MHB Logarithmic Integral on Stack Exchange - author unknown

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An alternative solution to the integral $$\int_0^1\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx$$ is presented, utilizing a generalized parametric case $$\mathcal{I}(z)$$. The solution involves substitutions and integration by parts, leading to a closed form that includes logarithmic and polylogarithmic functions. The discussion highlights the complexity of higher-order integrals of similar forms, which can be approached using comparable methods. Participants express interest in further developments on logarithmic integrals and the potential for more intricate solutions. The conversation emphasizes the mathematical intricacies involved in evaluating these types of integrals.
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Consider the generalized parametric case where $$0 < z \le 1$$:$$\mathcal{I}(z)=\int_0^z\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx$$Substitute $$y=1+x$$ to obtain:$$\int_1^{1+z}\frac{\log y\log[1-(y-1)]}{y}\,dy=\int_1^{1+z}\frac{\log y\log(2-y)}{y}\,dy=$$$$\int_1^{1+z}\frac{\log y\log\left[2 \left(1-\frac{y}{2} \right) \right]}{y}\,dy=$$$$\log 2\, \int_1^{1+z}\frac{\log y}{y}\,dy+\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$Performing an integration by parts on that first integral gives:$$\frac{1}{2}\log 2\, (\log y)^2\, \Bigg|_1^{1+z}=\frac{1}{2}\log 2\log^2(1+z)$$So$$\mathcal{I}(z)=\frac{1}{2}\log 2\log^2(1+z)+\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$Next, we split that last integral into two:$$\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy=
\int_0^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy-\int_0^1\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$For $$0 < z \le 1$$, Polylogarithms of order $$m \ge 1$$ have the integral representation:$$\text{Li}_m(z)=\frac{(-1)^{m-1}}{(m-2)!}\int_0^1\frac{(\log x)^{m-2}\log(1-zx)}{x}\,dx$$Hence$$\int_0^1\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy=\text{Li}_3\left( \tfrac{1}{2}\right) $$So$$\mathcal{I}(z)=\frac{1}{2}\log 2\log^2(1+z)-\text{Li}_3\left( \tfrac{1}{2}\right) +\int_0^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$For that final integral, apply the substitution $$y=(1+z)\, x$$ to change it into:$$\int_0^1\frac{\log[(1+z)\, x]\, \log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx=$$$$\log(1+z)\, \int_0^1\frac{\log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx+\int_0^1\frac{\log x\, \log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx$$By the integral representation for arbitrary (non-zero) order Polylogs given above, this equates to:$$-\log(1+z) \, \text{Li}_2\left(\frac{1+z}{2}\right)+ \text{Li}_3 \left(\frac{1+z}{2}\right)$$
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General parametric solution:

$$\int_0^z\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx=$$$$\frac{1}{2}\log 2\log^2(1+z)-\text{Li}_3\left( \tfrac{1}{2}\right) - \log(1+z) \, \text{Li}_2\left(\frac{1+z}{2}\right)+ \text{Li}_3 \left(\frac{1+z}{2}\right)$$For $$\, \, 0 < z \le 1$$
 
Nice solution . I always consider differentiating the hypergeometric function to be an alternative solution even though not easy .
 
ZaidAlyafey said:
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 $$\int_0^z\frac{\log^m(1+x)\log(1-x)}{(1+x)},dx$$and $$\int_0^z\frac{\log^m(1-x)\log(1+x)}{(1-x)},dx$$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... ;)
 
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That seems interesting , I'll be waiting to see that .
 
DreamWeaver said:
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 $$\int_0^z\frac{\log^m(1+x)\log(1-x)}{(1+x)},dx$$and $$\int_0^z\frac{\log^m(1-x)\log(1+x)}{(1-x)},dx$$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... ;)

ZaidAlyafey said:
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
 
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