<s>[INST] Evaluating $\int_0^{\pi/2} \log(1 - x \cot x) \, dx\;$ (Is there a closed form?) 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.
 [/INST] Approximating the integral using rational approximants
Using integration by parts, we easily get
$$I = \int_0^{\pi/2} \ln(1 - x\cot x) \mathrm{d} x
= - \frac{\pi^3}{24} + \frac{\pi}{2}\ln 2 - \int_0^{\pi/2} \frac{x^3}{\tan x - x} \mathrm{d} x.$$
$I \approx -3.353337262889472...$
We focus on
$$I_1 = \int_0^{\pi/2} \frac{x^3}{\tan x - x} \mathrm{d} x.$$
We may approximate $I_1$ by rational approximants to $\tan x$ at $0$.

*

*Pade $(5, 4)$ approximant $\tan x \approx \frac{x(x^4 - 105x^2 + 945)}{15x^4 - 420x^2 + 945}
\triangleq g(x)$
We can bound the error by
$$0 \le \frac{x^3}{g(x) - x} - \frac{x^3}{\tan x - x} \le g(\pi/2) \approx 0.000045.$$
We have
$$I_1 \approx \int_0^{\pi/2} \frac{x^3}{g(x) - x} \mathrm{d} x = 
- \frac{5}{112}\pi^3 + \frac{165}{56}\pi - \frac{243}{56}\sqrt{10} 
\operatorname{arctanh} \frac{\pi\sqrt{10}}{30}.$$
Then we have $I \approx -3.353344689$ with relative error $< 0.000003$.
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