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.

  • 4
    $\begingroup$ I suggest you to post this stuff in Mathoverflow. $\endgroup$
    – David
    Jan 17, 2021 at 8:18
  • 5
    $\begingroup$ 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, $a=\frac{\pi}2$. $\endgroup$
    – Mr Pie
    Jul 21, 2021 at 15:04
  • 3
    $\begingroup$ For the record I crossposted this on MathOverflow here. $\endgroup$
    – Zacky
    Oct 4, 2021 at 13:15
  • $\begingroup$ Note that $$\int_0^\pi x\,f(\sin x)\mathrm dx=\frac{\pi}{2}\int_0^\pi f(\sin x)\mathrm dx=\pi\int_0^{\pi/2}f(\sin 2x)\mathrm dx$$ $\endgroup$
    – user730361
    Oct 5, 2021 at 5:56

1 Answer 1


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}

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.