Prove 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 web, but it appears that wikipedia is the only place that it's mentioned, of course no proof is given. I would appreciate some help!

Edit. I have found here on AoPS a way using real methods that solves my problem.

  • $\begingroup$ This integral becomes much harder than it should be if you eschew complex analysis techniques. $\endgroup$ – pisco Oct 5 '18 at 15:28
  • $\begingroup$ You are right! I edited it to leave more space, but I will wait for an approach that doesn't use contour integration. $\endgroup$ – Zacky Oct 5 '18 at 16:40
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    $\begingroup$ There is a very sweet derivation of this result here arxiv.org/pdf/1612.03292.pdf ... Using residue calculus around a key hole contour. $\endgroup$ – Donald Splutterwit Jun 25 '19 at 19:20

For an approach using contour integration:

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.

  • $\begingroup$ 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? $\endgroup$ – Zacky Oct 5 '18 at 16:36
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    $\begingroup$ @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 $g(z+2\pi i) = g(z)$. $\endgroup$ – pisco Oct 5 '18 at 16:41

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