What are the steps to solving this definite integral?What steps do I take to solve this integral?Solving for...

What are the steps to solving this definite integral?

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What are the steps to solving this definite integral?


What steps do I take to solve this integral?Solving for a variable inside a definite integralExtremely short definite integral question here?another definite integralWhat are the steps to solving this average distance problem?Help solving definite integralCalculate the integral of thisStuck on definite integral problem due to inappropriate $log$Solving definite integral in two variables.Evaluate the definite integral $int^{infty }_{0}frac{x ,dx}{e^{x} -1}$ using contour integration













2












$begingroup$


I am curious to know how to solve this problem; I know the answer comes out to be $1.20205$ (Apéry's constant).



$$int_0^1 frac{ln(1-x)ln(x)}{x} dx= ? $$










share|cite|improve this question









New contributor




Coalition Coal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    $begingroup$
    Did you try expanding $ln(1-x)$?
    $endgroup$
    – Clayton
    3 hours ago










  • $begingroup$
    How would you do that?
    $endgroup$
    – Dr. Sonnhard Graubner
    3 hours ago










  • $begingroup$
    Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
    $endgroup$
    – GEdgar
    3 hours ago






  • 1




    $begingroup$
    Welcome to math.stackexchange please include any attempts you've made to solve the problem in your post. This will help us better help you.
    $endgroup$
    – Digitalis
    3 hours ago












  • $begingroup$
    Did you try to integrate it by parts? By taking $u=ln(1-x)$ and $dv=frac{ln x}{x}dx$
    $endgroup$
    – Fareed AF
    3 hours ago
















2












$begingroup$


I am curious to know how to solve this problem; I know the answer comes out to be $1.20205$ (Apéry's constant).



$$int_0^1 frac{ln(1-x)ln(x)}{x} dx= ? $$










share|cite|improve this question









New contributor




Coalition Coal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Did you try expanding $ln(1-x)$?
    $endgroup$
    – Clayton
    3 hours ago










  • $begingroup$
    How would you do that?
    $endgroup$
    – Dr. Sonnhard Graubner
    3 hours ago










  • $begingroup$
    Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
    $endgroup$
    – GEdgar
    3 hours ago






  • 1




    $begingroup$
    Welcome to math.stackexchange please include any attempts you've made to solve the problem in your post. This will help us better help you.
    $endgroup$
    – Digitalis
    3 hours ago












  • $begingroup$
    Did you try to integrate it by parts? By taking $u=ln(1-x)$ and $dv=frac{ln x}{x}dx$
    $endgroup$
    – Fareed AF
    3 hours ago














2












2








2


4



$begingroup$


I am curious to know how to solve this problem; I know the answer comes out to be $1.20205$ (Apéry's constant).



$$int_0^1 frac{ln(1-x)ln(x)}{x} dx= ? $$










share|cite|improve this question









New contributor




Coalition Coal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I am curious to know how to solve this problem; I know the answer comes out to be $1.20205$ (Apéry's constant).



$$int_0^1 frac{ln(1-x)ln(x)}{x} dx= ? $$







calculus integration logarithms polylogarithm






share|cite|improve this question









New contributor




Coalition Coal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Coalition Coal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Bernard

125k743119




125k743119






New contributor




Coalition Coal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









Coalition CoalCoalition Coal

163




163




New contributor




Coalition Coal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Coalition Coal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Coalition Coal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Did you try expanding $ln(1-x)$?
    $endgroup$
    – Clayton
    3 hours ago










  • $begingroup$
    How would you do that?
    $endgroup$
    – Dr. Sonnhard Graubner
    3 hours ago










  • $begingroup$
    Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
    $endgroup$
    – GEdgar
    3 hours ago






  • 1




    $begingroup$
    Welcome to math.stackexchange please include any attempts you've made to solve the problem in your post. This will help us better help you.
    $endgroup$
    – Digitalis
    3 hours ago












  • $begingroup$
    Did you try to integrate it by parts? By taking $u=ln(1-x)$ and $dv=frac{ln x}{x}dx$
    $endgroup$
    – Fareed AF
    3 hours ago














  • 1




    $begingroup$
    Did you try expanding $ln(1-x)$?
    $endgroup$
    – Clayton
    3 hours ago










  • $begingroup$
    How would you do that?
    $endgroup$
    – Dr. Sonnhard Graubner
    3 hours ago










  • $begingroup$
    Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
    $endgroup$
    – GEdgar
    3 hours ago






  • 1




    $begingroup$
    Welcome to math.stackexchange please include any attempts you've made to solve the problem in your post. This will help us better help you.
    $endgroup$
    – Digitalis
    3 hours ago












  • $begingroup$
    Did you try to integrate it by parts? By taking $u=ln(1-x)$ and $dv=frac{ln x}{x}dx$
    $endgroup$
    – Fareed AF
    3 hours ago








1




1




$begingroup$
Did you try expanding $ln(1-x)$?
$endgroup$
– Clayton
3 hours ago




$begingroup$
Did you try expanding $ln(1-x)$?
$endgroup$
– Clayton
3 hours ago












$begingroup$
How would you do that?
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago




$begingroup$
How would you do that?
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago












$begingroup$
Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
$endgroup$
– GEdgar
3 hours ago




$begingroup$
Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
$endgroup$
– GEdgar
3 hours ago




1




1




$begingroup$
Welcome to math.stackexchange please include any attempts you've made to solve the problem in your post. This will help us better help you.
$endgroup$
– Digitalis
3 hours ago






$begingroup$
Welcome to math.stackexchange please include any attempts you've made to solve the problem in your post. This will help us better help you.
$endgroup$
– Digitalis
3 hours ago














$begingroup$
Did you try to integrate it by parts? By taking $u=ln(1-x)$ and $dv=frac{ln x}{x}dx$
$endgroup$
– Fareed AF
3 hours ago




$begingroup$
Did you try to integrate it by parts? By taking $u=ln(1-x)$ and $dv=frac{ln x}{x}dx$
$endgroup$
– Fareed AF
3 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them



I: Taylor Series Expansion of $log(1-x)$



As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain



begin{align*}
int_0^1frac{log(1-x)log(x)}xmathrm dx&=int_0^1frac{log(x)}xleft[-sum_{n=1}^inftyfrac{x^n}nright]mathrm dx\
&=-sum_{n=1}^inftyfrac1nint_0^1x^{n-1}log(x)mathrm dx\
&=-sum_{n=1}^inftyfrac1nleft[-frac1{n^2}right]\
&=sum_{n=1}^inftyfrac1{n^3}\
&=zeta(3)
end{align*}



This might be the most straightforward approach possible.



II: Integration By Parts



Choosing $u=log(1-x)$ and $mathrm dv=frac{log(x)}x$ we can apply Integration By Parts which gives



begin{align*}
int_0^1frac{log(1-x)log(x)}x&=underbrace{left[log(1-x)frac{log^2(x)}2right]_0^1}_{to0}+frac12int_0^1frac{log^2(x)}{1-x}mathrm dx\
&=frac12int_0^1log^2(x)left[sum_{n=0}^infty x^nright]mathrm dx\
&=frac12sum_{n=0}^inftyint_0^1x^nlog^2(x)mathrm dx\
&=frac12sum_{n=0}^inftyleft[frac2{(n+1)^3}right]\
&=sum_{n=1}^inftyfrac1{n^3}\
&=zeta(3)
end{align*}



Again, we utilized a series expansion, this time the one of the geometric series.



III: Integral Representation of the Zeta Function



To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find



begin{align*}
int_0^1frac{log(1-x)log(x)}xmathrm dx&=-int_infty^0(-x)log(1-e^{-x})mathrm dx\
&=-int_0^infty xlog(1-e^{-x})mathrm dx\
&=underbrace{left[frac{x^2}2log(1-e^{-x})right]_0^infty}_{to0}+frac12int_0^inftyfrac{x^2}{1-e^{-x}}e^{-x}mathrm dx\
&=frac1{Gamma(3)}int_0^inftyfrac{x^{3-1}}{e^x-1}mathrm dx\
&=zeta(3)
end{align*}



Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.



IV: The Trilogarithm $operatorname{Li}_3(1)$



Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=frac{log(1-x)}x$ to get



begin{align*}
int_0^1frac{log(1-x)log(x)}xmathrm dx&=underbrace{left[log(x)(-operatorname{Li}_2(x))right]_0^1}_{to0}+int_0^1frac{operatorname{Li}_2(x)}xmathrm dx\
&=[operatorname{Li}_3(x)]_0^1\
&=zeta(3)
end{align*}



A quick look at the series representation of the Trilogarithm verifies the last line.






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    begin{align}J&=int_0^1 frac{ln(1-x)ln x}{x} dx\
    &=-int_0^1 left(sum_{n=1}^infty frac{x^{n-1}}{n}right)ln x,dx\
    &=-sum_{n=1}^infty frac{1}{n}int_0^1 x^{n-1}ln x,dx\
    &=sum_{n=1}^infty frac{1}{n^3}\
    &=zeta(3)
    end{align}






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them



      I: Taylor Series Expansion of $log(1-x)$



      As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain



      begin{align*}
      int_0^1frac{log(1-x)log(x)}xmathrm dx&=int_0^1frac{log(x)}xleft[-sum_{n=1}^inftyfrac{x^n}nright]mathrm dx\
      &=-sum_{n=1}^inftyfrac1nint_0^1x^{n-1}log(x)mathrm dx\
      &=-sum_{n=1}^inftyfrac1nleft[-frac1{n^2}right]\
      &=sum_{n=1}^inftyfrac1{n^3}\
      &=zeta(3)
      end{align*}



      This might be the most straightforward approach possible.



      II: Integration By Parts



      Choosing $u=log(1-x)$ and $mathrm dv=frac{log(x)}x$ we can apply Integration By Parts which gives



      begin{align*}
      int_0^1frac{log(1-x)log(x)}x&=underbrace{left[log(1-x)frac{log^2(x)}2right]_0^1}_{to0}+frac12int_0^1frac{log^2(x)}{1-x}mathrm dx\
      &=frac12int_0^1log^2(x)left[sum_{n=0}^infty x^nright]mathrm dx\
      &=frac12sum_{n=0}^inftyint_0^1x^nlog^2(x)mathrm dx\
      &=frac12sum_{n=0}^inftyleft[frac2{(n+1)^3}right]\
      &=sum_{n=1}^inftyfrac1{n^3}\
      &=zeta(3)
      end{align*}



      Again, we utilized a series expansion, this time the one of the geometric series.



      III: Integral Representation of the Zeta Function



      To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find



      begin{align*}
      int_0^1frac{log(1-x)log(x)}xmathrm dx&=-int_infty^0(-x)log(1-e^{-x})mathrm dx\
      &=-int_0^infty xlog(1-e^{-x})mathrm dx\
      &=underbrace{left[frac{x^2}2log(1-e^{-x})right]_0^infty}_{to0}+frac12int_0^inftyfrac{x^2}{1-e^{-x}}e^{-x}mathrm dx\
      &=frac1{Gamma(3)}int_0^inftyfrac{x^{3-1}}{e^x-1}mathrm dx\
      &=zeta(3)
      end{align*}



      Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.



      IV: The Trilogarithm $operatorname{Li}_3(1)$



      Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=frac{log(1-x)}x$ to get



      begin{align*}
      int_0^1frac{log(1-x)log(x)}xmathrm dx&=underbrace{left[log(x)(-operatorname{Li}_2(x))right]_0^1}_{to0}+int_0^1frac{operatorname{Li}_2(x)}xmathrm dx\
      &=[operatorname{Li}_3(x)]_0^1\
      &=zeta(3)
      end{align*}



      A quick look at the series representation of the Trilogarithm verifies the last line.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them



        I: Taylor Series Expansion of $log(1-x)$



        As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain



        begin{align*}
        int_0^1frac{log(1-x)log(x)}xmathrm dx&=int_0^1frac{log(x)}xleft[-sum_{n=1}^inftyfrac{x^n}nright]mathrm dx\
        &=-sum_{n=1}^inftyfrac1nint_0^1x^{n-1}log(x)mathrm dx\
        &=-sum_{n=1}^inftyfrac1nleft[-frac1{n^2}right]\
        &=sum_{n=1}^inftyfrac1{n^3}\
        &=zeta(3)
        end{align*}



        This might be the most straightforward approach possible.



        II: Integration By Parts



        Choosing $u=log(1-x)$ and $mathrm dv=frac{log(x)}x$ we can apply Integration By Parts which gives



        begin{align*}
        int_0^1frac{log(1-x)log(x)}x&=underbrace{left[log(1-x)frac{log^2(x)}2right]_0^1}_{to0}+frac12int_0^1frac{log^2(x)}{1-x}mathrm dx\
        &=frac12int_0^1log^2(x)left[sum_{n=0}^infty x^nright]mathrm dx\
        &=frac12sum_{n=0}^inftyint_0^1x^nlog^2(x)mathrm dx\
        &=frac12sum_{n=0}^inftyleft[frac2{(n+1)^3}right]\
        &=sum_{n=1}^inftyfrac1{n^3}\
        &=zeta(3)
        end{align*}



        Again, we utilized a series expansion, this time the one of the geometric series.



        III: Integral Representation of the Zeta Function



        To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find



        begin{align*}
        int_0^1frac{log(1-x)log(x)}xmathrm dx&=-int_infty^0(-x)log(1-e^{-x})mathrm dx\
        &=-int_0^infty xlog(1-e^{-x})mathrm dx\
        &=underbrace{left[frac{x^2}2log(1-e^{-x})right]_0^infty}_{to0}+frac12int_0^inftyfrac{x^2}{1-e^{-x}}e^{-x}mathrm dx\
        &=frac1{Gamma(3)}int_0^inftyfrac{x^{3-1}}{e^x-1}mathrm dx\
        &=zeta(3)
        end{align*}



        Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.



        IV: The Trilogarithm $operatorname{Li}_3(1)$



        Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=frac{log(1-x)}x$ to get



        begin{align*}
        int_0^1frac{log(1-x)log(x)}xmathrm dx&=underbrace{left[log(x)(-operatorname{Li}_2(x))right]_0^1}_{to0}+int_0^1frac{operatorname{Li}_2(x)}xmathrm dx\
        &=[operatorname{Li}_3(x)]_0^1\
        &=zeta(3)
        end{align*}



        A quick look at the series representation of the Trilogarithm verifies the last line.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them



          I: Taylor Series Expansion of $log(1-x)$



          As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain



          begin{align*}
          int_0^1frac{log(1-x)log(x)}xmathrm dx&=int_0^1frac{log(x)}xleft[-sum_{n=1}^inftyfrac{x^n}nright]mathrm dx\
          &=-sum_{n=1}^inftyfrac1nint_0^1x^{n-1}log(x)mathrm dx\
          &=-sum_{n=1}^inftyfrac1nleft[-frac1{n^2}right]\
          &=sum_{n=1}^inftyfrac1{n^3}\
          &=zeta(3)
          end{align*}



          This might be the most straightforward approach possible.



          II: Integration By Parts



          Choosing $u=log(1-x)$ and $mathrm dv=frac{log(x)}x$ we can apply Integration By Parts which gives



          begin{align*}
          int_0^1frac{log(1-x)log(x)}x&=underbrace{left[log(1-x)frac{log^2(x)}2right]_0^1}_{to0}+frac12int_0^1frac{log^2(x)}{1-x}mathrm dx\
          &=frac12int_0^1log^2(x)left[sum_{n=0}^infty x^nright]mathrm dx\
          &=frac12sum_{n=0}^inftyint_0^1x^nlog^2(x)mathrm dx\
          &=frac12sum_{n=0}^inftyleft[frac2{(n+1)^3}right]\
          &=sum_{n=1}^inftyfrac1{n^3}\
          &=zeta(3)
          end{align*}



          Again, we utilized a series expansion, this time the one of the geometric series.



          III: Integral Representation of the Zeta Function



          To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find



          begin{align*}
          int_0^1frac{log(1-x)log(x)}xmathrm dx&=-int_infty^0(-x)log(1-e^{-x})mathrm dx\
          &=-int_0^infty xlog(1-e^{-x})mathrm dx\
          &=underbrace{left[frac{x^2}2log(1-e^{-x})right]_0^infty}_{to0}+frac12int_0^inftyfrac{x^2}{1-e^{-x}}e^{-x}mathrm dx\
          &=frac1{Gamma(3)}int_0^inftyfrac{x^{3-1}}{e^x-1}mathrm dx\
          &=zeta(3)
          end{align*}



          Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.



          IV: The Trilogarithm $operatorname{Li}_3(1)$



          Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=frac{log(1-x)}x$ to get



          begin{align*}
          int_0^1frac{log(1-x)log(x)}xmathrm dx&=underbrace{left[log(x)(-operatorname{Li}_2(x))right]_0^1}_{to0}+int_0^1frac{operatorname{Li}_2(x)}xmathrm dx\
          &=[operatorname{Li}_3(x)]_0^1\
          &=zeta(3)
          end{align*}



          A quick look at the series representation of the Trilogarithm verifies the last line.






          share|cite|improve this answer









          $endgroup$



          There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them



          I: Taylor Series Expansion of $log(1-x)$



          As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain



          begin{align*}
          int_0^1frac{log(1-x)log(x)}xmathrm dx&=int_0^1frac{log(x)}xleft[-sum_{n=1}^inftyfrac{x^n}nright]mathrm dx\
          &=-sum_{n=1}^inftyfrac1nint_0^1x^{n-1}log(x)mathrm dx\
          &=-sum_{n=1}^inftyfrac1nleft[-frac1{n^2}right]\
          &=sum_{n=1}^inftyfrac1{n^3}\
          &=zeta(3)
          end{align*}



          This might be the most straightforward approach possible.



          II: Integration By Parts



          Choosing $u=log(1-x)$ and $mathrm dv=frac{log(x)}x$ we can apply Integration By Parts which gives



          begin{align*}
          int_0^1frac{log(1-x)log(x)}x&=underbrace{left[log(1-x)frac{log^2(x)}2right]_0^1}_{to0}+frac12int_0^1frac{log^2(x)}{1-x}mathrm dx\
          &=frac12int_0^1log^2(x)left[sum_{n=0}^infty x^nright]mathrm dx\
          &=frac12sum_{n=0}^inftyint_0^1x^nlog^2(x)mathrm dx\
          &=frac12sum_{n=0}^inftyleft[frac2{(n+1)^3}right]\
          &=sum_{n=1}^inftyfrac1{n^3}\
          &=zeta(3)
          end{align*}



          Again, we utilized a series expansion, this time the one of the geometric series.



          III: Integral Representation of the Zeta Function



          To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find



          begin{align*}
          int_0^1frac{log(1-x)log(x)}xmathrm dx&=-int_infty^0(-x)log(1-e^{-x})mathrm dx\
          &=-int_0^infty xlog(1-e^{-x})mathrm dx\
          &=underbrace{left[frac{x^2}2log(1-e^{-x})right]_0^infty}_{to0}+frac12int_0^inftyfrac{x^2}{1-e^{-x}}e^{-x}mathrm dx\
          &=frac1{Gamma(3)}int_0^inftyfrac{x^{3-1}}{e^x-1}mathrm dx\
          &=zeta(3)
          end{align*}



          Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.



          IV: The Trilogarithm $operatorname{Li}_3(1)$



          Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=frac{log(1-x)}x$ to get



          begin{align*}
          int_0^1frac{log(1-x)log(x)}xmathrm dx&=underbrace{left[log(x)(-operatorname{Li}_2(x))right]_0^1}_{to0}+int_0^1frac{operatorname{Li}_2(x)}xmathrm dx\
          &=[operatorname{Li}_3(x)]_0^1\
          &=zeta(3)
          end{align*}



          A quick look at the series representation of the Trilogarithm verifies the last line.







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          answered 2 hours ago









          mrtaurhomrtaurho

          6,31071742




          6,31071742























              5












              $begingroup$

              begin{align}J&=int_0^1 frac{ln(1-x)ln x}{x} dx\
              &=-int_0^1 left(sum_{n=1}^infty frac{x^{n-1}}{n}right)ln x,dx\
              &=-sum_{n=1}^infty frac{1}{n}int_0^1 x^{n-1}ln x,dx\
              &=sum_{n=1}^infty frac{1}{n^3}\
              &=zeta(3)
              end{align}






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                begin{align}J&=int_0^1 frac{ln(1-x)ln x}{x} dx\
                &=-int_0^1 left(sum_{n=1}^infty frac{x^{n-1}}{n}right)ln x,dx\
                &=-sum_{n=1}^infty frac{1}{n}int_0^1 x^{n-1}ln x,dx\
                &=sum_{n=1}^infty frac{1}{n^3}\
                &=zeta(3)
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  begin{align}J&=int_0^1 frac{ln(1-x)ln x}{x} dx\
                  &=-int_0^1 left(sum_{n=1}^infty frac{x^{n-1}}{n}right)ln x,dx\
                  &=-sum_{n=1}^infty frac{1}{n}int_0^1 x^{n-1}ln x,dx\
                  &=sum_{n=1}^infty frac{1}{n^3}\
                  &=zeta(3)
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  begin{align}J&=int_0^1 frac{ln(1-x)ln x}{x} dx\
                  &=-int_0^1 left(sum_{n=1}^infty frac{x^{n-1}}{n}right)ln x,dx\
                  &=-sum_{n=1}^infty frac{1}{n}int_0^1 x^{n-1}ln x,dx\
                  &=sum_{n=1}^infty frac{1}{n^3}\
                  &=zeta(3)
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  FDPFDP

                  6,29211931




                  6,29211931






















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