Proving that $(-1)^{n+1}G_n =\int_0^\infty \frac{1}{(1+x)^n (\pi^2+\ln^2 x)} dx$

The goal of this question is to find a proof for Schroder's Integral formula: $$(-1)^{n+1}G_n =\int_0^\infty \frac{1}{(1+x)^n (\pi^2+\ln^2 x)}\mathrm dx$$ where $$G_n$$ are Gregory coefficients. This is into my interest because I received an answer to this question: Integral $$\int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{\left(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\right)^2}dx$$ that uses the above formula. I thought of this for some time already, but I have no idea how to start, I also searched the internet, but it appears that wikipedia is the only place that it's mentioned, of course no proof is given. I would appreciate some help!

• This integral becomes much harder than it should be if you eschew complex analysis techniques. – pisco Oct 5 '18 at 15:28
• You are right! I edited it to leave more space, but I will wait for an approach that doesn't use contour integration. – Zacky Oct 5 '18 at 16:40

By definition, $$G_n$$ are coefficients of $$z^n$$ of $$z/\ln(1+z)$$, it suffices to prove
For $$a<1$$:$$\int_0^\infty {\frac{1}{{(x + 1 - a)({{\ln }^2}x + {\pi ^2})}}dx} = \frac{1}{a} + \frac{1}{{\ln (1 - a)}}$$
Let $$f(z) = \frac{{{e^z}}}{{({e^z} + 1 - a)(z - \pi i)}}$$ integrate it along rectangular contour with vertices $$-R, R, R+2\pi i, -R+2\pi i$$, $$R$$ being very large. The only poles inside are $$z=i\pi, \ln(1-a)+i\pi$$, with residues $$1/a, 1/\ln(1-a)$$ respectively. Hence $$\int_{ - \infty }^\infty {\frac{{{e^x}}}{{({e^x} + 1 - a)(x - \pi i)}}dx} - \int_{ - \infty + 2\pi i}^{\infty + 2\pi i} {\frac{{{e^x}}}{{({e^x} + 1 - a)(x - \pi i)}}dx} = 2\pi i\left[ {\frac{1}{a} + \frac{1}{{\ln (1 - a)}}} \right]$$ Combining them (they individually diverge, but this is not an issue): $$\int_{ - \infty }^\infty {\frac{{{e^x}}}{{{e^x} + 1 - a}}\left[ {\frac{1}{{x - \pi i}} - \frac{1}{{x + \pi i}}} \right]dx} = 2\pi i\left[ {\frac{1}{a} + \frac{1}{{\ln (1 - a)}}} \right]$$ the result follows via a simple substitution.
• Thanks for this! Isn't easier to consider the first integral $I(a)$, then differentiate it $n-1$ times w. r. t. a? Also can you explain how did you know to choose this contour? – Zacky Oct 5 '18 at 16:36
• @Dahaka Yes, actually differentiation under the integral is exactly what I wanted to convey. Regarding the choice of contour, whenever you see $x^2+\pi^2$ in the denominator, it can probably be tackled via integrating something like $g(z)/(z-\pi i)$, with $g(z+2\pi i) = g(z)$. – pisco Oct 5 '18 at 16:41