TITLE: Evaluating $\int_0^{\pi/2} \log(1 - x \cot x) \, dx\;$ (Is there a closed form?)
QUESTION [12 upvotes]: 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.

REPLY [2 votes]: This is a long comment on an intriguing approximation to the integral most easily accessible by CAS calculation.
Taking the integral partition
$$I=-\int_0^{\frac{\pi }{2}} \log (\cos (x)) \, dx-\int_0^{\frac{\pi }{2}} \frac{x^2 \sin (x)}{\sin (x)-x \cos (x)} \, dx$$
we work with the second integral.
First note that $\frac{d }{dx}(\sin (x)-x \cos (x))=x \sin (x)$. Using the substitution $y=(\sin (x)-x \cos (x))$ we are left to compute the inverse trigonometric function $x= arc[\sin x-x\cos x](y)$, since $dy=x \sin x \,dx$ and $y=(\sin (x)-x \cos (x))$ is a one-to-one function between $0$ and $\pi/2$.
(I write $x= arc[\sin x-x\cos x](y)$ for want of better terminology).
Therefore
$$I=-\int_0^1 \frac{arc[\sin x-x\cos x](y)}{y} \,dy$$
Using CAS to determine $arc[\sin x-x\cos x](y)$ in series form, we have
$$arc[\sin x-x\cos x](y)=\sqrt[3]{3} \sqrt[3]{y}+\frac{y}{10}+\frac{41 y^{5/3}}{1400 \sqrt[3]{3}}+\frac{97 y^{7/3}}{8400\ 3^{2/3}}+\frac{15063 y^3}{8624000}+\frac{6199421 y^{11/3}}{7207200000 \sqrt[3]{3}}+\frac{1894073471 y^{13/3}}{4237833600000\ 3^{2/3}}+\frac{1532283399 y^5}{19059040000000}+\frac{11288018107137217 y^{17/3}}{252957982717440000000 \sqrt[3]{3}}+\frac{1496852049518257 y^{19/3}}{59133034920960000000\ 3^{2/3}}+\frac{31831777776715674057 y^7}{6536309355896320000000000}+...$$
with the general term of the term by term integral being
$$-a_n\int_0^1 \frac{y^{(2n-1)/3}}{y} \,dy=-\frac{3\, a_n}{2n-1}$$
and with the general form of the resulting approximate sum being
$$I\approx a+3^{-1/3}b+3^{-2/3}c$$
where $a,b,c$ are rational, with $a\approx-0.10059903$, $b\approx-3.0016854$ and $c\approx-0.0059380965$
Obviously in the summation limit as $n \to \infty$ $a,b$ and $c$ may no longer be rational.