I am trying to show that $$I=\int_{0}^\infty \frac{1}{(1+x-u)(\pi^2+\ln^2(x))}dx=\frac{1}{u}+\frac{1}{\ln(1-u)}$$

My first thought was to follow the $m=\frac{1}{x}$ which led to $$I=1+(u-1)\int_{-\infty}^\infty \frac{1}{(e^{-m}+(1-u))(\pi^2+m^2)}dm $$ But that unfortunately did not seem to go anywhere. I original saw this integral in the Laplace Transform section of Jack D'Aurizio's notes, but I, for the life of me, could not seem to use the Laplace Transform to evaluate the integral. I'm also aware that this is very similar to the integral involving the Gregory Coefficients, but I want to avoid those if I can. I've attempted to use differentiating under the integral but couldn't find any reasonable way of doing it. I've tried series, but that did not work either. Any push in the right direction would be much appreciated.

  • $\begingroup$ How does manipulating $u$, a variable independent of $x$, alter the integral and its boundaries? Please edit your question. $\endgroup$ – NestorV S Jun 24 '19 at 3:54
  • $\begingroup$ My mistake. Fixed. $\endgroup$ – Tom Himler Jun 24 '19 at 3:57
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    $\begingroup$ Hint: rewrite your integral as a contour integral $\frac{1}{2\pi i}\int_C \frac{dz}{\log(-z)(1+z-u)}$ over the keyhole contour $C$ appeared here. $\endgroup$ – achille hui Jun 24 '19 at 4:07
  • $\begingroup$ Ah thank you. Didn't even think about contour integration. Do you think there is a way of evaluating it without complex analysis? $\endgroup$ – Tom Himler Jun 24 '19 at 4:11
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    $\begingroup$ See this answer: math.stackexchange.com/a/2943531/515527. $\endgroup$ – Zacky Jun 24 '19 at 6:59

Not an answer, just some progress on the integral

We define $$P(u)=\int_0^\infty \frac{dx}{(x+u)(\pi^2+\ln^2x)}$$ which is $$P(u)=1-u\int_{-\infty}^{\infty}\frac{dx}{(e^x+u)(\pi^2+x^2)}.$$ Then preform the sub $x\mapsto \pi\tan x$ so that $$\begin{align} P(u)&=1-\frac{u}\pi\int_{-\pi/2}^{\pi/2}\frac{dx}{u+e^{\pi\tan x}}\\ &=1-\frac{u}\pi\left[\int_{0}^{\pi/2}\frac{dx}{u+e^{\pi\tan x}}+\int_{-\pi/2}^0\frac{dx}{u+e^{\pi\tan x}}\right]\\ &=1-\frac{u}\pi\left[\int_{0}^{\pi/2}\frac{dx}{u+e^{\pi\tan x}}+\int_0^{\pi/2}\frac{dx}{u+e^{-\pi\tan x}}\right]\\ &=1-\frac{u}\pi\int_{0}^{\pi/2}\frac{2u+2\cosh(\pi\tan x)}{1+u^2+2u\cosh(\pi\tan x)}dx\\ &=1-\frac{1}\pi\int_0^{\pi/2}\left(1+\frac{u^2-1}{1+u^2+2u\cosh(\pi\tan x)}\right)dx\\ &=\frac12+\frac{1-u^2}\pi\int_0^{\pi/2}\frac{dx}{1+u^2+2u\cosh(\pi\tan x)} \end{align}$$ Which is as far as I could get. Perhaps we may be able to use $$\frac1{1+2x\cos\phi+x^2}=\sum_{n\ge0}(-1)^n\frac{\sin(n+1)\phi}{\sin\phi}x^n$$ but it sort of seems like a shot in the dark.

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  • $\begingroup$ The $P(u)$ should contain $(1-u+x)$ $\endgroup$ – Archis Welankar Jun 24 '19 at 5:57
  • $\begingroup$ @ArchisWelankar No, we just note that $I=P(1-u)$ $\endgroup$ – clathratus Jun 24 '19 at 8:12
  • $\begingroup$ Okay so now we just need the ans to summation $\endgroup$ – Archis Welankar Jun 24 '19 at 8:42

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