Logarithmic Integral on Stack Exchange - author unknown

[DreamWeaver]

Here's an alternative solution to solving the integral:\(\displaystyle \int_0^1\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx\)[Original post: calculus - A challenging logarithmic integral $\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx$ - Mathematics Stack Exchange]
I'll post a link on there so the OP can see my work...

 

[DreamWeaver]

Consider the generalized parametric case where \(\displaystyle 0 < z \le 1\):\(\displaystyle \mathcal{I}(z)=\int_0^z\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx\)Substitute \(\displaystyle y=1+x\) to obtain:\(\displaystyle \int_1^{1+z}\frac{\log y\log[1-(y-1)]}{y}\,dy=\int_1^{1+z}\frac{\log y\log(2-y)}{y}\,dy=\)\(\displaystyle \int_1^{1+z}\frac{\log y\log\left[2 \left(1-\frac{y}{2} \right) \right]}{y}\,dy=\)\(\displaystyle \log 2\, \int_1^{1+z}\frac{\log y}{y}\,dy+\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy\)Performing an integration by parts on that first integral gives:\(\displaystyle \frac{1}{2}\log 2\, (\log y)^2\, \Bigg|_1^{1+z}=\frac{1}{2}\log 2\log^2(1+z)\)So\(\displaystyle \mathcal{I}(z)=\frac{1}{2}\log 2\log^2(1+z)+\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy\)Next, we split that last integral into two:\(\displaystyle \int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy=
\int_0^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy-\int_0^1\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy\)For \(\displaystyle 0 < z \le 1\), Polylogarithms of order \(\displaystyle m \ge 1\) have the integral representation:\(\displaystyle \text{Li}_m(z)=\frac{(-1)^{m-1}}{(m-2)!}\int_0^1\frac{(\log x)^{m-2}\log(1-zx)}{x}\,dx\)Hence\(\displaystyle \int_0^1\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy=\text{Li}_3\left( \tfrac{1}{2}\right) \)So\(\displaystyle \mathcal{I}(z)=\frac{1}{2}\log 2\log^2(1+z)-\text{Li}_3\left( \tfrac{1}{2}\right) +\int_0^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy\)For that final integral, apply the substitution \(\displaystyle y=(1+z)\, x\) to change it into:\(\displaystyle \int_0^1\frac{\log[(1+z)\, x]\, \log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx=\)\(\displaystyle \log(1+z)\, \int_0^1\frac{\log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx+\int_0^1\frac{\log x\, \log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx\)By the integral representation for arbitrary (non-zero) order Polylogs given above, this equates to:\(\displaystyle -\log(1+z) \, \text{Li}_2\left(\frac{1+z}{2}\right)+ \text{Li}_3 \left(\frac{1+z}{2}\right)\)
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General parametric solution:

\(\displaystyle \int_0^z\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx=\)\(\displaystyle \frac{1}{2}\log 2\log^2(1+z)-\text{Li}_3\left( \tfrac{1}{2}\right) - \log(1+z) \, \text{Li}_2\left(\frac{1+z}{2}\right)+ \text{Li}_3 \left(\frac{1+z}{2}\right)\)For \(\displaystyle \, \, 0 < z \le 1\)

 

[alyafey22]

Nice solution . I always consider differentiating the hypergeometric function to be an alternative solution even though not easy .

 

[DreamWeaver]

Nice solution . I always consider differentiating the hypergeometric function to be an alternative solution even though not easy .

Thanks Zaid! (Sun)

It can get a bit - erm - 'hairy' when your dealing with higher order integrals of the type above, but all integrals of the form \(\displaystyle \int_0^z\frac{\log^m(1+x)\log(1-x)}{(1+x)},dx\)and \(\displaystyle \int_0^z\frac{\log^m(1-x)\log(1+x)}{(1-x)},dx\)can be done in the same way. The closed form involves some pretty annoying (finite!) double sums (possibly a triple sum too)... I'll add this onto the Logarithmic Integrals thread at some point (soon). It's been "in the post" for a while... ;)

 

[alyafey22]

That seems interesting , I'll be waiting to see that .

 

[anemone]

Thanks Zaid! (Sun)

It can get a bit - erm - 'hairy' when your dealing with higher order integrals of the type above, but all integrals of the form \(\displaystyle \int_0^z\frac{\log^m(1+x)\log(1-x)}{(1+x)},dx\)and \(\displaystyle \int_0^z\frac{\log^m(1-x)\log(1+x)}{(1-x)},dx\)can be done in the same way. The closed form involves some pretty annoying (finite!) double sums (possibly a triple sum too)... I'll add this onto the Logarithmic Integrals thread at some point (soon). It's been "in the post" for a while... ;)

That seems interesting , I'll be waiting to see that .

Hi DreamWeaver,

I see it now that you have got a fan, congrats! (Sun)And I envy you!:p