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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?

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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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