Closed form of $\int_0^{\pi/2}x^{n}\ln{(\sin x)}dx $

Question

Does a closed form of the above integral exists? $$\displaystyle \int_0^{\pi/2}x^{n}\ln{(\sin x)}dx $$

where $n$ is a positive integer.

Answer 1

I want to add a proof which uses complex analysis:

First observe that our integral can be split as follows:

$$ I_n=-\underbrace{\int_0^{\pi/2}dx\left(ix^{n+1}+\log(\frac{1}{2i})x^n\right)}_{A_n}+\underbrace{\int_0^{\pi/2}dxx^n\log(1-e^{-2ix})}_{B_n} $$

The first part is trivial and yields $A_n=\log(\frac{1}{2i})\frac{x^{n+1}}{n+1}+i\frac{x^{n+2}}{n+2}$+

To calculate $B_n$, define the complex valued function

$$ f(z)=\log(1-e^{2z}) $$

Now we integrate this function over a rectangle in the complex plane with vertices $\{0,\frac{\pi}{2},\frac{\pi}{2}-iR,-iR\}$

Because $f(z)$ is analytic in the chosen domain, the contour integral yields zero (Note that we have taken the limit $R\rightarrow\infty$,where the upper vertical line vanishs)

$$ \oint f(z)=\underbrace{\int_0^{\pi/2}dxx^n\log(1-e^{-2ix})}_{B_n}+i\underbrace{\int_0^{\infty}dy(\frac{\pi}{2}-iy)^n\log(1+e^{-2y})}_{K_n}+i\underbrace{\int_0^{\infty}dy(-iy)^n\log(1-e^{-2y})}_{J_n}=0 $$

$K_n$ and $J_n$ are now straightforwardly calculated by using the Taylor expansion of $\log(1+z)$ and the binomial theorem

$$ K_n=\int_0^{\infty}dy(\frac{\pi}{2}-iy)^n\log(1+e^{-2y})=\sum_{k=1}^n\binom{n}{k}\left(\frac{\pi}{2}\right)^{n-k}(-i)^k\int_0^{\infty}dy\sum_{q=1}^{\infty}(-1)^{q+1}\frac{y^ke^{-2yq}}{q}=\\ \sum_{k=0}^n\binom{n}{k}\left(\frac{\pi}{2}\right)^{n-k}(-i)^k\frac{k!}{2^{k+1}}\sum_{q=1}^{\infty}\frac{(-1)^{q+1}}{q^{k+2}}=\\ \sum_{k=0}^n\binom{n}{k}\left(\frac{\pi}{2}\right)^{n-k}(-i)^k\frac{k!}{2^{k+2}}\left(1-\frac{1}{2^{k+1}}\right)\zeta(k+2) $$

by the same technique $$ J_n=\int_0^{\infty}dy(-iy)^n\log(1+e^{-2y})=(-i)^n \frac{n!}{2^{n+2}} \zeta(n+2) $$

yielding:

$$ I_n=B_n+A_n=J_n+K_n+A_n=\\ (-i)^{n+1} \frac{n!}{2^{n+1}} \zeta(n+2)+\sum_{k=0}^n\binom{n}{k}\left(\frac{\pi}{2}\right)^{n-k}(-i)^{k+1}\frac{k!}{2^{k+2}}\left(1-\frac{1}{2^{k+1}}\right)\zeta(k+2)\\ -i\frac{\pi^{n+2}}{2^{n+2}}\frac{2n+1}{(n+2)(n+1)}-\log(2)\frac{\pi^{n+1}}{2^{n+1}} $$

Please note, that the imaginary part of the right hand side has to be zero (because our integral is real), which gives us a disturbing summation formula:

$$ \Im\left[\sum_{k=0}^n\binom{n}{k}\left(\frac{\pi}{2}\right)^{n-k}(-i)^{k+1}\frac{k!}{2^{k+2}}\left(1-\frac{1}{2^{k+1}}\right)\zeta(k+2)\right]=\\ (-1)^{n} \frac{n!}{2^{n+2}} \zeta(n+1)\delta_{n+1,2j}+\frac{\pi^{n+2}}{2^{n+2}}\frac{2n+1}{(n+2)(n+1)} $$

Answer 2

Using integration by parts the problem boils down to finding a closed form for: $$ I_n = \int_{0}^{\pi/2} x^n\cot(x)\,dx. \tag{1}$$ By considering the logarithmic derivative of the Weierstrass product for the sine function we have: $$ \cot(x) = \frac{1}{x}+\sum_{n\geq 1}\frac{2x}{x^2-\pi^2 n^2}=\frac{1}{x}+\sum_{n\geq 1}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right) \tag{2}$$ and: $$ \frac{1-\pi x\cot(\pi x)}{2}=\sum_{n\geq 1}\zeta(2n)\,x^{2n}\tag{3} $$ hence: $$ I_n = \pi^{n}\int_{0}^{1/2}x^{n-1}\cdot \pi x\cot(\pi x)\,dx =\pi^n\left(\frac{1}{n 2^n}-2\sum_{n\geq 1}\frac{\zeta(2n)}{3n 8^n}\right).\tag{4}$$ On the other hand, by setting $z=e^{ix}$ in $(1)$ we may see that $I_n$ is given by a countour integral: $$ I_n = \int_{C} (-i\log(z))^n \frac{z^2+1}{z^3-z}\,dz \tag{5}$$ or simply by: $$ I_n = \int_{0}^{+\infty}\left(\frac{\pi}{2}-\arctan(u)\right)^n \frac{u}{1+u^2}\,dx = \int_{0}^{+\infty}\frac{\left(\arctan u\right)^n}{u(1+u^2)}\,du\tag{6}$$ so if we apply partial fraction decomposition and integration by parts again, we may see that $I_n$ is given by a finite combination of $\pi^n \log(2)$ and values of the zeta function at the odd integers multiplied by powers of $\pi$.

Answer 3

Sketch of the proof: Using the Fourier series $$\log\left(\sin\left(x\right)\right)=-\log\left(2\right)-\sum_{k\geq1}\frac{\cos\left(2kx\right)}{k} $$ we have $$\int_{0}^{\pi/2}x^{n}\log\left(\sin\left(x\right)\right)dx=-\frac{\log\left(2\right)}{n+1}\left(\frac{\pi}{2}\right)^{n+1}-\sum_{k\geq1}\frac{1}{k}\int_{0}^{\pi/2}x^{n}\cos\left(2kx\right)dx $$ and the second integral can be computated using integration by parts. You will get combinations of zeta function multiplied by powers of $\pi/2 $.