I am trying to prove that
$$\int_{0}^1 \frac{1}{x}\log{(1-x^2)}\cos{(p \log{x})}dx = \frac{1}{p^{2}}-\frac{\pi}{2p}\coth\left(\frac{p\pi}{2}\right),\quad p \in \mathbb{R}$$
where $\coth {x} = \frac{e^{2x}+1}{e^{2x}-1}$. At $p=0$, we take the limiting value. It is taken from "Table of Integrals, series, and products" 7th edition by Gradshteyn and Ryzhik. Where they state that on page 595, $\int_{0}^1 \frac{1}{x}\log{(1-x^2)}\cos{(p \log{x})}dx = \frac{1}{2p^{2}}+\frac{\pi}{2p}\coth\left(\frac{p\pi}{2}\right)$ which seemed incorrect to me as the right side is mostly positive. I don't have a complete proof, but I think the correct approach would be integration by parts, $$\int_{0}^1 \frac{1}{x}\log{(1-x^2)}\cos{(p \log{x})}dx = \left. \ln(1-x^2)\frac{\sin(p\ln(x))}{x}\right|_0^1-\frac{1}{P}\int_0^{1}\frac{-2x}{1-x^2}\sin{(p\ln{x})}dx\\ = \frac{2}{P}\int_0^{1} \frac{x}{1-x^2}\sin{(p\ln{x})} \overset{x=e^{-t}}{=} -\frac{2}{p}\int_0^{\infty} \frac{\sin{pt}}{e^{2t}-1}dt.$$
where I am currently stuck at the last integral. Feynman's trick is not gonna work as the integral is gonna be unbounded. How should I proceed from here?
Edit: Maybe the approach is related to this solution which uses contour integration.
Best Answer
Note that
\begin{eqnarray*} \int_{0}^1 \frac{1}{x}\log{(1-x^2)}\cos{(p \log{x})}dx &=& \int_{0}^{\infty}\ln(1-e^{-2w})\cos(pw)dw \quad \left(w\mapsto -\ln(x)\right)\\ &=& \Re \int_{0}^{\infty}\ln(1-e^{-2w})e^{ipw}dw \\ &=& -\Re \sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{\infty}e^{-2wn}e^{ipw}dw\\ &=& -\Re \sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{\infty}e^{-w(2n-ip)}dw \\ &=& -\Re \sum_{n=1}^{\infty}\frac{1}{n(2n-ip)} \\ &=& -2\sum_{n=1}^{\infty}\frac{1}{4n^2+p^2}\\ &=& \frac{1}{p^2} -\frac{\pi\coth\left(\frac{p\pi}{2}\right)}{2p} \end{eqnarray*}
The last series is the partial fractions decomposition of the hyperbolic cotangent.