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Recently I run into this integral

$$\mathcal{J} = \int_{0}^{\pi/2} \frac{x \log \left ( 1-\sin x \right )}{\sin x} \, \mathrm{d}x$$

I don't know to what it evaluates. I tried several approaches.

1st: Differentiation under the integral sign

Consider the function $\displaystyle f(\alpha)= \int_{0}^{\pi/2} \frac{x \log \left ( 1-\alpha\sin x \right )}{\sin x} \, \mathrm{d}x$. Hence

\begin{align*} \frac{\mathrm{d} }{\mathrm{d} \alpha} f(\alpha) &= \frac{\mathrm{d} }{\mathrm{d} \alpha} \int_{0}^{\pi/2} \frac{x \log \left ( 1-\alpha\sin x \right )}{\sin x} \, \mathrm{d}x \\ &= \int_{0}^{\pi/2} \frac{\partial }{\partial \alpha} \frac{x \log \left ( 1-\alpha\sin x \right )}{\sin x} \, \mathrm{d}x \\ &= -\int_{0}^{\pi/2} \frac{x \sin x}{\sin x \left ( 1- \alpha \sin x \right )} \, \mathrm{d}x\\ &=- \int_{0}^{\pi/2} \frac{x}{1- \alpha \sin x} \, \mathrm{d}x \end{align*}

And the last integral equals?

2nd: Taylor series expansion

Lemma: It holds that

$$x \sin^n x = \left\{\begin{matrix} 2^{1-n}\displaystyle\mathop{\sum}\limits_{k=0}^{\frac{n-1}{2}}(-1)^{\frac{n-1}{2}-k}\binom{n}{k}\,x\sin\big((n-2k)x\big) & , & n \;\; \text{odd} \\\\ 2^{-n}\displaystyle\binom{n}{\frac{n}{2}}\,x+2^{1-n}\mathop{\sum}\limits_{k=0}^{\frac{n}{2}-1}(-1)^{\frac{n}{2}-k}\binom{n}{k}\,x\cos\big((n-2k)x\big) & , & n \;\; \text{even} \end{matrix}\right.$$

Hence,

\begin{align*} \int_{0}^{\pi/2} \frac{x \log \left ( 1-\sin x \right )}{\sin x} \, \mathrm{d}x &= -\int_{0}^{\pi/2} \frac{x}{\sin x} \sum_{n=1}^{\infty} \frac{\sin^n x}{n} \, \mathrm{d}x \\ &=-\sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\pi/2} x \sin^{n-1} x \, \mathrm{d}x \end{align*}

However the lemma does not help at all. In fact, if someone substitutes the RHS what it seems to be in there is an $\arcsin $ Taylor expansion. The series that remains to be evaluated is very daunting.

To sum up, I don't know to what this integral evaluates. I don't even know if a nice closed form exists neither do I expect one. But , I still hope.

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  • 1
    $\begingroup$ Nice closed form exists: $\int_0^{\frac{\pi }{2}} \frac{x \log (1-\sin (x))}{\sin (x)} \, dx=-\frac{\pi ^3}{8}$ found by lookup: isc.carma.newcastle.edu.au/advancedCalc $\endgroup$ – Mariusz Iwaniuk Jun 30 '18 at 8:53
  • $\begingroup$ Sorry, try:isc.carma.newcastle.edu.au $\endgroup$ – Mariusz Iwaniuk Jun 30 '18 at 9:00
  • $\begingroup$ @MariuszIwaniuk I'm curious to see where that number comes from. It's interesting. $\endgroup$ – Tolaso Jun 30 '18 at 9:06
  • $\begingroup$ By numeric integration of yours integral.I used CAS. $\endgroup$ – Mariusz Iwaniuk Jun 30 '18 at 9:07
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The given problem is equivalent to the evaluation of $$ \int_{0}^{1}\frac{\arcsin(x)}{\sqrt{1-x^2}}\cdot\frac{\log(1-x)}{x}\,dx =\sum_{n\geq 1}\frac{4^n}{2n\binom{2n}{n}}\int_{0}^{1}x^{2n-2}\log(1-x)\,dx=\sum_{n\geq 1}\frac{4^n H_{2n-1}}{2n\binom{2n}{n}(1-2n)}$$ which is a twisted hypergeometric series. On the other hand $$ \mathcal{J}= 2\int_{0}^{\pi/4}\frac{2x \log(1-\sin(2x))}{\sin(2x)}\,dx=2\int_{0}^{1}\frac{\arctan(t)}{t}\log\left(\frac{(1-t)^2}{1+t^2}\right)\,dt $$ appears to be manageable through the polylogarithms machinery.
Indeed $\arctan t=\text{Im}\log(1+it)$ and the integrals $$ \int \frac{\log(1+it)\log(1\pm it)}{t}\,dt, \qquad \int \frac{\log(1+it)\log(1-t)}{t}\,dt $$ have closed forms in terms of $\text{Li}_2$ and $\text{Li}_3$. However the simplest way to recover $\mathcal{J}=-\frac{\pi^3}{8}$ might be to exploit complex analysis and contour integration, as it often happens when integrating multiples of $\frac{x}{\sin x}$.

Through the Fourier series of $\log\sin$ we have $$ \log(1-\cos x)=-\log(2)-2\sum_{n\geq 1}\frac{\cos(nx)}{n} $$ pointwise on $(0,\pi/2)$. We have that $\int_{0}^{\pi/2}\frac{x}{\sin x}\,dx $ equals $2K$, with $K$ being Catalan's constant, and by induction

$$ \int_{0}^{\pi/2}\frac{x}{\sin x}\cos\left[n\left(\frac{\pi}{2}-x\right)\right]\,dx $$ up to the sign, equals $\sum_{m>n/2}\frac{2(-1)^m}{(2m+1)^2}$ or $\sum_{m> n/2}\frac{1}{(2m+1)^2}$, according to the parity of $n$. This allows to write the original twisted sum in terms of standard Euler sums. $K$ disappears from the outcome after some simplification and $$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3} = \frac{\pi^3}{32} $$ is well-known.

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  • $\begingroup$ (+1) Yes, real analysis seems to stuck in Euler sums and plenty of other twisted things!! Let's see what you or other members can additionally come up with. $\endgroup$ – Tolaso Jun 30 '18 at 20:17

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