How to prove $int_0^1xln^2(1+x)ln(frac{x^2}{1+x})frac{dx}{1+x^2}$












10














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







share|cite|improve this question




















  • 2




    Hi! Where did you found this integral?
    – Zacky
    2 days ago








  • 2




    $$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
    – clathratus
    2 days ago










  • Do any of you know the Maclaurin series for $ln^2(1+x)$? Maybe we can solve it this way.
    – Larry
    2 days ago








  • 4




    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.
    – Zacky
    2 days ago








  • 1




    @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.
    – Frpzzd
    2 days ago
















10














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







share|cite|improve this question




















  • 2




    Hi! Where did you found this integral?
    – Zacky
    2 days ago








  • 2




    $$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
    – clathratus
    2 days ago










  • Do any of you know the Maclaurin series for $ln^2(1+x)$? Maybe we can solve it this way.
    – Larry
    2 days ago








  • 4




    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.
    – Zacky
    2 days ago








  • 1




    @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.
    – Frpzzd
    2 days ago














10












10








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7





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Don Thousand

4,219734




4,219734










asked 2 days ago









user178256

2,1301832




2,1301832








  • 2




    Hi! Where did you found this integral?
    – Zacky
    2 days ago








  • 2




    $$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
    – clathratus
    2 days ago










  • Do any of you know the Maclaurin series for $ln^2(1+x)$? Maybe we can solve it this way.
    – Larry
    2 days ago








  • 4




    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.
    – Zacky
    2 days ago








  • 1




    @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.
    – Frpzzd
    2 days ago














  • 2




    Hi! Where did you found this integral?
    – Zacky
    2 days ago








  • 2




    $$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
    – clathratus
    2 days ago










  • Do any of you know the Maclaurin series for $ln^2(1+x)$? Maybe we can solve it this way.
    – Larry
    2 days ago








  • 4




    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.
    – Zacky
    2 days ago








  • 1




    @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.
    – Frpzzd
    2 days ago








2




2




Hi! Where did you found this integral?
– Zacky
2 days ago






Hi! Where did you found this integral?
– Zacky
2 days ago






2




2




$$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
– clathratus
2 days ago




$$lnbigg(frac{x^2}{x+1}bigg)=ln(x^2)-ln(1+x)=2ln(x)-ln(1+x)$$ don't know if that helps
– clathratus
2 days ago












Do any of you know the Maclaurin series for $ln^2(1+x)$? Maybe we can solve it this way.
– Larry
2 days ago






Do any of you know the Maclaurin series for $ln^2(1+x)$? Maybe we can solve it this way.
– Larry
2 days ago






4




4




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.
– Zacky
2 days ago






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.
– Zacky
2 days ago






1




1




@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.
– Frpzzd
2 days ago




@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.
– Frpzzd
2 days ago















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