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?

Who was the lone kid in the line of people at the lake at the end of Avengers: Endgame?

Critique of timeline aesthetic

Multiple options vs single option UI

Don’t seats that recline flat defeat the purpose of having seatbelts?

What does the integral of a function times a function of a random variable represent, conceptually?

How can Republicans who favour free markets, consistently express anger when they don't like the outcome of that choice?

Why did C use the -> operator instead of reusing the . operator?

Which big number is bigger?

How can I print the prosodic symbols in LaTeX?

Like totally amazing interchangeable sister outfits II: The Revenge

Is it idiomatic to construct against `this`

How did Captain America manage to do this?

Constructions of PRF (Pseudo Random Function)

Overlay of two functions leaves gaps

Two field separators (colon and space) in awk

Aligning equation numbers vertically

Is the claim "Employers won't employ people with no 'social media presence'" realistic?

Can an Area of Effect spell cast outside a Prismatic Wall extend inside it?

Rivers without rain

What is the philosophical significance of speech acts/implicature?

How to write a column outside the braces in a matrix?

Was there a shared-world project before "Thieves World"?

Pulling the rope with one hand is as heavy as with two hands?



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.
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












$begingroup$


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









    $endgroup$














      Your Answer








      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });






      Coalition Coal is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3203711%2fwhat-are-the-steps-to-solving-this-definite-integral%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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






















                      Coalition Coal is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      Coalition Coal is a new contributor. Be nice, and check out our Code of Conduct.













                      Coalition Coal is a new contributor. Be nice, and check out our Code of Conduct.












                      Coalition Coal is a new contributor. Be nice, and check out our Code of Conduct.
















                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3203711%2fwhat-are-the-steps-to-solving-this-definite-integral%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

                      Castillo d'Acher Características Menú de navegación

                      Puerta de Hutt Referencias Enlaces externos Menú de navegación15°58′00″S 5°42′00″O /...