Evaluate $int_0^1 dfrac{ln (1 - x) ln (1 + x)}{x} , dx$












5












$begingroup$


I was playing around with trying to prove the following alternating Euler sum:
$$sum_{n = 1}^infty frac{(-1)^n H_n}{n^2} = -frac{5}{8} zeta (3).$$
Here $H_n$ is the Harmonic number.



At least two different proofs for this result that I could find can be seen here and here. Embarking on an alternative proof I did the following. From the integral representation for the Harmonic number of
$$H_n = int_0^1 frac{1 - x^n}{1 - x} , dx$$
I rewrote the alternating Euler sum as
begin{align}
sum_{n = 1}^infty frac{(-1)^n H_n}{n^2} &= int_0^1 frac{1}{1 - x} left [sum_{n = 1}^infty frac{(-1)^n}{n^2} - sum_{n = 1}^infty frac{(-1)^n x^n}{n^2} right ]\
&= int_0^1 frac{-pi^2/12 - operatorname{Li}_2 (-x)}{1 - x} , dx\
&= int_0^1 frac{ln (1 - x) ln (1 + x)}{x} , dx,
end{align}
where integration by parts has been used. I expect this last integral can be knocked over with relative easy but I have been going around in circles now for quite some time without any success.




So my question is, how can this last integral be evaluated (a real method is preferred) that does not rely on the alternating Euler sum I started out with?







integration definite-integrals improper-integrals euler-sums







share|cite|improve this question









$endgroup$

















    5












    $begingroup$


    I was playing around with trying to prove the following alternating Euler sum:
    $$sum_{n = 1}^infty frac{(-1)^n H_n}{n^2} = -frac{5}{8} zeta (3).$$
    Here $H_n$ is the Harmonic number.



    At least two different proofs for this result that I could find can be seen here and here. Embarking on an alternative proof I did the following. From the integral representation for the Harmonic number of
    $$H_n = int_0^1 frac{1 - x^n}{1 - x} , dx$$
    I rewrote the alternating Euler sum as
    begin{align}
    sum_{n = 1}^infty frac{(-1)^n H_n}{n^2} &= int_0^1 frac{1}{1 - x} left [sum_{n = 1}^infty frac{(-1)^n}{n^2} - sum_{n = 1}^infty frac{(-1)^n x^n}{n^2} right ]\
    &= int_0^1 frac{-pi^2/12 - operatorname{Li}_2 (-x)}{1 - x} , dx\
    &= int_0^1 frac{ln (1 - x) ln (1 + x)}{x} , dx,
    end{align}
    where integration by parts has been used. I expect this last integral can be knocked over with relative easy but I have been going around in circles now for quite some time without any success.




    So my question is, how can this last integral be evaluated (a real method is preferred) that does not rely on the alternating Euler sum I started out with?







    integration definite-integrals improper-integrals euler-sums







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      I was playing around with trying to prove the following alternating Euler sum:
      $$sum_{n = 1}^infty frac{(-1)^n H_n}{n^2} = -frac{5}{8} zeta (3).$$
      Here $H_n$ is the Harmonic number.



      At least two different proofs for this result that I could find can be seen here and here. Embarking on an alternative proof I did the following. From the integral representation for the Harmonic number of
      $$H_n = int_0^1 frac{1 - x^n}{1 - x} , dx$$
      I rewrote the alternating Euler sum as
      begin{align}
      sum_{n = 1}^infty frac{(-1)^n H_n}{n^2} &= int_0^1 frac{1}{1 - x} left [sum_{n = 1}^infty frac{(-1)^n}{n^2} - sum_{n = 1}^infty frac{(-1)^n x^n}{n^2} right ]\
      &= int_0^1 frac{-pi^2/12 - operatorname{Li}_2 (-x)}{1 - x} , dx\
      &= int_0^1 frac{ln (1 - x) ln (1 + x)}{x} , dx,
      end{align}
      where integration by parts has been used. I expect this last integral can be knocked over with relative easy but I have been going around in circles now for quite some time without any success.




      So my question is, how can this last integral be evaluated (a real method is preferred) that does not rely on the alternating Euler sum I started out with?







      integration definite-integrals improper-integrals euler-sums







      share|cite|improve this question









      $endgroup$




      I was playing around with trying to prove the following alternating Euler sum:
      $$sum_{n = 1}^infty frac{(-1)^n H_n}{n^2} = -frac{5}{8} zeta (3).$$
      Here $H_n$ is the Harmonic number.



      At least two different proofs for this result that I could find can be seen here and here. Embarking on an alternative proof I did the following. From the integral representation for the Harmonic number of
      $$H_n = int_0^1 frac{1 - x^n}{1 - x} , dx$$
      I rewrote the alternating Euler sum as
      begin{align}
      sum_{n = 1}^infty frac{(-1)^n H_n}{n^2} &= int_0^1 frac{1}{1 - x} left [sum_{n = 1}^infty frac{(-1)^n}{n^2} - sum_{n = 1}^infty frac{(-1)^n x^n}{n^2} right ]\
      &= int_0^1 frac{-pi^2/12 - operatorname{Li}_2 (-x)}{1 - x} , dx\
      &= int_0^1 frac{ln (1 - x) ln (1 + x)}{x} , dx,
      end{align}
      where integration by parts has been used. I expect this last integral can be knocked over with relative easy but I have been going around in circles now for quite some time without any success.




      So my question is, how can this last integral be evaluated (a real method is preferred) that does not rely on the alternating Euler sum I started out with?







      integration definite-integrals improper-integrals euler-sums




      integration definite-integrals improper-integrals euler-sums






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 24 at 10:21









      omegadotomegadot

      6,3972829




      6,3972829






















          1 Answer
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          8












          $begingroup$

          Similarly to this post. We have:
          $$4ab=(a+b)^2-(a-b)^2$$
          Thus we can take $a=ln(1-x)$ and $b=ln(1+x)$ to get:
          $$I=int_0^1 frac{ln(1-x)ln(1+x)}{x}dx=frac14 int_0^1 frac{ln^2(1-x^2)}{x}dx-frac14 int_0^1 frac{ln^2left(frac{1-x}{1+x}right)}{x}dx$$
          By setting $x^2 =t$ in the first integral and $frac{1-x}{1+x}=t$ in the second one we get:
          $$I=frac18 int_0^1 frac{ln^2(1-t)}{t}dt -frac12 int_0^1 frac{ln^2 t}{1-t^2}dt=frac18 int_0^1 frac{ln^2t}{1-t}dt-frac12 int_0^1 frac{ln^2 t}{1-t^2}dt$$
          $$=frac18sum_{nge 0}int_0^1 t^n ln^2 tdt-frac12 sum_{nge 0}int_0^1 t^{2n}ln^2 tdt$$
          Using the following fact:
          $$int_0^1 t^a dt=frac{1}{a+1}Rightarrow int_0^1 t^a ln^k tdt=frac{d^k}{da^k} left(frac{1}{a+1}right)=frac{(-1)^k k!}{(a+1)^{k+1}}$$
          $$Rightarrow I=frac18sum_{nge 0}frac{2}{(n+1)^3} -frac12 sum_{nge 0}frac{2}{(2n+1)^3}=frac28 zeta(3) -frac78 zeta(3)=-frac58zeta(3)$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Now that's a nice trick! Thanks.
            $endgroup$
            – omegadot
            Jan 24 at 11:08












          • $begingroup$
            @omegadot I have edited my answer, it didn't update?
            $endgroup$
            – Zacky
            Jan 24 at 11:10






          • 1




            $begingroup$
            Yes. Now it has and thanks again.
            $endgroup$
            – omegadot
            Jan 24 at 11:13











          Your Answer





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          1 Answer
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          active

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          active

          oldest

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          8












          $begingroup$

          Similarly to this post. We have:
          $$4ab=(a+b)^2-(a-b)^2$$
          Thus we can take $a=ln(1-x)$ and $b=ln(1+x)$ to get:
          $$I=int_0^1 frac{ln(1-x)ln(1+x)}{x}dx=frac14 int_0^1 frac{ln^2(1-x^2)}{x}dx-frac14 int_0^1 frac{ln^2left(frac{1-x}{1+x}right)}{x}dx$$
          By setting $x^2 =t$ in the first integral and $frac{1-x}{1+x}=t$ in the second one we get:
          $$I=frac18 int_0^1 frac{ln^2(1-t)}{t}dt -frac12 int_0^1 frac{ln^2 t}{1-t^2}dt=frac18 int_0^1 frac{ln^2t}{1-t}dt-frac12 int_0^1 frac{ln^2 t}{1-t^2}dt$$
          $$=frac18sum_{nge 0}int_0^1 t^n ln^2 tdt-frac12 sum_{nge 0}int_0^1 t^{2n}ln^2 tdt$$
          Using the following fact:
          $$int_0^1 t^a dt=frac{1}{a+1}Rightarrow int_0^1 t^a ln^k tdt=frac{d^k}{da^k} left(frac{1}{a+1}right)=frac{(-1)^k k!}{(a+1)^{k+1}}$$
          $$Rightarrow I=frac18sum_{nge 0}frac{2}{(n+1)^3} -frac12 sum_{nge 0}frac{2}{(2n+1)^3}=frac28 zeta(3) -frac78 zeta(3)=-frac58zeta(3)$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Now that's a nice trick! Thanks.
            $endgroup$
            – omegadot
            Jan 24 at 11:08












          • $begingroup$
            @omegadot I have edited my answer, it didn't update?
            $endgroup$
            – Zacky
            Jan 24 at 11:10






          • 1




            $begingroup$
            Yes. Now it has and thanks again.
            $endgroup$
            – omegadot
            Jan 24 at 11:13
















          8












          $begingroup$

          Similarly to this post. We have:
          $$4ab=(a+b)^2-(a-b)^2$$
          Thus we can take $a=ln(1-x)$ and $b=ln(1+x)$ to get:
          $$I=int_0^1 frac{ln(1-x)ln(1+x)}{x}dx=frac14 int_0^1 frac{ln^2(1-x^2)}{x}dx-frac14 int_0^1 frac{ln^2left(frac{1-x}{1+x}right)}{x}dx$$
          By setting $x^2 =t$ in the first integral and $frac{1-x}{1+x}=t$ in the second one we get:
          $$I=frac18 int_0^1 frac{ln^2(1-t)}{t}dt -frac12 int_0^1 frac{ln^2 t}{1-t^2}dt=frac18 int_0^1 frac{ln^2t}{1-t}dt-frac12 int_0^1 frac{ln^2 t}{1-t^2}dt$$
          $$=frac18sum_{nge 0}int_0^1 t^n ln^2 tdt-frac12 sum_{nge 0}int_0^1 t^{2n}ln^2 tdt$$
          Using the following fact:
          $$int_0^1 t^a dt=frac{1}{a+1}Rightarrow int_0^1 t^a ln^k tdt=frac{d^k}{da^k} left(frac{1}{a+1}right)=frac{(-1)^k k!}{(a+1)^{k+1}}$$
          $$Rightarrow I=frac18sum_{nge 0}frac{2}{(n+1)^3} -frac12 sum_{nge 0}frac{2}{(2n+1)^3}=frac28 zeta(3) -frac78 zeta(3)=-frac58zeta(3)$$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Now that's a nice trick! Thanks.
            $endgroup$
            – omegadot
            Jan 24 at 11:08












          • $begingroup$
            @omegadot I have edited my answer, it didn't update?
            $endgroup$
            – Zacky
            Jan 24 at 11:10






          • 1




            $begingroup$
            Yes. Now it has and thanks again.
            $endgroup$
            – omegadot
            Jan 24 at 11:13














          8












          8








          8





          $begingroup$

          Similarly to this post. We have:
          $$4ab=(a+b)^2-(a-b)^2$$
          Thus we can take $a=ln(1-x)$ and $b=ln(1+x)$ to get:
          $$I=int_0^1 frac{ln(1-x)ln(1+x)}{x}dx=frac14 int_0^1 frac{ln^2(1-x^2)}{x}dx-frac14 int_0^1 frac{ln^2left(frac{1-x}{1+x}right)}{x}dx$$
          By setting $x^2 =t$ in the first integral and $frac{1-x}{1+x}=t$ in the second one we get:
          $$I=frac18 int_0^1 frac{ln^2(1-t)}{t}dt -frac12 int_0^1 frac{ln^2 t}{1-t^2}dt=frac18 int_0^1 frac{ln^2t}{1-t}dt-frac12 int_0^1 frac{ln^2 t}{1-t^2}dt$$
          $$=frac18sum_{nge 0}int_0^1 t^n ln^2 tdt-frac12 sum_{nge 0}int_0^1 t^{2n}ln^2 tdt$$
          Using the following fact:
          $$int_0^1 t^a dt=frac{1}{a+1}Rightarrow int_0^1 t^a ln^k tdt=frac{d^k}{da^k} left(frac{1}{a+1}right)=frac{(-1)^k k!}{(a+1)^{k+1}}$$
          $$Rightarrow I=frac18sum_{nge 0}frac{2}{(n+1)^3} -frac12 sum_{nge 0}frac{2}{(2n+1)^3}=frac28 zeta(3) -frac78 zeta(3)=-frac58zeta(3)$$






          share|cite|improve this answer











          $endgroup$



          Similarly to this post. We have:
          $$4ab=(a+b)^2-(a-b)^2$$
          Thus we can take $a=ln(1-x)$ and $b=ln(1+x)$ to get:
          $$I=int_0^1 frac{ln(1-x)ln(1+x)}{x}dx=frac14 int_0^1 frac{ln^2(1-x^2)}{x}dx-frac14 int_0^1 frac{ln^2left(frac{1-x}{1+x}right)}{x}dx$$
          By setting $x^2 =t$ in the first integral and $frac{1-x}{1+x}=t$ in the second one we get:
          $$I=frac18 int_0^1 frac{ln^2(1-t)}{t}dt -frac12 int_0^1 frac{ln^2 t}{1-t^2}dt=frac18 int_0^1 frac{ln^2t}{1-t}dt-frac12 int_0^1 frac{ln^2 t}{1-t^2}dt$$
          $$=frac18sum_{nge 0}int_0^1 t^n ln^2 tdt-frac12 sum_{nge 0}int_0^1 t^{2n}ln^2 tdt$$
          Using the following fact:
          $$int_0^1 t^a dt=frac{1}{a+1}Rightarrow int_0^1 t^a ln^k tdt=frac{d^k}{da^k} left(frac{1}{a+1}right)=frac{(-1)^k k!}{(a+1)^{k+1}}$$
          $$Rightarrow I=frac18sum_{nge 0}frac{2}{(n+1)^3} -frac12 sum_{nge 0}frac{2}{(2n+1)^3}=frac28 zeta(3) -frac78 zeta(3)=-frac58zeta(3)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 24 at 10:53

























          answered Jan 24 at 10:35









          ZackyZacky

          7,76011061




          7,76011061








          • 1




            $begingroup$
            Now that's a nice trick! Thanks.
            $endgroup$
            – omegadot
            Jan 24 at 11:08












          • $begingroup$
            @omegadot I have edited my answer, it didn't update?
            $endgroup$
            – Zacky
            Jan 24 at 11:10






          • 1




            $begingroup$
            Yes. Now it has and thanks again.
            $endgroup$
            – omegadot
            Jan 24 at 11:13














          • 1




            $begingroup$
            Now that's a nice trick! Thanks.
            $endgroup$
            – omegadot
            Jan 24 at 11:08












          • $begingroup$
            @omegadot I have edited my answer, it didn't update?
            $endgroup$
            – Zacky
            Jan 24 at 11:10






          • 1




            $begingroup$
            Yes. Now it has and thanks again.
            $endgroup$
            – omegadot
            Jan 24 at 11:13








          1




          1




          $begingroup$
          Now that's a nice trick! Thanks.
          $endgroup$
          – omegadot
          Jan 24 at 11:08






          $begingroup$
          Now that's a nice trick! Thanks.
          $endgroup$
          – omegadot
          Jan 24 at 11:08














          $begingroup$
          @omegadot I have edited my answer, it didn't update?
          $endgroup$
          – Zacky
          Jan 24 at 11:10




          $begingroup$
          @omegadot I have edited my answer, it didn't update?
          $endgroup$
          – Zacky
          Jan 24 at 11:10




          1




          1




          $begingroup$
          Yes. Now it has and thanks again.
          $endgroup$
          – omegadot
          Jan 24 at 11:13




          $begingroup$
          Yes. Now it has and thanks again.
          $endgroup$
          – omegadot
          Jan 24 at 11:13


















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