Prove without evaluating the integrals that:$$2\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x}dx=\int_\frac{\pi}{2}^\pi\frac{x\ln(1-\sin x)}{\sin x}dx\label{*}\tag{*}$$

Or equivalently: $$\boxed{\int_0^\pi\frac{x\ln(1-\sin x)}{\sin x}dx=3\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x}dx}$$ In contrast we have: $$\boxed{\int_0^\pi\frac{\ln(1-\sin x)}{\sin x}dx=2\int_0^\frac{\pi}{2}\frac{\ln(1-\sin x)}{\sin x}dx}$$ This is of course easily provable by splitting the integral as $\int_0^\frac{\pi}{2}+\int_\frac{\pi}{2}^\pi$ and letting $x\to \pi-x$ in the second part, unfortunately this method doesn't work for the other one.

I am already aware how to evaluate the integrals as we have: $$\mathcal I= \int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x} dx\overset{\tan \frac{x}{2}\to x}=-2\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=-\frac{\pi^3}{8}$$ And the latter integral is evaluated in many ways here, so if you have other approaches please add them there.

Here's how I came up with $\eqref{*}$:
I knew from here that: $$I\left(\frac{3\pi}{2}\right)=\int_0^\frac{\pi}{2}\frac{\ln(1-\sin x)}{\sin x}dx=-\frac{3\pi^2}{8}$$ And since this result is very similar to the one from above, I tried to show that $\mathcal I=\frac{\pi}{3} I\left(\frac{3\pi}{2}\right)$, equivalent to: $$\boxed{\int_0^\frac{\pi}{2}\left(\frac{\pi}{3}-x\right)\frac{\ln(1-\sin x)}{\sin x}dx=0}$$
I also noticed that we have: $$\mathcal J=\int_\frac{\pi}{2}^\pi\frac{x\ln(1-\sin x)}{\sin x}dx\overset{x\to \pi-x}=\int_0^\frac{\pi}{2}\frac{(\pi-x)\ln(1-\sin x)}{\sin x}dx=\pi I\left(\frac{3\pi}{2}\right)-\mathcal I$$ $$\Rightarrow \mathcal I+\mathcal J=\int_0^\pi \frac{x\ln(1-\sin x)}{\sin x}dx=\pi I\left(\frac{3\pi}{2}\right)=-\frac{3\pi^3}{8}$$ Of course now it's trivial to deduce that $2\mathcal I=\mathcal J$ as we know the result for $\mathcal I$, but I'm interested to show that relationship without making use of the result or by calculating any of the integrals. If possible showing $\eqref{*}$ using only integral manipulation (elementary tools such as substitution/integration by parts etc). I hope there's a nice slick way to do it as it will give an easy evaluation of the main integral.

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    I suggest you to post this stuff in Mathoverflow. – David Jan 17 '21 at 08:18
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    Just want to point out btw, $$\int_0^\pi\frac{\ln(1-\sin x)}{\sin x}dx=2\int_0^\frac{\pi}{2}\frac{\ln(1-\sin x)}{\sin x}\,\mathrm dx$$ by the fact that $$\int_0^{2a}f(x)\,\mathrm dx = 2\int_0^a f(x)\,\mathrm dx$$ for any $a$ iff $f(2a-x)=f(x)$. Here, $a=\frac{\pi}2$. – Mr Pie Jul 21 '21 at 15:04
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    For the record I crossposted this on MathOverflow [here](https://mathoverflow.net/q/403395/145342). – Zacky Oct 04 '21 at 13:15
  • Note that $$\int_0^\pi x\,f(\sin x)\mathrm dx=\frac{\pi}{2}\int_0^\pi f(\sin x)\mathrm dx=\pi\int_0^{\pi/2}f(\sin 2x)\mathrm dx$$ –  Oct 05 '21 at 05:56

1 Answers1


It suffices to show the vanishing integral below \begin{align}I=& \int^\frac{\pi}{2}_0\frac{(3x-\pi)\ln(1-\sin x)}{\sin x}dx\\ =& \int^\frac{\pi}{2}_0\int^\frac{\pi}{2}_0 \frac{(\pi-3x)\cos y}{1-\sin y \sin x}dy\>dx\\ =& \int^\frac{\pi}{2}_0\int^\frac{\pi}{2}_0 (\pi-3x)\frac{d}{dx} \left(2\tan^{-1}\frac{\sin\frac{x-y}2}{\cos\frac{x+y}2} \right) \overset{ibp}{dx}\> dy\\ =& \int^\frac{\pi}{2}_0\int^\frac{\pi}{2}_0 6\tan^{-1}\frac{\sin\frac{x-y}2}{\cos\frac{x+y}2}\>\overset{x \leftrightarrows y}{dxdy}=-I=0 \end{align}

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  • Awesome! I think you have two sign mistakes (which cancel out and doesn't influence anything anyway). $$\frac{\ln(1-\sin x)}{\sin x}dx = -\int_0^\frac{\pi}{2} \frac{\cos y}{1-\sin y \sin x}dy$$ $$\frac{d}{dx}\left(-2\arctan\left(\frac{\sin((y-x)/2}{\cos((y+x)/2)}\right)\right)=- \frac{\cos y}{1-\sin y \sin x}$$ – Zacky Apr 21 '22 at 19:40
  • @Zacky - I’ll fix it; thanks for spoting it – Quanto Apr 21 '22 at 20:01