Integral involving composition of cosine and log

Question

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.

Answer 1

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.

Answer 2

$\textbf{Hint:}$

$$\int_0^1\frac{1}{x}\log(1-x^2)\cos(p\log x)dx = \operatorname{Re}\left\{\int_0^1x^{ip-1}\log(1-x^2)dx\right\}$$

and

$$\int_0^1x^{ip-1}\log(1-x^2)dx = \frac{x^{ip}-1}{ip}\log(1-x^2)\Biggr|_0^1+\frac{2}{ip}\int_0^1\frac{x^{ip+1}-x}{1-x^2}dx$$

$$= \frac{i}{p}\int_0^1\frac{1-x^{\frac{ip}{2}}}{1-x}dx$$

by choosing the antiderivative of the monomial which is $0$ at $x=1$. Can you complete the computation from here using Taylor series or any method of your choice?

Answer 3

Observing that $$ \int_0^1 \frac{1}{x} \ln \left(1-x^2\right) \cos (p \ln x) d x = \Re\int_0^1 \frac{1}{x} \ln \left(1-x^2\right) e^{i p \ln x} d x = \Re \int_0^1 \frac{1}{x} \ln \left(1-x^2\right) x^{p i} d x $$ Putting $y=x^2$ and expanding $\ln(1-y)$ yields

$$ \begin{aligned} \int_0^1 \frac{1}{x} \ln \left(1-x^2\right) \cos (p \ln x) d x & =-\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} \Re \int_0^1 y^{n+\frac{p i}{2}-1} d y \\ & =-\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} \Re\left(\frac{1}{n+\frac{pi}{2}} \cdot \frac{n-\frac{p i}{2}}{n-\frac{pi}{2}}\right) \\ & =-\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{n}{n^2+\frac{p^2}{4}}\right)\\&=-2 \sum_{n=1}^{\infty} \frac{1}{4 n^2+p^2}\\&= \frac{1}{p^2} -\frac{\pi\coth\left(\frac{p\pi}{2}\right)}{2p} \end{aligned} $$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\left.\int_{0}^{1}{1 \over x} \ln\pars{1 - x^{2}} \cos\pars{p\ln\pars{x}}\,\dd x\right\vert_{\, p\ \in\ \mathbb{R}}} \\[5mm] = & \ \Re\int_{0}^{1}{\ln\pars{1 - x^{2}} \over x} \expo{\ic\,\verts{p}\ln{x}}\,\dd x \\[5mm] & \sr{x\ \mapsto\ x^{1/2}}{=} {1 \over 2}\,\Re\int_{0}^{1}x^{\ic\verts{p}/2\ -\ 1}\,\, \ln\pars{1 - x}\,\dd x \\[5mm] = & \ {1 \over 2}\,\Re\lim_{\pars{\epsilon,\nu}\ \to\ \pars{0^{+}\ ,\ 0}}\,\, \partiald{}{\nu}\int_{0}^{1}x^{\epsilon\ -\ 1\ +\ \verts{p}\ic/2}\,\,\, \bracks{\pars{1 - x}^{\nu} - 1}\dd x \\[5mm] = & \ {1 \over 2}\,\Re\lim_{\pars{\epsilon,\nu}\ \to\ \pars{0^{+}\ ,\ 0}}\,\,\,\partiald{}{\nu}\left[\Gamma\pars{\epsilon + \verts{p}\ic/2}\Gamma\pars{\nu + 1} \over \Gamma\pars{\epsilon + \verts{p}\ic/2 + \nu + 1}\right. \\ & \left.\phantom{AAAAAAAAAAAAAA} -\, {1 \over \epsilon + \verts{p}\ic/2}\right] \\[5mm] = & \ -\,{1 \over \verts{p}}\,\Im\Psi\pars{1 + {\verts{p} \over 2}\ic} \\[5mm] & \sr{\color{red}{\LARGE\S}}{=} -\,{1 \over \verts{p}}\bracks{-\,{1 \over 2\pars{\verts{p}/2}} + {\pi \over 2}\coth\pars{\pi{\verts{p} \over 2}}} \\[5mm] = & \ \bbx{\color{#44f}{{1 \over p^{2}} - {\pi \over 2p}\coth\pars{\pi p \over 2}}}\qquad \left\vert\rule{0pt}{1cm}\quad\Psi:\ Digamma\ Function\right. \\ & \end{align} $\ds{\color{red}{\Large\S}:}$ See ${\bf{\mbox{6.3.13}}}$ in A & S