1. definite integral #5

Show that $\int_{0}^{1} \frac{x \ln x}{(1-x^{2})(\pi^2+\ln^{2} x )} \ dx = \frac{1}{4} + \frac{\psi_{0}(1)}{2} = \frac{1}{4} - \frac{\gamma}{2}$

where $\psi_{0}(x)$ is the digamma function and $\gamma$ is Euler's constant.

2. Re: definite integral #5

It doesn't look like anyone is going to post a solution.

The solution requires two facts:

1) $\sum_{k=1}^{\infty} \frac{1}{a^{2}+k^{2}} = \frac{\pi}{2a} \coth {\pi a} - \frac{1}{2a^{2}}$

2) $\int_{0}^{\infty} \Big(\frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}} \Big) \ dt = \psi_{0}(z)$

$\int_{0}^{1} \frac{\ln x}{\pi^{2} + \ln^{2} x} \frac{x}{1-x^{2}} \ dx = \int_{0}^{1} \int_{0}^{\infty} \sin (t \ln x) e^{-\pi t} \frac{x}{1-x^{2}} \ dt \ dx$

assuming the integral converges absolutely, change the order of integration and let $u = -\ln x$

$= \int_{0}^{\infty} \int_{0}^{\infty} e^{-\pi t} \sin (-tu) \frac{e^{-2u}}{1-e^{-2u}} \ du \ dt$

$= - \int_{0}^{\infty} \int_{0}^{\infty} e^{-\pi t} \sin (tu) \sum_{n=1}^{\infty} e^{-2nu} \ du \ dt$

$= - \int_{0}^{\infty} e^{-\pi t} \sum_{n=1}^{\infty} \int_{0}^{\infty} \sin (tu) e^{-2nu} \ du \ dt$

$- \int_{0}^{\infty} e^{-\pi t} \sum_{n=1}^{\infty} \frac{t}{t^{2}+4n^{2}}} \ dt$

$= -\frac{1}{4} \int_{0}^{\infty} e^{-\pi t } t \sum_{n=1}^{\infty} \frac{1}{(\frac{t}{2})^{2} + n^{2}} \ dt$

$= \frac{1}{4} \int_{0}^{\infty} e^{-\pi t} \Bigg( \frac{2}{t} - \pi \coth \Big(\frac{\pi t}{2}\Big) \Bigg) \ dt$

continued in next post

3. Re: definite integral #5

let $w = \pi t$

$= \frac{1}{4 \pi} \int_{0}^{\infty} e^{-w} \Bigg( \frac{2 \pi}{w} - \pi \coth \Big(\frac{w}{2} \Big) \Bigg) \ dw$

$= \frac{1}{4} \int_{0}^{\infty} e^{-w} \Big( \frac{2}{w} - \frac{1+e^{-w}}{1-e^{-w}} \Big) \ dw$

$= \frac{1}{4} \int^{\infty}_{0} e^{-w} \Big(\frac{2}{w} - \frac{2}{1-e^{-w}} + 1 \Big) \ dw$

$= \frac{1}{4} \int_{0}^{\infty} e^{-w} \ dw + \frac{1}{2} \int_{0}^{\infty} \Big( \frac{e^{-w}}{w} - \frac{e^{-w}}{1-e^{-w}} \Big) \ dw$

$= \frac{1}{4} + \frac{\psi_{0}(1)}{2} = \frac{1}{4} - \frac{\gamma}{2}$