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I would like to see how you evaluate $$S(n)=\int_0^{\pi/2} \log(\sin x)^n\mathrm dx,\qquad n\in\Bbb N_0$$ Here's how I do it.

Start with $$\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx$$ Use $t=\sin(x)^2$ to see that $$\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx=\frac12\int_0^1t^{\frac{a-1}2}(1-t)^{\frac{b-1}2}\mathrm dt=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ So $$\int_0^{\pi/2}\sin(x)^{2a}\mathrm dx=\frac{\sqrt\pi}2\frac{\Gamma(a+\frac12)}{\Gamma(a+1)}$$ Taking $\left(\frac{d}{da}\right)^n$ on both sides $$2^{n}\int_0^{\pi/2}\sin(x)^{2a}\log(\sin x)^n\mathrm dx=\frac{\sqrt\pi}2\left(\frac{d}{da}\right)^n\,\frac{\Gamma(a+\frac12)}{\Gamma(a+1)}$$ And evaluating at $a=0$, $$S(n)=\frac{\sqrt{\pi}}{2^{n+1}}\left[\left(\frac{d}{da}\right)^n\,\frac{\Gamma(a+\frac12)}{\Gamma(a+1)}\right]_{a=0}$$ Which is a closed form.

Enjoy!

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    $\begingroup$ I love the use of $f(x)^a$ in solving $ln(f(x))^a$! I've used it when $f(x) = x$ but never thought to extend. Excellent stuff here! $\endgroup$ – user150203 Feb 6 at 5:31
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Let $$S_n = \int_0^{\frac{\pi}{2}} \log^n (\sin x) \, dx.$$ Setting $t = \sin x$ one has $$S_n = \int_0^1 \frac{\log^n t}{\sqrt{1 - t^2}} \, dt.\tag1$$ Making use of the generalised binomial expansion, we have: $$\frac{1}{\sqrt{1 - t^2}} = \sum_{k = 0}^\infty \binom{-\frac{1}{2}}{k} t^{2k} = \sum_{k = 0}^\infty \binom{2k}{k} \frac{t^{2k}}{2^{2k}}, \qquad |t| < 1.$$ The integral in (1) can thus be rewritten as $$S_n = \sum_{k = 0}^\infty \binom{2k}{k} \frac{1}{2^{2k}} \int_0^1 t^{2k} \log^n t \, dt, \tag2$$ where the summation and the integration have been interchanged.

Since $n$ is a positive integer, by integrating by parts $n$ times it can be readily shown that $$\int_0^1 t^{2k} \log^n t \, dt = \frac{(-1)^n n!}{(2k + 1)^{n + 1}}.$$

So (2) becomes $$S_n = (-1)^n n! \sum_{k = 0}^\infty \binom{2k}{k} \frac{1}{4^{k} (2k + 1)^{n + 1}}\tag3$$ which is an infinite sum. (3) however can be written in "closed form" in terms of a generalised hypergeometric function. Observing that $$\sum_{k = 0}^\infty \binom{2k}{k} \frac{x^k}{(2k + 1)^{n + 1}}, = \, _{n+2}F_{n + 1} \left (\frac{1}{2},\ldots ,\frac{1}{2}; \frac{3}{2}, \ldots ,\frac{3}{2}; 4x \right ), \qquad |x| < \frac{1}{4},$$ setting $x = 1/4$, in terms of this function the expression for $S_n$ in (3) becomes $$S_n = (-1)^n n! \, _{n+2}F_{n + 1} \left (\frac{1}{2},\ldots ,\frac{1}{2}; \frac{3}{2}, \ldots ,\frac{3}{2}; 1 \right ).$$

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  • $\begingroup$ Really cool! (+1) that's a really nice series representation $\endgroup$ – clathratus Feb 6 at 4:31
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Enforcing the substitution $x\mapsto \arcsin(x)$ reveals

$$\begin{align} S(n)&=\int_0^{\pi/2}\log^n(\sin(x))\,dx\\\\ &=\int_0^1 \frac{\log^n(x)}{\sqrt{1-x^2}}\,dx\\\\ &=\left.\left(\frac{d^n}{dy^n}\int_0^1 \frac{x^y}{\sqrt{1-x^2}}\,dx\right)\right|_{y=0}\\\\ &=\frac12\left.\left(\frac{d^n}{dy^n}\int_0^1 x^{(y-1)/2}(1-x)^{-1/2}\,dx\right)\right|_{y=0}\\\\ &=\frac12\left.\left(\frac{d^n}{dy^n}B\left(\frac{y+1}{2},\frac12\right)\right)\right|_{y=0}\\\\ &=\frac{\Gamma(1/2)}2\left.\left(\frac{d^n}{dy^n}\left(\frac{\Gamma\left(\frac{y+1}{2}\right)}{\Gamma\left(\frac{y}{2}+1\right)}\right)\right)\right|_{y=0} \end{align}$$

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