Let $n \in \mathbb{N}$. May we have a closed form for the integral:

$$\mathcal{J}=\int_0^{\pi/2} \ln^n (\sin x) \, {\rm d}x$$

One obvious approach would be to go through beta functions and differentiate $n$ times but that sounds very tedious. My guess is, that is doable but a long shot.

Well, a relation that might help is the following:

\begin{equation} \int_0^{\pi/2} (\log \cos x+ \log 2 + ix)^n \, {\rm d}x + \int_0^{\pi/2} (\log \cos x + \log 2 - ix)^n \, {\rm d}x=0 \end{equation}

To prove this just consider the function $f(z)=\log^n z$ that is analytic in the open disk $\mathbb{D}=\{z \in \mathbb{C} \mid |z-1|<1\}$ and just apply the Gauss Mean Value Theorem. Now, since $f(1)=0$ the result will follow because of:

\begin{align*} 0 &=\int_{0}^{2\pi} \log^n \left ( 1+\cos x +i \sin x \right )\, {\rm d}x \\ &= \int_{0}^{2\pi} \log^n \left ( 2\cos \frac{x}{2} e^{ix/2} \right )\, {\rm d}x\\ &\overset{u=x/2}{=\! =\! =\! =\!}\int_{0}^{\pi} \log^n \left ( 2\cos y \cdot e^{iy} \right )\, {\rm d}y \\ &= \int_{0}^{\pi/2} \log^n \left ( 2\cos x \cdot e^{iy} \right ) \, {\rm d}y + \int_{\pi/2}^{\pi} \log^n \left ( 2\cos y \cdot e^{iy} \right )\, {\rm d}y \\ &\overset{y =\pi-x }{=\! =\! =\! =\!} \int_{0}^{\pi/2}\log^n \left ( 2\cos x \cdot e^{ix} \right )\, {\rm d}y + \int_{0}^{\pi/2}\log^n \left ( -2 \cos x \cdot e^{i(\pi-x)} \right ) \, {\rm d}x \\ &= \int_{0}^{\pi/2}\log^n \left ( 2\cos x \cdot e^{ix} \right )\, {\rm d}x + \int_0^{\pi/2}\log^n \left ( 2\cos x \cdot e^{-ix} \right )\, {\rm d}x \end{align*}

Now, if we were to expand using binomial theorem (do not do it) we were to get the value of the requested integral but expansion seems rather tedious again.

Any other ideas?


There is a closed form of this integral:

$$\int_0^{\pi/2} (\log \sin x)^n\text{ d}x=\frac{1}{2^{n+1}}B^{(n)}\left(\frac{1}{2},\frac{1}{2}\right)$$


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