How to prove $int_0^1xln^2(1+x)ln(frac{x^2}{1+x})frac{dx}{1+x^2}$
11
$begingroup$
How to prove$$int_0^1xln^2(1+x)lnleft(frac{x^2}{1+x}right)frac{dx}{1+x^2}=-frac{7}{32}cdotzeta{(3)}ln2+frac{3pi^2}{128}cdotln^22-frac{1}{64}cdotln^42-frac{13pi^4}{46080}$$
The substitution $$x=frac{1-y}{1+y}$$leads to calculate the integrals that are unknow.
$$int_0^1yln(1-y)ln^2(1+y)frac{dy}{1+y^2}, int_0^1frac{yln^3(1+y)}{1+y^2}dy$$.
For the moment,Ido not see how to calculate this integral
integration definite-integrals polylogarithm
edited Dec 25 '18 at 23:00
Don Thousand
4,574734
asked Dec 25 '18 at 22:42
user178256user178256
2,1401832
$endgroup$
|
show 15 more comments
11
$begingroup$
How to prove$$int_0^1xln^2(1+x)lnleft(frac{x^2}{1+x}right)frac{dx}{1+x^2}=-frac{7}{32}cdotzeta{(3)}ln2+frac{3pi^2}{128}cdotln^22-frac{1}{64}cdotln^42-frac{13pi^4}{46080}$$
The substitution $$x=frac{1-y}{1+y}$$leads to calculate the integrals that are unknow.
$$int_0^1yln(1-y)ln^2(1+y)frac{dy}{1+y^2}, int_0^1frac{yln^3(1+y)}{1+y^2}dy$$.
For the moment,Ido not see how to calculate this integral
integration definite-integrals polylogarithm
edited Dec 25 '18 at 23:00
Don Thousand
4,574734
asked Dec 25 '18 at 22:42
user178256user178256
2,1401832
$endgroup$
2
$begingroup$
Hi! Where did you found this integral?
$endgroup$
– Zacky
Dec 25 '18 at 22:50
2
$begingroup$
$$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
$endgroup$
– clathratus
Dec 25 '18 at 23:08
4
$begingroup$
Here is an ideea, we have: $$a^2 b=frac16left((a+b)^3-(a-b)^3-2b^3right)$$ Thus the integral is equal to: $$I=2cdot frac16left( int_0^1frac{xln^3(1-x^2)}{1+x^2}dx+ int_0^1frac{xln^3left(frac{1-x}{1+x}right)}{1+x^2}dx -2int_0^1 frac{xln^3(1-x)}{1+x^2}right)-int_0^1 frac{xln^3(1+x)}{1+x^2}dx$$ Thus let $x^2=t$ for the first one and $frac{1-x}{1+x}=t$ for the second one and try to simplify some. This way we don't have any product between the logarithms.
$endgroup$
– Zacky
Dec 25 '18 at 23:22
1
$begingroup$
@Zacky If it helps, making the substitutions $xto -frac{x}{1+x}$ and $xto frac{x-1}{2}$ shows that the integral in question is equal to $$int_0^1 ln^2bigg(frac{1+x}{2}bigg)lnbigg(frac{1}{2}frac{(1-x)^2}{1+x}bigg)frac{x-1}{x+1}frac{dx}{1+x^2}$$ Perhaps using your $a^2 b$ trick on this integral and combining it with the original representation can cause some sort of desirable cancellation to happen.
$endgroup$
– Frpzzd
Dec 25 '18 at 23:30
1
$begingroup$
@Diger: I will be glad to see your solution using polylog identities.
$endgroup$
– FDP
Dec 28 '18 at 10:32
|
show 15 more comments
11
11
11
7
$begingroup$
How to prove$$int_0^1xln^2(1+x)lnleft(frac{x^2}{1+x}right)frac{dx}{1+x^2}=-frac{7}{32}cdotzeta{(3)}ln2+frac{3pi^2}{128}cdotln^22-frac{1}{64}cdotln^42-frac{13pi^4}{46080}$$
The substitution $$x=frac{1-y}{1+y}$$leads to calculate the integrals that are unknow.
$$int_0^1yln(1-y)ln^2(1+y)frac{dy}{1+y^2}, int_0^1frac{yln^3(1+y)}{1+y^2}dy$$.
For the moment,Ido not see how to calculate this integral
integration definite-integrals polylogarithm
edited Dec 25 '18 at 23:00
Don Thousand
4,574734
asked Dec 25 '18 at 22:42
user178256user178256
2,1401832
$endgroup$
How to prove$$int_0^1xln^2(1+x)lnleft(frac{x^2}{1+x}right)frac{dx}{1+x^2}=-frac{7}{32}cdotzeta{(3)}ln2+frac{3pi^2}{128}cdotln^22-frac{1}{64}cdotln^42-frac{13pi^4}{46080}$$
The substitution $$x=frac{1-y}{1+y}$$leads to calculate the integrals that are unknow.
$$int_0^1yln(1-y)ln^2(1+y)frac{dy}{1+y^2}, int_0^1frac{yln^3(1+y)}{1+y^2}dy$$.
For the moment,Ido not see how to calculate this integral
integration definite-integrals polylogarithm
integration definite-integrals polylogarithm
edited Dec 25 '18 at 23:00
Don Thousand
4,574734
asked Dec 25 '18 at 22:42
user178256user178256
2,1401832
edited Dec 25 '18 at 23:00
Don Thousand
4,574734
asked Dec 25 '18 at 22:42
user178256user178256
2,1401832
edited Dec 25 '18 at 23:00
Don Thousand
4,574734
edited Dec 25 '18 at 23:00
Don Thousand
4,574734
edited Dec 25 '18 at 23:00
Don Thousand
4,574734
4,574734
asked Dec 25 '18 at 22:42
user178256user178256
2,1401832
asked Dec 25 '18 at 22:42
user178256user178256
2,1401832
asked Dec 25 '18 at 22:42
user178256user178256
2,1401832
2,1401832
2
$begingroup$
Hi! Where did you found this integral?
$endgroup$
– Zacky
Dec 25 '18 at 22:50
2
$begingroup$
$$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
$endgroup$
– clathratus
Dec 25 '18 at 23:08
4
$begingroup$
Here is an ideea, we have: $$a^2 b=frac16left((a+b)^3-(a-b)^3-2b^3right)$$ Thus the integral is equal to: $$I=2cdot frac16left( int_0^1frac{xln^3(1-x^2)}{1+x^2}dx+ int_0^1frac{xln^3left(frac{1-x}{1+x}right)}{1+x^2}dx -2int_0^1 frac{xln^3(1-x)}{1+x^2}right)-int_0^1 frac{xln^3(1+x)}{1+x^2}dx$$ Thus let $x^2=t$ for the first one and $frac{1-x}{1+x}=t$ for the second one and try to simplify some. This way we don't have any product between the logarithms.
$endgroup$
– Zacky
Dec 25 '18 at 23:22
1
$begingroup$
@Zacky If it helps, making the substitutions $xto -frac{x}{1+x}$ and $xto frac{x-1}{2}$ shows that the integral in question is equal to $$int_0^1 ln^2bigg(frac{1+x}{2}bigg)lnbigg(frac{1}{2}frac{(1-x)^2}{1+x}bigg)frac{x-1}{x+1}frac{dx}{1+x^2}$$ Perhaps using your $a^2 b$ trick on this integral and combining it with the original representation can cause some sort of desirable cancellation to happen.
$endgroup$
– Frpzzd
Dec 25 '18 at 23:30
1
$begingroup$
@Diger: I will be glad to see your solution using polylog identities.
$endgroup$
– FDP
Dec 28 '18 at 10:32
|
show 15 more comments
2
$begingroup$
Hi! Where did you found this integral?
$endgroup$
– Zacky
Dec 25 '18 at 22:50
2
$begingroup$
$$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
$endgroup$
– clathratus
Dec 25 '18 at 23:08
4
$begingroup$
Here is an ideea, we have: $$a^2 b=frac16left((a+b)^3-(a-b)^3-2b^3right)$$ Thus the integral is equal to: $$I=2cdot frac16left( int_0^1frac{xln^3(1-x^2)}{1+x^2}dx+ int_0^1frac{xln^3left(frac{1-x}{1+x}right)}{1+x^2}dx -2int_0^1 frac{xln^3(1-x)}{1+x^2}right)-int_0^1 frac{xln^3(1+x)}{1+x^2}dx$$ Thus let $x^2=t$ for the first one and $frac{1-x}{1+x}=t$ for the second one and try to simplify some. This way we don't have any product between the logarithms.
$endgroup$
– Zacky
Dec 25 '18 at 23:22
1
$begingroup$
@Zacky If it helps, making the substitutions $xto -frac{x}{1+x}$ and $xto frac{x-1}{2}$ shows that the integral in question is equal to $$int_0^1 ln^2bigg(frac{1+x}{2}bigg)lnbigg(frac{1}{2}frac{(1-x)^2}{1+x}bigg)frac{x-1}{x+1}frac{dx}{1+x^2}$$ Perhaps using your $a^2 b$ trick on this integral and combining it with the original representation can cause some sort of desirable cancellation to happen.
$endgroup$
– Frpzzd
Dec 25 '18 at 23:30
1
$begingroup$
@Diger: I will be glad to see your solution using polylog identities.
$endgroup$
– FDP
Dec 28 '18 at 10:32
2
2
$begingroup$
Hi! Where did you found this integral?
$endgroup$
– Zacky
Dec 25 '18 at 22:50
$begingroup$
Hi! Where did you found this integral?
$endgroup$
– Zacky
Dec 25 '18 at 22:50
2
2
$begingroup$
$$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
$endgroup$
– clathratus
Dec 25 '18 at 23:08
$begingroup$
$$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
$endgroup$
– clathratus
Dec 25 '18 at 23:08
4
4
$begingroup$
Here is an ideea, we have: $$a^2 b=frac16left((a+b)^3-(a-b)^3-2b^3right)$$ Thus the integral is equal to: $$I=2cdot frac16left( int_0^1frac{xln^3(1-x^2)}{1+x^2}dx+ int_0^1frac{xln^3left(frac{1-x}{1+x}right)}{1+x^2}dx -2int_0^1 frac{xln^3(1-x)}{1+x^2}right)-int_0^1 frac{xln^3(1+x)}{1+x^2}dx$$ Thus let $x^2=t$ for the first one and $frac{1-x}{1+x}=t$ for the second one and try to simplify some. This way we don't have any product between the logarithms.
$endgroup$
– Zacky
Dec 25 '18 at 23:22
$begingroup$
Here is an ideea, we have: $$a^2 b=frac16left((a+b)^3-(a-b)^3-2b^3right)$$ Thus the integral is equal to: $$I=2cdot frac16left( int_0^1frac{xln^3(1-x^2)}{1+x^2}dx+ int_0^1frac{xln^3left(frac{1-x}{1+x}right)}{1+x^2}dx -2int_0^1 frac{xln^3(1-x)}{1+x^2}right)-int_0^1 frac{xln^3(1+x)}{1+x^2}dx$$ Thus let $x^2=t$ for the first one and $frac{1-x}{1+x}=t$ for the second one and try to simplify some. This way we don't have any product between the logarithms.
$endgroup$
– Zacky
Dec 25 '18 at 23:22
1
1
$begingroup$
@Zacky If it helps, making the substitutions $xto -frac{x}{1+x}$ and $xto frac{x-1}{2}$ shows that the integral in question is equal to $$int_0^1 ln^2bigg(frac{1+x}{2}bigg)lnbigg(frac{1}{2}frac{(1-x)^2}{1+x}bigg)frac{x-1}{x+1}frac{dx}{1+x^2}$$ Perhaps using your $a^2 b$ trick on this integral and combining it with the original representation can cause some sort of desirable cancellation to happen.
$endgroup$
– Frpzzd
Dec 25 '18 at 23:30
$begingroup$
@Zacky If it helps, making the substitutions $xto -frac{x}{1+x}$ and $xto frac{x-1}{2}$ shows that the integral in question is equal to $$int_0^1 ln^2bigg(frac{1+x}{2}bigg)lnbigg(frac{1}{2}frac{(1-x)^2}{1+x}bigg)frac{x-1}{x+1}frac{dx}{1+x^2}$$ Perhaps using your $a^2 b$ trick on this integral and combining it with the original representation can cause some sort of desirable cancellation to happen.
$endgroup$
– Frpzzd
Dec 25 '18 at 23:30
1
1
$begingroup$
@Diger: I will be glad to see your solution using polylog identities.
$endgroup$
– FDP
Dec 28 '18 at 10:32
$begingroup$
@Diger: I will be glad to see your solution using polylog identities.
$endgroup$
– FDP
Dec 28 '18 at 10:32
|
show 15 more comments
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2
$begingroup$
Hi! Where did you found this integral?
$endgroup$
– Zacky
Dec 25 '18 at 22:50
2
$begingroup$
$$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
$endgroup$
– clathratus
Dec 25 '18 at 23:08
4
$begingroup$
Here is an ideea, we have: $$a^2 b=frac16left((a+b)^3-(a-b)^3-2b^3right)$$ Thus the integral is equal to: $$I=2cdot frac16left( int_0^1frac{xln^3(1-x^2)}{1+x^2}dx+ int_0^1frac{xln^3left(frac{1-x}{1+x}right)}{1+x^2}dx -2int_0^1 frac{xln^3(1-x)}{1+x^2}right)-int_0^1 frac{xln^3(1+x)}{1+x^2}dx$$ Thus let $x^2=t$ for the first one and $frac{1-x}{1+x}=t$ for the second one and try to simplify some. This way we don't have any product between the logarithms.
$endgroup$
– Zacky
Dec 25 '18 at 23:22
1
$begingroup$
@Zacky If it helps, making the substitutions $xto -frac{x}{1+x}$ and $xto frac{x-1}{2}$ shows that the integral in question is equal to $$int_0^1 ln^2bigg(frac{1+x}{2}bigg)lnbigg(frac{1}{2}frac{(1-x)^2}{1+x}bigg)frac{x-1}{x+1}frac{dx}{1+x^2}$$ Perhaps using your $a^2 b$ trick on this integral and combining it with the original representation can cause some sort of desirable cancellation to happen.
$endgroup$
– Frpzzd
Dec 25 '18 at 23:30
1
$begingroup$
@Diger: I will be glad to see your solution using polylog identities.
$endgroup$
– FDP
Dec 28 '18 at 10:32