-x+x^2)}x dx =int_0^1frac{ln(1+x^3)-ln(1+x)}{x}dx$$
$$int_0^1frac{ln(1+x^3)}{x}dxoverset{x=t^{1/3}}=frac13int_0^1 frac{ln(1+t)}{t^{1/3}},t^{1/3-1}dtoverset{t=x}=frac13int_0^1frac{ln(1+x)}{x}dx$$
$$sum_{n=1}^infty frac1{n^2{2nchoose n}}
=frac23 int_0^1 frac{ln(1+x)}{x}dx=frac23sum_{n=1}^infty int_0^1frac{(-1)^{n-1}x^{n-1}}{n}dx$$$$=frac23sum_{n=1}^infty frac{(-1)^{n-1}}{n^2}=frac23cdotfrac{pi^2}{12}=frac{pi^2}{18}$$
Above I used: $ displaystyle{ln(1+x)=sum_{n=1}^infty frac{(-1)^{n-1}x^n}{n}}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What was the $1$ you multiplied by in the first step?
    $endgroup$
    – clathratus
    Dec 4 '18 at 0:02






  • 1




    $begingroup$
    $frac{1+x}{1+x}$
    $endgroup$
    – Zacky
    Dec 4 '18 at 0:03








  • 1




    $begingroup$
    Perfect! thank you!
    $endgroup$
    – clathratus
    Dec 4 '18 at 0:04
















4












$begingroup$

Let's multiply by $1$ the integral found in @Federico's answer.
$$int_0^1frac{log(1-x+x^2)}x dx =int_0^1frac{ln(1+x^3)-ln(1+x)}{x}dx$$
$$int_0^1frac{ln(1+x^3)}{x}dxoverset{x=t^{1/3}}=frac13int_0^1 frac{ln(1+t)}{t^{1/3}},t^{1/3-1}dtoverset{t=x}=frac13int_0^1frac{ln(1+x)}{x}dx$$
$$sum_{n=1}^infty frac1{n^2{2nchoose n}}
=frac23 int_0^1 frac{ln(1+x)}{x}dx=frac23sum_{n=1}^infty int_0^1frac{(-1)^{n-1}x^{n-1}}{n}dx$$$$=frac23sum_{n=1}^infty frac{(-1)^{n-1}}{n^2}=frac23cdotfrac{pi^2}{12}=frac{pi^2}{18}$$
Above I used: $ displaystyle{ln(1+x)=sum_{n=1}^infty frac{(-1)^{n-1}x^n}{n}}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What was the $1$ you multiplied by in the first step?
    $endgroup$
    – clathratus
    Dec 4 '18 at 0:02






  • 1




    $begingroup$
    $frac{1+x}{1+x}$
    $endgroup$
    – Zacky
    Dec 4 '18 at 0:03








  • 1




    $begingroup$
    Perfect! thank you!
    $endgroup$
    – clathratus
    Dec 4 '18 at 0:04














4












4








4





$begingroup$

Let's multiply by $1$ the integral found in @Federico's answer.
$$int_0^1frac{log(1-x+x^2)}x dx =int_0^1frac{ln(1+x^3)-ln(1+x)}{x}dx$$
$$int_0^1frac{ln(1+x^3)}{x}dxoverset{x=t^{1/3}}=frac13int_0^1 frac{ln(1+t)}{t^{1/3}},t^{1/3-1}dtoverset{t=x}=frac13int_0^1frac{ln(1+x)}{x}dx$$
$$sum_{n=1}^infty frac1{n^2{2nchoose n}}
=frac23 int_0^1 frac{ln(1+x)}{x}dx=frac23sum_{n=1}^infty int_0^1frac{(-1)^{n-1}x^{n-1}}{n}dx$$$$=frac23sum_{n=1}^infty frac{(-1)^{n-1}}{n^2}=frac23cdotfrac{pi^2}{12}=frac{pi^2}{18}$$
Above I used: $ displaystyle{ln(1+x)=sum_{n=1}^infty frac{(-1)^{n-1}x^n}{n}}$






share|cite|improve this answer











$endgroup$



Let's multiply by $1$ the integral found in @Federico's answer.
$$int_0^1frac{log(1-x+x^2)}x dx =int_0^1frac{ln(1+x^3)-ln(1+x)}{x}dx$$
$$int_0^1frac{ln(1+x^3)}{x}dxoverset{x=t^{1/3}}=frac13int_0^1 frac{ln(1+t)}{t^{1/3}},t^{1/3-1}dtoverset{t=x}=frac13int_0^1frac{ln(1+x)}{x}dx$$
$$sum_{n=1}^infty frac1{n^2{2nchoose n}}
=frac23 int_0^1 frac{ln(1+x)}{x}dx=frac23sum_{n=1}^infty int_0^1frac{(-1)^{n-1}x^{n-1}}{n}dx$$$$=frac23sum_{n=1}^infty frac{(-1)^{n-1}}{n^2}=frac23cdotfrac{pi^2}{12}=frac{pi^2}{18}$$
Above I used: $ displaystyle{ln(1+x)=sum_{n=1}^infty frac{(-1)^{n-1}x^n}{n}}$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 4 '18 at 0:55

























answered Dec 3 '18 at 23:42









ZackyZacky

5,5081856




5,5081856












  • $begingroup$
    What was the $1$ you multiplied by in the first step?
    $endgroup$
    – clathratus
    Dec 4 '18 at 0:02






  • 1




    $begingroup$
    $frac{1+x}{1+x}$
    $endgroup$
    – Zacky
    Dec 4 '18 at 0:03








  • 1




    $begingroup$
    Perfect! thank you!
    $endgroup$
    – clathratus
    Dec 4 '18 at 0:04


















  • $begingroup$
    What was the $1$ you multiplied by in the first step?
    $endgroup$
    – clathratus
    Dec 4 '18 at 0:02






  • 1




    $begingroup$
    $frac{1+x}{1+x}$
    $endgroup$
    – Zacky
    Dec 4 '18 at 0:03








  • 1




    $begingroup$
    Perfect! thank you!
    $endgroup$
    – clathratus
    Dec 4 '18 at 0:04
















$begingroup$
What was the $1$ you multiplied by in the first step?
$endgroup$
– clathratus
Dec 4 '18 at 0:02




$begingroup$
What was the $1$ you multiplied by in the first step?
$endgroup$
– clathratus
Dec 4 '18 at 0:02




1




1




$begingroup$
$frac{1+x}{1+x}$
$endgroup$
– Zacky
Dec 4 '18 at 0:03






$begingroup$
$frac{1+x}{1+x}$
$endgroup$
– Zacky
Dec 4 '18 at 0:03






1




1




$begingroup$
Perfect! thank you!
$endgroup$
– clathratus
Dec 4 '18 at 0:04




$begingroup$
Perfect! thank you!
$endgroup$
– clathratus
Dec 4 '18 at 0:04











3












$begingroup$

I start from
$$
int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt-int_0^1frac{log(t^2-t+1)}tmathrm dt .
$$
In the first integral, substitute $t=1-u$. Then
$$
int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt
=-int_0^1frac{log(u^2-u+1)}{u}mathrm du .
$$



So you get
$$
sum_{kgeq1}frac1{k^2{2kchoose k}}
= -int_0^1frac{log(t^2-t+1)}tmathrm dt .
$$



ADDENDUM



After some sleep, I managed to compute the integral with the help of polylogarithm.
For $ninmathbb R$, define
$$
mathrm{Li}_n(z) = sum_{k=1}^infty frac{z^k}{k^n}.
$$



Some simple facts.





  • $mathrm{Li}_1(z)=-log(1-z)$.


  • $mathrm{Li}_2(-1)=sum_{k=1}^infty frac{(-1)^k}{k^2} = -frac{pi^2}{12}$.


  • $zfrac{d}{dz} mathrm{Li}_n(z) = mathrm{Li}_{n-1}(z)$.


  • $mathrm{Li}_n(z) = int_0^z frac{mathrm{Li}_{n-1}(s)}{s},ds$.


Then
$$
begin{split}
-int_0^1frac{log(t^2-t+1)}tmathrm dt
&= -int_0^1frac{log(1+t^3)-log(1+t)}tmathrm dt \
&= frac13int_0^1frac{mathrm{Li}_1(-t^3)}{t^3}3t^2mathrm dt
- int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= frac13 int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt
- int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= -frac23 int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= -frac23 int_0^{-1}frac{mathrm{Li}_1(t)}{t}mathrm dt \
&= -frac23 mathrm{Li}_2(-1)
= -frac23 left(-frac{pi^2}{12}right)
= frac{pi^2}{18}.
end{split}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    can I have a teeny tiny hint on how to start with this remaining integral?
    $endgroup$
    – clathratus
    Dec 3 '18 at 18:25






  • 2




    $begingroup$
    Oh well, I don't have the slightest idea right now :D
    $endgroup$
    – Federico
    Dec 3 '18 at 18:32






  • 2




    $begingroup$
    But at least you have only one integral and not two!
    $endgroup$
    – Federico
    Dec 3 '18 at 18:32










  • $begingroup$
    Thanks for this addendum, it's pretty neat.
    $endgroup$
    – clathratus
    Dec 4 '18 at 17:44
















3












$begingroup$

I start from
$$
int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt-int_0^1frac{log(t^2-t+1)}tmathrm dt .
$$
In the first integral, substitute $t=1-u$. Then
$$
int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt
=-int_0^1frac{log(u^2-u+1)}{u}mathrm du .
$$



So you get
$$
sum_{kgeq1}frac1{k^2{2kchoose k}}
= -int_0^1frac{log(t^2-t+1)}tmathrm dt .
$$



ADDENDUM



After some sleep, I managed to compute the integral with the help of polylogarithm.
For $ninmathbb R$, define
$$
mathrm{Li}_n(z) = sum_{k=1}^infty frac{z^k}{k^n}.
$$



Some simple facts.





  • $mathrm{Li}_1(z)=-log(1-z)$.


  • $mathrm{Li}_2(-1)=sum_{k=1}^infty frac{(-1)^k}{k^2} = -frac{pi^2}{12}$.


  • $zfrac{d}{dz} mathrm{Li}_n(z) = mathrm{Li}_{n-1}(z)$.


  • $mathrm{Li}_n(z) = int_0^z frac{mathrm{Li}_{n-1}(s)}{s},ds$.


Then
$$
begin{split}
-int_0^1frac{log(t^2-t+1)}tmathrm dt
&= -int_0^1frac{log(1+t^3)-log(1+t)}tmathrm dt \
&= frac13int_0^1frac{mathrm{Li}_1(-t^3)}{t^3}3t^2mathrm dt
- int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= frac13 int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt
- int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= -frac23 int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= -frac23 int_0^{-1}frac{mathrm{Li}_1(t)}{t}mathrm dt \
&= -frac23 mathrm{Li}_2(-1)
= -frac23 left(-frac{pi^2}{12}right)
= frac{pi^2}{18}.
end{split}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    can I have a teeny tiny hint on how to start with this remaining integral?
    $endgroup$
    – clathratus
    Dec 3 '18 at 18:25






  • 2




    $begingroup$
    Oh well, I don't have the slightest idea right now :D
    $endgroup$
    – Federico
    Dec 3 '18 at 18:32






  • 2




    $begingroup$
    But at least you have only one integral and not two!
    $endgroup$
    – Federico
    Dec 3 '18 at 18:32










  • $begingroup$
    Thanks for this addendum, it's pretty neat.
    $endgroup$
    – clathratus
    Dec 4 '18 at 17:44














3












3








3





$begingroup$

I start from
$$
int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt-int_0^1frac{log(t^2-t+1)}tmathrm dt .
$$
In the first integral, substitute $t=1-u$. Then
$$
int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt
=-int_0^1frac{log(u^2-u+1)}{u}mathrm du .
$$



So you get
$$
sum_{kgeq1}frac1{k^2{2kchoose k}}
= -int_0^1frac{log(t^2-t+1)}tmathrm dt .
$$



ADDENDUM



After some sleep, I managed to compute the integral with the help of polylogarithm.
For $ninmathbb R$, define
$$
mathrm{Li}_n(z) = sum_{k=1}^infty frac{z^k}{k^n}.
$$



Some simple facts.





  • $mathrm{Li}_1(z)=-log(1-z)$.


  • $mathrm{Li}_2(-1)=sum_{k=1}^infty frac{(-1)^k}{k^2} = -frac{pi^2}{12}$.


  • $zfrac{d}{dz} mathrm{Li}_n(z) = mathrm{Li}_{n-1}(z)$.


  • $mathrm{Li}_n(z) = int_0^z frac{mathrm{Li}_{n-1}(s)}{s},ds$.


Then
$$
begin{split}
-int_0^1frac{log(t^2-t+1)}tmathrm dt
&= -int_0^1frac{log(1+t^3)-log(1+t)}tmathrm dt \
&= frac13int_0^1frac{mathrm{Li}_1(-t^3)}{t^3}3t^2mathrm dt
- int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= frac13 int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt
- int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= -frac23 int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= -frac23 int_0^{-1}frac{mathrm{Li}_1(t)}{t}mathrm dt \
&= -frac23 mathrm{Li}_2(-1)
= -frac23 left(-frac{pi^2}{12}right)
= frac{pi^2}{18}.
end{split}
$$






share|cite|improve this answer











$endgroup$



I start from
$$
int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt-int_0^1frac{log(t^2-t+1)}tmathrm dt .
$$
In the first integral, substitute $t=1-u$. Then
$$
int_0^1frac{log(t^2-t+1)}{t-1}mathrm dt
=-int_0^1frac{log(u^2-u+1)}{u}mathrm du .
$$



So you get
$$
sum_{kgeq1}frac1{k^2{2kchoose k}}
= -int_0^1frac{log(t^2-t+1)}tmathrm dt .
$$



ADDENDUM



After some sleep, I managed to compute the integral with the help of polylogarithm.
For $ninmathbb R$, define
$$
mathrm{Li}_n(z) = sum_{k=1}^infty frac{z^k}{k^n}.
$$



Some simple facts.





  • $mathrm{Li}_1(z)=-log(1-z)$.


  • $mathrm{Li}_2(-1)=sum_{k=1}^infty frac{(-1)^k}{k^2} = -frac{pi^2}{12}$.


  • $zfrac{d}{dz} mathrm{Li}_n(z) = mathrm{Li}_{n-1}(z)$.


  • $mathrm{Li}_n(z) = int_0^z frac{mathrm{Li}_{n-1}(s)}{s},ds$.


Then
$$
begin{split}
-int_0^1frac{log(t^2-t+1)}tmathrm dt
&= -int_0^1frac{log(1+t^3)-log(1+t)}tmathrm dt \
&= frac13int_0^1frac{mathrm{Li}_1(-t^3)}{t^3}3t^2mathrm dt
- int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= frac13 int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt
- int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= -frac23 int_0^1frac{mathrm{Li}_1(-t)}{t}mathrm dt \
&= -frac23 int_0^{-1}frac{mathrm{Li}_1(t)}{t}mathrm dt \
&= -frac23 mathrm{Li}_2(-1)
= -frac23 left(-frac{pi^2}{12}right)
= frac{pi^2}{18}.
end{split}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 4 '18 at 14:11

























answered Dec 3 '18 at 18:11









FedericoFederico

4,919514




4,919514












  • $begingroup$
    can I have a teeny tiny hint on how to start with this remaining integral?
    $endgroup$
    – clathratus
    Dec 3 '18 at 18:25






  • 2




    $begingroup$
    Oh well, I don't have the slightest idea right now :D
    $endgroup$
    – Federico
    Dec 3 '18 at 18:32






  • 2




    $begingroup$
    But at least you have only one integral and not two!
    $endgroup$
    – Federico
    Dec 3 '18 at 18:32










  • $begingroup$
    Thanks for this addendum, it's pretty neat.
    $endgroup$
    – clathratus
    Dec 4 '18 at 17:44


















  • $begingroup$
    can I have a teeny tiny hint on how to start with this remaining integral?
    $endgroup$
    – clathratus
    Dec 3 '18 at 18:25






  • 2




    $begingroup$
    Oh well, I don't have the slightest idea right now :D
    $endgroup$
    – Federico
    Dec 3 '18 at 18:32






  • 2




    $begingroup$
    But at least you have only one integral and not two!
    $endgroup$
    – Federico
    Dec 3 '18 at 18:32










  • $begingroup$
    Thanks for this addendum, it's pretty neat.
    $endgroup$
    – clathratus
    Dec 4 '18 at 17:44
















$begingroup$
can I have a teeny tiny hint on how to start with this remaining integral?
$endgroup$
– clathratus
Dec 3 '18 at 18:25




$begingroup$
can I have a teeny tiny hint on how to start with this remaining integral?
$endgroup$
– clathratus
Dec 3 '18 at 18:25




2




2




$begingroup$
Oh well, I don't have the slightest idea right now :D
$endgroup$
– Federico
Dec 3 '18 at 18:32




$begingroup$
Oh well, I don't have the slightest idea right now :D
$endgroup$
– Federico
Dec 3 '18 at 18:32




2




2




$begingroup$
But at least you have only one integral and not two!
$endgroup$
– Federico
Dec 3 '18 at 18:32




$begingroup$
But at least you have only one integral and not two!
$endgroup$
– Federico
Dec 3 '18 at 18:32












$begingroup$
Thanks for this addendum, it's pretty neat.
$endgroup$
– clathratus
Dec 4 '18 at 17:44




$begingroup$
Thanks for this addendum, it's pretty neat.
$endgroup$
– clathratus
Dec 4 '18 at 17:44











2












$begingroup$

Another Approach is to employ Feynman's Trick:



Let



$$I(x) = int_{0}^{1} frac{lnleft| x^2left(t^2 - tright) + 1right|}{t^2 - t}:dt$$



Note $I = I(1)$ and $I(0) = 0$



Thus



begin{align}
I'(x) &= int_{0}^{1} frac{2xleft(t^2 - tright)}{left(x^2left(t^2 - tright) + 1right)left( t^2 - tright)}:dt = frac{2}{x}int_{0}^{1} frac{1}{left(t - frac{1}{2}right)^2 + frac{4 - x^2}{4x^2}}:dt\
&= frac{4}{x}int_{0}^{frac{1}{2}} frac{1}{t^2 + frac{4 - x^2}{4x^2}}:dt = frac{8}{sqrt{4 - x^2}}arctanleft(frac{x}{sqrt{4 -x^2}} right)
end{align}



We now integrate to solve $I(x)$



$$I(x) = intfrac{8}{sqrt{4 - x^2}}arctanleft(frac{x}{sqrt{4 -x^2}} right) :dx = 4left[arctanleft( frac{x}{sqrt{4 - x^2}}right) right]^2 + C $$



Where $C$ is a constant of integration. As $I(0) = 0$ we find $C = 0$ and so:



$$I(x) = 4left[arctanleft( frac{x}{sqrt{4 - x^2}}right) right]^2$$



And finally



$$ I = I(1) = 4left[arctanleft( frac{1}{sqrt{3}}right) right]^2 = frac{pi^2}{9}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a really cool approach! In retrospect, seeing the $$frac{log(t^2-t+1)}{t^2-t}$$ really begs for the parameterization which you introduced. Great job.
    $endgroup$
    – clathratus
    Dec 5 '18 at 3:57






  • 1




    $begingroup$
    I'm glad you enjoyed it. I've been on a Feynman 'kick' lately and have started to develop pattern matching skills. One element that may seem odd on the surface is the use of $x^2$ rather than just $x$. If you are interested, try both. You will find the $x$ creates an incredibly nasty indefinite integral to solve at the end and yet $x^2$ makes the process incredibly easy.
    $endgroup$
    – DavidG
    Dec 5 '18 at 4:33










  • $begingroup$
    I've done quite a bit with series, but not much with the 'Feynman' trick. Any resources that you've found especially helpful in learning about when and how to use it?
    $endgroup$
    – clathratus
    Dec 5 '18 at 4:37






  • 1




    $begingroup$
    @clathratus - Please refer to my page. I've almost exclusively being posting questions/answers in this area. In some parts I combine both Feynman's Trick and Laplace Transforms. I've also used multiple introduced parameters to solve systems. I would love to have a 'Feynman Trick' mate (/buddy for the US Folk) so please contact me at any time. I want to put together a simple LaTeX document with examples of this method and would love contributors.
    $endgroup$
    – DavidG
    Dec 5 '18 at 4:58










  • $begingroup$
    How can I contact you? I'm afraid that I may not have much to offer with the Feynman trick, but I do have more experience with generalized series, factorials/products, and I have been recently investigating special functions and finding closed forms, which may be useful. We could combine forces and do some cool stuff. By the way, you may contact me via the email address joverton2020@gmail.com
    $endgroup$
    – clathratus
    Dec 5 '18 at 5:22
















2












$begingroup$

Another Approach is to employ Feynman's Trick:



Let



$$I(x) = int_{0}^{1} frac{lnleft| x^2left(t^2 - tright) + 1right|}{t^2 - t}:dt$$



Note $I = I(1)$ and $I(0) = 0$



Thus



begin{align}
I'(x) &= int_{0}^{1} frac{2xleft(t^2 - tright)}{left(x^2left(t^2 - tright) + 1right)left( t^2 - tright)}:dt = frac{2}{x}int_{0}^{1} frac{1}{left(t - frac{1}{2}right)^2 + frac{4 - x^2}{4x^2}}:dt\
&= frac{4}{x}int_{0}^{frac{1}{2}} frac{1}{t^2 + frac{4 - x^2}{4x^2}}:dt = frac{8}{sqrt{4 - x^2}}arctanleft(frac{x}{sqrt{4 -x^2}} right)
end{align}



We now integrate to solve $I(x)$



$$I(x) = intfrac{8}{sqrt{4 - x^2}}arctanleft(frac{x}{sqrt{4 -x^2}} right) :dx = 4left[arctanleft( frac{x}{sqrt{4 - x^2}}right) right]^2 + C $$



Where $C$ is a constant of integration. As $I(0) = 0$ we find $C = 0$ and so:



$$I(x) = 4left[arctanleft( frac{x}{sqrt{4 - x^2}}right) right]^2$$



And finally



$$ I = I(1) = 4left[arctanleft( frac{1}{sqrt{3}}right) right]^2 = frac{pi^2}{9}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a really cool approach! In retrospect, seeing the $$frac{log(t^2-t+1)}{t^2-t}$$ really begs for the parameterization which you introduced. Great job.
    $endgroup$
    – clathratus
    Dec 5 '18 at 3:57






  • 1




    $begingroup$
    I'm glad you enjoyed it. I've been on a Feynman 'kick' lately and have started to develop pattern matching skills. One element that may seem odd on the surface is the use of $x^2$ rather than just $x$. If you are interested, try both. You will find the $x$ creates an incredibly nasty indefinite integral to solve at the end and yet $x^2$ makes the process incredibly easy.
    $endgroup$
    – DavidG
    Dec 5 '18 at 4:33










  • $begingroup$
    I've done quite a bit with series, but not much with the 'Feynman' trick. Any resources that you've found especially helpful in learning about when and how to use it?
    $endgroup$
    – clathratus
    Dec 5 '18 at 4:37






  • 1




    $begingroup$
    @clathratus - Please refer to my page. I've almost exclusively being posting questions/answers in this area. In some parts I combine both Feynman's Trick and Laplace Transforms. I've also used multiple introduced parameters to solve systems. I would love to have a 'Feynman Trick' mate (/buddy for the US Folk) so please contact me at any time. I want to put together a simple LaTeX document with examples of this method and would love contributors.
    $endgroup$
    – DavidG
    Dec 5 '18 at 4:58










  • $begingroup$
    How can I contact you? I'm afraid that I may not have much to offer with the Feynman trick, but I do have more experience with generalized series, factorials/products, and I have been recently investigating special functions and finding closed forms, which may be useful. We could combine forces and do some cool stuff. By the way, you may contact me via the email address joverton2020@gmail.com
    $endgroup$
    – clathratus
    Dec 5 '18 at 5:22














2












2








2





$begingroup$

Another Approach is to employ Feynman's Trick:



Let



$$I(x) = int_{0}^{1} frac{lnleft| x^2left(t^2 - tright) + 1right|}{t^2 - t}:dt$$



Note $I = I(1)$ and $I(0) = 0$



Thus



begin{align}
I'(x) &= int_{0}^{1} frac{2xleft(t^2 - tright)}{left(x^2left(t^2 - tright) + 1right)left( t^2 - tright)}:dt = frac{2}{x}int_{0}^{1} frac{1}{left(t - frac{1}{2}right)^2 + frac{4 - x^2}{4x^2}}:dt\
&= frac{4}{x}int_{0}^{frac{1}{2}} frac{1}{t^2 + frac{4 - x^2}{4x^2}}:dt = frac{8}{sqrt{4 - x^2}}arctanleft(frac{x}{sqrt{4 -x^2}} right)
end{align}



We now integrate to solve $I(x)$



$$I(x) = intfrac{8}{sqrt{4 - x^2}}arctanleft(frac{x}{sqrt{4 -x^2}} right) :dx = 4left[arctanleft( frac{x}{sqrt{4 - x^2}}right) right]^2 + C $$



Where $C$ is a constant of integration. As $I(0) = 0$ we find $C = 0$ and so:



$$I(x) = 4left[arctanleft( frac{x}{sqrt{4 - x^2}}right) right]^2$$



And finally



$$ I = I(1) = 4left[arctanleft( frac{1}{sqrt{3}}right) right]^2 = frac{pi^2}{9}$$






share|cite|improve this answer









$endgroup$



Another Approach is to employ Feynman's Trick:



Let



$$I(x) = int_{0}^{1} frac{lnleft| x^2left(t^2 - tright) + 1right|}{t^2 - t}:dt$$



Note $I = I(1)$ and $I(0) = 0$



Thus



begin{align}
I'(x) &= int_{0}^{1} frac{2xleft(t^2 - tright)}{left(x^2left(t^2 - tright) + 1right)left( t^2 - tright)}:dt = frac{2}{x}int_{0}^{1} frac{1}{left(t - frac{1}{2}right)^2 + frac{4 - x^2}{4x^2}}:dt\
&= frac{4}{x}int_{0}^{frac{1}{2}} frac{1}{t^2 + frac{4 - x^2}{4x^2}}:dt = frac{8}{sqrt{4 - x^2}}arctanleft(frac{x}{sqrt{4 -x^2}} right)
end{align}



We now integrate to solve $I(x)$



$$I(x) = intfrac{8}{sqrt{4 - x^2}}arctanleft(frac{x}{sqrt{4 -x^2}} right) :dx = 4left[arctanleft( frac{x}{sqrt{4 - x^2}}right) right]^2 + C $$



Where $C$ is a constant of integration. As $I(0) = 0$ we find $C = 0$ and so:



$$I(x) = 4left[arctanleft( frac{x}{sqrt{4 - x^2}}right) right]^2$$



And finally



$$ I = I(1) = 4left[arctanleft( frac{1}{sqrt{3}}right) right]^2 = frac{pi^2}{9}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 3:29









DavidGDavidG

2,090720




2,090720












  • $begingroup$
    This is a really cool approach! In retrospect, seeing the $$frac{log(t^2-t+1)}{t^2-t}$$ really begs for the parameterization which you introduced. Great job.
    $endgroup$
    – clathratus
    Dec 5 '18 at 3:57






  • 1




    $begingroup$
    I'm glad you enjoyed it. I've been on a Feynman 'kick' lately and have started to develop pattern matching skills. One element that may seem odd on the surface is the use of $x^2$ rather than just $x$. If you are interested, try both. You will find the $x$ creates an incredibly nasty indefinite integral to solve at the end and yet $x^2$ makes the process incredibly easy.
    $endgroup$
    – DavidG
    Dec 5 '18 at 4:33










  • $begingroup$
    I've done quite a bit with series, but not much with the 'Feynman' trick. Any resources that you've found especially helpful in learning about when and how to use it?
    $endgroup$
    – clathratus
    Dec 5 '18 at 4:37






  • 1




    $begingroup$
    @clathratus - Please refer to my page. I've almost exclusively being posting questions/answers in this area. In some parts I combine both Feynman's Trick and Laplace Transforms. I've also used multiple introduced parameters to solve systems. I would love to have a 'Feynman Trick' mate (/buddy for the US Folk) so please contact me at any time. I want to put together a simple LaTeX document with examples of this method and would love contributors.
    $endgroup$
    – DavidG
    Dec 5 '18 at 4:58










  • $begingroup$
    How can I contact you? I'm afraid that I may not have much to offer with the Feynman trick, but I do have more experience with generalized series, factorials/products, and I have been recently investigating special functions and finding closed forms, which may be useful. We could combine forces and do some cool stuff. By the way, you may contact me via the email address joverton2020@gmail.com
    $endgroup$
    – clathratus
    Dec 5 '18 at 5:22


















  • $begingroup$
    This is a really cool approach! In retrospect, seeing the $$frac{log(t^2-t+1)}{t^2-t}$$ really begs for the parameterization which you introduced. Great job.
    $endgroup$
    – clathratus
    Dec 5 '18 at 3:57






  • 1




    $begingroup$
    I'm glad you enjoyed it. I've been on a Feynman 'kick' lately and have started to develop pattern matching skills. One element that may seem odd on the surface is the use of $x^2$ rather than just $x$. If you are interested, try both. You will find the $x$ creates an incredibly nasty indefinite integral to solve at the end and yet $x^2$ makes the process incredibly easy.
    $endgroup$
    – DavidG
    Dec 5 '18 at 4:33










  • $begingroup$
    I've done quite a bit with series, but not much with the 'Feynman' trick. Any resources that you've found especially helpful in learning about when and how to use it?
    $endgroup$
    – clathratus
    Dec 5 '18 at 4:37






  • 1




    $begingroup$
    @clathratus - Please refer to my page. I've almost exclusively being posting questions/answers in this area. In some parts I combine both Feynman's Trick and Laplace Transforms. I've also used multiple introduced parameters to solve systems. I would love to have a 'Feynman Trick' mate (/buddy for the US Folk) so please contact me at any time. I want to put together a simple LaTeX document with examples of this method and would love contributors.
    $endgroup$
    – DavidG
    Dec 5 '18 at 4:58










  • $begingroup$
    How can I contact you? I'm afraid that I may not have much to offer with the Feynman trick, but I do have more experience with generalized series, factorials/products, and I have been recently investigating special functions and finding closed forms, which may be useful. We could combine forces and do some cool stuff. By the way, you may contact me via the email address joverton2020@gmail.com
    $endgroup$
    – clathratus
    Dec 5 '18 at 5:22
















$begingroup$
This is a really cool approach! In retrospect, seeing the $$frac{log(t^2-t+1)}{t^2-t}$$ really begs for the parameterization which you introduced. Great job.
$endgroup$
– clathratus
Dec 5 '18 at 3:57




$begingroup$
This is a really cool approach! In retrospect, seeing the $$frac{log(t^2-t+1)}{t^2-t}$$ really begs for the parameterization which you introduced. Great job.
$endgroup$
– clathratus
Dec 5 '18 at 3:57




1




1




$begingroup$
I'm glad you enjoyed it. I've been on a Feynman 'kick' lately and have started to develop pattern matching skills. One element that may seem odd on the surface is the use of $x^2$ rather than just $x$. If you are interested, try both. You will find the $x$ creates an incredibly nasty indefinite integral to solve at the end and yet $x^2$ makes the process incredibly easy.
$endgroup$
– DavidG
Dec 5 '18 at 4:33




$begingroup$
I'm glad you enjoyed it. I've been on a Feynman 'kick' lately and have started to develop pattern matching skills. One element that may seem odd on the surface is the use of $x^2$ rather than just $x$. If you are interested, try both. You will find the $x$ creates an incredibly nasty indefinite integral to solve at the end and yet $x^2$ makes the process incredibly easy.
$endgroup$
– DavidG
Dec 5 '18 at 4:33












$begingroup$
I've done quite a bit with series, but not much with the 'Feynman' trick. Any resources that you've found especially helpful in learning about when and how to use it?
$endgroup$
– clathratus
Dec 5 '18 at 4:37




$begingroup$
I've done quite a bit with series, but not much with the 'Feynman' trick. Any resources that you've found especially helpful in learning about when and how to use it?
$endgroup$
– clathratus
Dec 5 '18 at 4:37




1




1




$begingroup$
@clathratus - Please refer to my page. I've almost exclusively being posting questions/answers in this area. In some parts I combine both Feynman's Trick and Laplace Transforms. I've also used multiple introduced parameters to solve systems. I would love to have a 'Feynman Trick' mate (/buddy for the US Folk) so please contact me at any time. I want to put together a simple LaTeX document with examples of this method and would love contributors.
$endgroup$
– DavidG
Dec 5 '18 at 4:58




$begingroup$
@clathratus - Please refer to my page. I've almost exclusively being posting questions/answers in this area. In some parts I combine both Feynman's Trick and Laplace Transforms. I've also used multiple introduced parameters to solve systems. I would love to have a 'Feynman Trick' mate (/buddy for the US Folk) so please contact me at any time. I want to put together a simple LaTeX document with examples of this method and would love contributors.
$endgroup$
– DavidG
Dec 5 '18 at 4:58












$begingroup$
How can I contact you? I'm afraid that I may not have much to offer with the Feynman trick, but I do have more experience with generalized series, factorials/products, and I have been recently investigating special functions and finding closed forms, which may be useful. We could combine forces and do some cool stuff. By the way, you may contact me via the email address joverton2020@gmail.com
$endgroup$
– clathratus
Dec 5 '18 at 5:22




$begingroup$
How can I contact you? I'm afraid that I may not have much to offer with the Feynman trick, but I do have more experience with generalized series, factorials/products, and I have been recently investigating special functions and finding closed forms, which may be useful. We could combine forces and do some cool stuff. By the way, you may contact me via the email address joverton2020@gmail.com
$endgroup$
– clathratus
Dec 5 '18 at 5:22











1












$begingroup$

Firstly observe that, for $x$ real,



begin{align}(1-x)^2-(1-x)+1&=(1-2x+x^2)-1+x+1\
&=x^2-x+1
end{align}



begin{align}J&=int_0^1 frac{ln(x^2-x+1)}{x(x-1)},dx\
&=-int_0^1 frac{ln(x^2-x+1)}{1-x},dx-int_0^1 frac{ln(x^2-x+1)}{x},dx
end{align}



In the first integral perform the change of variable $y=1-x$,



begin{align}J&=-2int_0^1 frac{ln(x^2-x+1)}{x},dx\
&=-2int_0^1 frac{lnleft(frac{x^3+1}{x+1}right)}{x},dx\
&=2int_0^1 frac{lnleft(x+1right)}{x},dx-2int_0^1 frac{lnleft(x^3+1right)}{x},dx\
&=2int_0^1 frac{lnleft(x+1right)}{x},dx-2int_0^1 frac{x^2lnleft(x^3+1right)}{x^3},dx\
end{align}



In the latter integral perform the change of variable $y=x^3$,



begin{align}J&=2int_0^1 frac{lnleft(x+1right)}{x},dx-frac{2}{3}int_0^1
frac{lnleft(x+1right)}{x},dx\
&=frac{4}{3}int_0^1 frac{lnleft(x+1right)}{x},dx\
&=frac{4}{3}int_0^1 frac{lnleft(1-x^2right)}{x},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=frac{4}{3}int_0^1 frac{xlnleft(1-x^2right)}{x^2},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx
end{align}



In the first integral perform the change of variable $y=x^2$,



begin{align}J&=frac{2}{3}int_0^1 frac{lnleft(1-xright)}{x},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=-frac{2}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=-frac{2}{3}Big[ln xln(1-x)Big]_0^1 -frac{2}{3}int_0^1 frac{ln x}{1-x},dx\
&=-frac{2}{3}int_0^1 frac{ln x}{1-x},dx\
&=frac{2}{3}zeta(2)\
&=frac{2}{3}times frac{pi^2}{6}\
&=boxed{frac{pi^2}{9}}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I really appreciate the detailed and aesthetically pleasing nature of this response, not to mention the elegance. Thanks for this great answer.
    $endgroup$
    – clathratus
    Dec 5 '18 at 17:47










  • $begingroup$
    Thanks, but after reading all the anwers i'm afraid i have only summed up computations already written by other people (with some details ok)
    $endgroup$
    – FDP
    Dec 5 '18 at 18:53
















1












$begingroup$

Firstly observe that, for $x$ real,



begin{align}(1-x)^2-(1-x)+1&=(1-2x+x^2)-1+x+1\
&=x^2-x+1
end{align}



begin{align}J&=int_0^1 frac{ln(x^2-x+1)}{x(x-1)},dx\
&=-int_0^1 frac{ln(x^2-x+1)}{1-x},dx-int_0^1 frac{ln(x^2-x+1)}{x},dx
end{align}



In the first integral perform the change of variable $y=1-x$,



begin{align}J&=-2int_0^1 frac{ln(x^2-x+1)}{x},dx\
&=-2int_0^1 frac{lnleft(frac{x^3+1}{x+1}right)}{x},dx\
&=2int_0^1 frac{lnleft(x+1right)}{x},dx-2int_0^1 frac{lnleft(x^3+1right)}{x},dx\
&=2int_0^1 frac{lnleft(x+1right)}{x},dx-2int_0^1 frac{x^2lnleft(x^3+1right)}{x^3},dx\
end{align}



In the latter integral perform the change of variable $y=x^3$,



begin{align}J&=2int_0^1 frac{lnleft(x+1right)}{x},dx-frac{2}{3}int_0^1
frac{lnleft(x+1right)}{x},dx\
&=frac{4}{3}int_0^1 frac{lnleft(x+1right)}{x},dx\
&=frac{4}{3}int_0^1 frac{lnleft(1-x^2right)}{x},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=frac{4}{3}int_0^1 frac{xlnleft(1-x^2right)}{x^2},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx
end{align}



In the first integral perform the change of variable $y=x^2$,



begin{align}J&=frac{2}{3}int_0^1 frac{lnleft(1-xright)}{x},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=-frac{2}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=-frac{2}{3}Big[ln xln(1-x)Big]_0^1 -frac{2}{3}int_0^1 frac{ln x}{1-x},dx\
&=-frac{2}{3}int_0^1 frac{ln x}{1-x},dx\
&=frac{2}{3}zeta(2)\
&=frac{2}{3}times frac{pi^2}{6}\
&=boxed{frac{pi^2}{9}}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I really appreciate the detailed and aesthetically pleasing nature of this response, not to mention the elegance. Thanks for this great answer.
    $endgroup$
    – clathratus
    Dec 5 '18 at 17:47










  • $begingroup$
    Thanks, but after reading all the anwers i'm afraid i have only summed up computations already written by other people (with some details ok)
    $endgroup$
    – FDP
    Dec 5 '18 at 18:53














1












1








1





$begingroup$

Firstly observe that, for $x$ real,



begin{align}(1-x)^2-(1-x)+1&=(1-2x+x^2)-1+x+1\
&=x^2-x+1
end{align}



begin{align}J&=int_0^1 frac{ln(x^2-x+1)}{x(x-1)},dx\
&=-int_0^1 frac{ln(x^2-x+1)}{1-x},dx-int_0^1 frac{ln(x^2-x+1)}{x},dx
end{align}



In the first integral perform the change of variable $y=1-x$,



begin{align}J&=-2int_0^1 frac{ln(x^2-x+1)}{x},dx\
&=-2int_0^1 frac{lnleft(frac{x^3+1}{x+1}right)}{x},dx\
&=2int_0^1 frac{lnleft(x+1right)}{x},dx-2int_0^1 frac{lnleft(x^3+1right)}{x},dx\
&=2int_0^1 frac{lnleft(x+1right)}{x},dx-2int_0^1 frac{x^2lnleft(x^3+1right)}{x^3},dx\
end{align}



In the latter integral perform the change of variable $y=x^3$,



begin{align}J&=2int_0^1 frac{lnleft(x+1right)}{x},dx-frac{2}{3}int_0^1
frac{lnleft(x+1right)}{x},dx\
&=frac{4}{3}int_0^1 frac{lnleft(x+1right)}{x},dx\
&=frac{4}{3}int_0^1 frac{lnleft(1-x^2right)}{x},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=frac{4}{3}int_0^1 frac{xlnleft(1-x^2right)}{x^2},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx
end{align}



In the first integral perform the change of variable $y=x^2$,



begin{align}J&=frac{2}{3}int_0^1 frac{lnleft(1-xright)}{x},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=-frac{2}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=-frac{2}{3}Big[ln xln(1-x)Big]_0^1 -frac{2}{3}int_0^1 frac{ln x}{1-x},dx\
&=-frac{2}{3}int_0^1 frac{ln x}{1-x},dx\
&=frac{2}{3}zeta(2)\
&=frac{2}{3}times frac{pi^2}{6}\
&=boxed{frac{pi^2}{9}}
end{align}






share|cite|improve this answer









$endgroup$



Firstly observe that, for $x$ real,



begin{align}(1-x)^2-(1-x)+1&=(1-2x+x^2)-1+x+1\
&=x^2-x+1
end{align}



begin{align}J&=int_0^1 frac{ln(x^2-x+1)}{x(x-1)},dx\
&=-int_0^1 frac{ln(x^2-x+1)}{1-x},dx-int_0^1 frac{ln(x^2-x+1)}{x},dx
end{align}



In the first integral perform the change of variable $y=1-x$,



begin{align}J&=-2int_0^1 frac{ln(x^2-x+1)}{x},dx\
&=-2int_0^1 frac{lnleft(frac{x^3+1}{x+1}right)}{x},dx\
&=2int_0^1 frac{lnleft(x+1right)}{x},dx-2int_0^1 frac{lnleft(x^3+1right)}{x},dx\
&=2int_0^1 frac{lnleft(x+1right)}{x},dx-2int_0^1 frac{x^2lnleft(x^3+1right)}{x^3},dx\
end{align}



In the latter integral perform the change of variable $y=x^3$,



begin{align}J&=2int_0^1 frac{lnleft(x+1right)}{x},dx-frac{2}{3}int_0^1
frac{lnleft(x+1right)}{x},dx\
&=frac{4}{3}int_0^1 frac{lnleft(x+1right)}{x},dx\
&=frac{4}{3}int_0^1 frac{lnleft(1-x^2right)}{x},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=frac{4}{3}int_0^1 frac{xlnleft(1-x^2right)}{x^2},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx
end{align}



In the first integral perform the change of variable $y=x^2$,



begin{align}J&=frac{2}{3}int_0^1 frac{lnleft(1-xright)}{x},dx-frac{4}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=-frac{2}{3}int_0^1 frac{lnleft(1-xright)}{x},dx\
&=-frac{2}{3}Big[ln xln(1-x)Big]_0^1 -frac{2}{3}int_0^1 frac{ln x}{1-x},dx\
&=-frac{2}{3}int_0^1 frac{ln x}{1-x},dx\
&=frac{2}{3}zeta(2)\
&=frac{2}{3}times frac{pi^2}{6}\
&=boxed{frac{pi^2}{9}}
end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 16:48









FDPFDP

5,17211424




5,17211424












  • $begingroup$
    I really appreciate the detailed and aesthetically pleasing nature of this response, not to mention the elegance. Thanks for this great answer.
    $endgroup$
    – clathratus
    Dec 5 '18 at 17:47










  • $begingroup$
    Thanks, but after reading all the anwers i'm afraid i have only summed up computations already written by other people (with some details ok)
    $endgroup$
    – FDP
    Dec 5 '18 at 18:53


















  • $begingroup$
    I really appreciate the detailed and aesthetically pleasing nature of this response, not to mention the elegance. Thanks for this great answer.
    $endgroup$
    – clathratus
    Dec 5 '18 at 17:47










  • $begingroup$
    Thanks, but after reading all the anwers i'm afraid i have only summed up computations already written by other people (with some details ok)
    $endgroup$
    – FDP
    Dec 5 '18 at 18:53
















$begingroup$
I really appreciate the detailed and aesthetically pleasing nature of this response, not to mention the elegance. Thanks for this great answer.
$endgroup$
– clathratus
Dec 5 '18 at 17:47




$begingroup$
I really appreciate the detailed and aesthetically pleasing nature of this response, not to mention the elegance. Thanks for this great answer.
$endgroup$
– clathratus
Dec 5 '18 at 17:47












$begingroup$
Thanks, but after reading all the anwers i'm afraid i have only summed up computations already written by other people (with some details ok)
$endgroup$
– FDP
Dec 5 '18 at 18:53




$begingroup$
Thanks, but after reading all the anwers i'm afraid i have only summed up computations already written by other people (with some details ok)
$endgroup$
– FDP
Dec 5 '18 at 18:53


















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