I am interested in knowing if it is possible to find a closed-form solution to the following challenging log-cotangent integral
$$\int_0^{\frac{\pi}{2}} \log(1 – x \cot x) \, dx$$
I very much doubt a closed form in terms of known mathematical constants can be found and have no reason to suspect one exists.
In the absence of a simple closed-form value being found, maybe the integral can be evaluated in terms of an infinite series. In this direction, since $0 < x \cot x < 1$ for all $x \in (0,\frac{\pi}{2})$ the log term appearing in the integrand can be expanded. Doing so produces
$$\int_0^{\frac{\pi}{2}} \log(1 – x \cot x) \, dx = -\sum_{n = 1}^\infty \frac{1}{n} \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx = -\sum_{n = 1}^\infty \frac{a_n}{n},$$
where
$$a_n = \int_0^{\frac{\pi}{2}} (x \cot x)^n \, dx. \tag1$$
Noting that
\begin{align*}
a_1 &= \int_0^{\frac{\pi}{2}} x \cot x \, dx = \frac{\pi}{2} \log (2)\\
a_2 &= \int_0^{\frac{\pi}{2}} (x \cot x)^2 \, dx = \pi \log (2) – \frac{\pi^3}{24}\\
a_3 &= \int_0^{\frac{\pi}{2}} (x \cot x)^3 \, dx = \frac{9\pi}{16} \zeta (3) – \frac{\pi^3}{16} + \frac{3\pi}{2} \log (2) – \frac{\pi^3}{8} \log (2),
\end{align*}
perhaps it is possible to find a general expression for $a_n$. Indeed, an attempt at finding (1) can be found here.
Any other approaches or suggestions to this challenging integral would be welcome.
Best Answer
Edited to make the answer self-contained
We introduce first a representation of the integral which makes use of Bessel functions: \begin{align} I&=\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx\\ &=\int_0^{\frac{\pi}{2}} \log\left( \frac{\sin x-x\cos x}{\sin x} \right) \, dx\\ &=\int_0^{\frac{\pi}{2}} \log\left( \sqrt{\frac{\pi}{2}}\frac{x^{3/2}J_{3/2}(x)}{\sin x} \right) \, dx\\ &=-\frac{3\pi}{4}+\pi\ln \pi-\frac{\pi}{4}\ln 2-\frac{\pi}{2}\ln 2 -\int_0^{\frac{\pi}{2}}\log\left( J_{3/2}(x) \right)\,dx \end{align}
In a paper by Dickinson a series expansion is given for the derivative of the logarithm of a Bessel function: \begin{equation} \frac{d}{dx}\log J_\nu(x)=\frac{\nu}{x}-\sum_{n=1}^\infty\sum_{k=0}^\infty \frac{2x^{2k+1}}{j_{\nu,n}^{2k+2}} \end{equation} $j_{\nu,n}$ is the $n$-th zero of the Bessel function of order $\nu$. This result holds provided $0<x<j_{\nu,1}$ (which is the case here). By integrating this result, \begin{equation} \log J_\nu(x)=\nu\log(x)-\sum_{n=1}^\infty\sum_{k=0}^\infty \frac{1}{k+1}\frac{x^{2k+2}}{j_{\nu,n}^{2k+2}}+C \end{equation} The constant is determined by adjusting the result for $x\to0$. As $\log J_\nu(x)= \nu\log x-\nu\log 2-\log\Gamma(\nu+1)+O(x^2)$ \begin{equation} C=-\nu\log 2-\log\Gamma(\nu+1) \end{equation} Then, \begin{equation} \log\left(J_{3/2}(x) \right)=\frac{3}{2}\log x-\frac{3}{2}\log 2-\log\left(\frac{3\sqrt{\pi}}{4}\right)-\sum_{k=0}^\infty \frac{x^{2k+2}}{k+1}\sum_{n=1}^\infty\frac{1}{j_{3/2,n}^{2k+2}} \end{equation} The integral reads thus \begin{equation} I=\frac{\pi}{2}\left( 3\ln\pi-3-\ln(12) \right)-\sum_{k=0}^\infty\frac{(\pi/2)^{2k+3}}{(k+1)(2k+3)}\sum_{n=1}^\infty \frac{1}{j_{3/2,n}^{2k+2}} \end{equation} The sum of inverse even powers of zeros of Bessel functions appears in problems involving the diffusion equation and a broad literature exists on the topic. The quantity \begin{equation} \sigma(p,\nu)=\sum_{n=1}^\infty\frac{1}{j_{\nu,n}^{2p}} \end{equation} is discussed here. A recent analysis can be found in a paper by Jorge L. deLyra. The first values for $\nu=3/2$ are \begin{align} &\sigma(1,3/2)=\frac{1}{10}\;;\;\sigma(2,3/2)=\frac{1}{350}\;;\;\sigma(3,3/2)=\frac{1}{7875}\\ &\sigma(4,3/2)=\frac{37}{6063750}\;;\;\sigma(5,3/2)=\frac{59}{197071875} \end{align} we find \begin{align} I&=\frac{\pi}{2}\left( 3\ln\pi-3-\ln(12) \right)-\sum_{k=0}^\infty\frac{\sigma(k+1,3/2)}{(k+1)(2k+3)}\left(\frac{\pi}{2}\right)^{2k+3}\\ &=\frac{\pi}{2}\left(3\ln\frac{\pi}{e}-\ln(12) \right)-\\ &\hspace{1cm}-\frac{\pi^3}{8}\left(\frac{1}{30}+\frac{(\pi/2)^2}{3500}+ \frac{(\pi/2)^4}{165375}+\frac{37(\pi/2)^6}{218295000}+\frac{59(\pi/2)^8}{10838953125}+\cdots \right) \end{align} which is identical to the series proposed by @ClaudeLeibovici.
As remarked by @JamesArathoon in a comment, it is also possible to perform first IBP on the integral. Then, the problem boils down to integrate $J_{1/2}(x)/J_{3/2}(x)$. By using the recurrence formula of the Bessel function, it requires to integrate $J_{5/2}(x)/J_{3/2}(x)$. A series expansion for the ratio of consecutive Bessel functions is also given in the above cited paper. It also makes use of the $\sigma(k,3/2)$, but the convergence is slower in this case.