Trying to evaluate $\int_{0}^{\pi/2}x^s\cot x\arctan(\cot x) dx$

Question

$$G(s)= \int_{0}^{\pi/2}x^s\cot x\arctan(\cot x)\,\mathrm dx$$

I am interested in evaluating $G(s)$ but I am unable to even make a start! Any help? Thank you

I have try a substitution of $u=\cot x$ but not right. I have try by parts but it is to complicate.

Answer 1

Thanks to Milan Ivanov in the comments, $$\int x^s\cot x\arctan(\cot x)\,\mathrm dx = \frac{\pi}{2}\int x^s\cot x\,\mathrm dx - \int x^{s+1}\cot x\,\mathrm dx$$

Let's do rough calculation on that common form:

$$\int x^n\cot x\,\mathrm dx \\= x^n\smallint \cot x\,\mathrm dx - \int\left(nx^{n-1}\cdot \smallint \cot x\,\mathrm dx\right)\mathrm dx \\= x^n\cdot\ln|\sin x\,| - n\int x^{n-1}\cdot \ln|\sin x\,|\,\mathrm dx$$

Mhh.. now, this $\int x^{n-1}\cdot \ln|\sin x\,|\,\mathrm dx$ is something you should try yourself.

Answer 2

The following is a short derivation of the formula claimed in my comment. It is based on the Fourier series $$-\ln(\sin(x)) = \ln(2) + \sum \limits_{k=1}^\infty \frac{\cos(2kx)}{k} \, , x \in (0,\pi) \, , \tag{1}$$ and the Dirichlet eta function $$ \eta (s) = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^s} = \left(1-2^{1-s}\right) \zeta(s) \, , \, \operatorname{Re}(s) > 0 \, , \tag{2}$$ with the limit $\eta(1) = \ln(2)$ .

We can integrate by parts and use $(1)$ to obtain \begin{align} K_n &\equiv \int \limits_0^{\pi/2} t^n \cot(t) \, \mathrm{d} t = - n \int \limits_0^{\pi/2} t^{n-1} \ln(\sin(t)) \, \mathrm{d} t\\ &= \left(\frac{\pi}{2}\right)^{n} \ln(2) + \sum \limits_{k=1}^\infty \frac{1}{k} n \int \limits_0^{\pi/2} t^{n-1} \cos(2kt)\, \mathrm{d} t \, . \end{align}

It is now useful to distinguish between even and odd $n$ . For $\nu \in \mathbb{N}$ we integrate by parts repeatedly to get \begin{align} K_{2\nu} &= \left(\frac{\pi}{2}\right)^{2 \nu} \ln(2) + \sum \limits_{k=1}^\infty \frac{1}{k} \frac{(2\nu)(2 \nu -1)}{4 k^2}\left[ (-1)^k \left( \frac{\pi}{2}\right)^{2\nu-2} - (2 \nu-2) \int \limits_0^{\pi/2} t^{2 \nu-3} \cos(2kt) \, \mathrm{d} t \right] \\ &= \frac{(2\nu)!}{2^{2 \nu}} \left[\frac{\pi^{2\nu}}{(2\nu)!} \eta(1) - \frac{\pi^{2(\nu-1)}}{(2(\nu-1))!} \eta(3) \right] - \frac{(2 \nu)!}{2^2 (2 \nu -3)!} \sum \limits_{k=1}^\infty \frac{1}{k^3} \int \limits_0^{\pi/2} t^{2\nu-3} \cos(2 k t) \, \mathrm{d} t \\ &= \frac{(2\nu)!}{2^{2 \nu}} \left[\frac{\pi^{2\nu}}{(2\nu)!} \eta(1) - \frac{\pi^{2(\nu-1)}}{(2(\nu-1))!} \eta(3) + \frac{\pi^{2(\nu-2)}}{(2(\nu-2))!} \eta(5) \right] \\ &\phantom{=}+ \frac{(2 \nu)!}{2^4 (2 \nu -5)!} \sum \limits_{k=1}^\infty \frac{1}{k^5} \int \limits_0^{\pi/2} t^{2\nu-5} \cos(2 k t) \, \mathrm{d} t \\ &= \dots \\ &= \frac{(2\nu)!}{2^{2 \nu}} \sum \limits_{l=0}^{\nu-1} (-1)^l \frac{\pi^{2(\nu-l)}}{(2(\nu-l))!} \eta(2l+1) + (-1)^{\nu-1} \frac{(2 \nu)!}{2^{2(\nu-1)}} \sum \limits_{k=1}^\infty \frac{1}{k^{2\nu-1}} \int \limits_0^{\pi/2} t \cos(2 k t) \, \mathrm{d} t \\ &= \frac{(2\nu)!}{2^{2 \nu}} \sum \limits_{l=0}^{\nu-1} (-1)^l \frac{\pi^{2(\nu-l)}}{(2(\nu-l))!} \eta(2l+1) + (-1)^\nu \frac{(2 \nu)!}{2^{2\nu}} \sum \limits_{k=1}^\infty \frac{1 + (-1)^{k-1}}{k^{2\nu+1}} \\ &= \frac{(2\nu)!}{2^{2 \nu}} \left[ \sum \limits_{l=0}^{\nu} (-1)^l \frac{\pi^{2(\nu-l)}}{(2(\nu-l))!} \eta(2l+1) + (-1)^\nu \zeta(2\nu+1)\right] \, . \end{align} Similarly, \begin{align} K_{2\nu-1} &= \frac{(2\nu-1)!}{2^{2 \nu-1}} \sum \limits_{l=0}^{\nu-1} (-1)^l \frac{\pi^{2(\nu-l)-1}}{(2(\nu-l)-1)!} \eta(2l+1) + (-1)^{\nu-1} \frac{(2 \nu-1)!}{2^{2\nu-3}} \sum \limits_{k=1}^\infty \frac{1}{k^{2\nu-2}} \int \limits_0^{\pi/2} \cos(2 k t) \, \mathrm{d} t \\ &= \frac{(2\nu-1)!}{2^{2 \nu-1}} \sum \limits_{l=0}^{\nu-1} (-1)^l \frac{\pi^{2(\nu-l)-1}}{(2(\nu-l)-1)!} \eta(2l+1) \, , \end{align} but here the final integral vanishes and there is no extra term.

Milen Ivanov's hint from the comments now yields $$G(n) = \frac{\pi}{2} K_n - K_{n+1}$$ for $n \in \mathbb{N}$ . Note that the term containing $\eta(1) = \ln(2)$ always cancels, so by the above calculation $G(n)$ is given by a sum of powers of $\pi$ times zeta values at the odd integers between $3$ and $2\left\lfloor\frac{n+1}{2}\right\rfloor+1$ with rational coefficients.