Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I used all I know to show that $$\int_0^\pi x\ln(\sin x)dx=-\ln(2) \pi^2/2$$ This is my homework but don't know where to start. I appreciate your help.

share|cite|improve this question
Nice question (+1) – OFFSHARING Sep 12 '12 at 19:42

3 Answers 3

up vote 7 down vote accepted

Let $I = \int_0^\pi x \ln(\sin x) \mathrm{d} x$. Then, changing variables $x \to \pi -x$: $$ I = \int_0^\pi \left(\pi -x \right) \ln( \sin x) \mathrm{d} x = \pi \int_0^\pi \ln(\sin x) \mathrm{d} x - I $$ Therefore: $$ I = \frac{\pi}{2} \int_0^\pi \ln(\sin x)\mathrm{d} x \stackrel{\text{symmetry}}{=} \pi \int_0^{\pi/2} \ln(\sin x) \mathrm{d} x $$ The latter integral had been solved elsewhere.

share|cite|improve this answer
Thank you sasha. – Nancy Rutkowskie Sep 12 '12 at 18:50
@Nancy R: for a fast, simple solution you may also use the variable change $x = 2u$ at the last integral from Sasha's work. – OFFSHARING Sep 12 '12 at 19:39
@Chris'ssister This is just simple and beautiful, simply beautiful! You ought to write it up. – Sasha Sep 12 '12 at 19:49

This answer is meant to offer an alternative from the last integral of Sasha's work

By variable change $x = 2u$ we have that $$I=\int_{0}^{\frac{\pi}{2}} \ln(\sin (x)) \ dx=$$ $$2\int_{0}^{\frac{\pi}{4}} \ln(\sin (2u)) \ du=$$ $$2\left(\int_{0}^{\frac{\pi}{4}} \ln (2) \ du + \int_{0}^{\frac{\pi}{4}} \ln(\sin (u)) \ du +\int_{0}^{\frac{\pi}{4}} \ln(\cos (u)) \ du \right)=$$ $$\frac{\pi}{2} \ln(2) + 2\left(\int_{0}^{\frac{\pi}{4}} \ln(\sin (u)) \ du+\int_{0}^{\frac{\pi}{4}} \ln(\cos (u)) \ du\right)$$ that may be rewritten as

$$I=\frac{\pi}{2} \ln(2)+2I$$ thus $$I=-\frac{\pi}{2} \ln(2)$$ Then the final result is $\displaystyle -\frac{\pi^2}{2} \ln(2).$

share|cite|improve this answer
Or we could start from the fact that $I=\int_{0}^{\frac{\pi}{2}} \ln(\sin (x)) \ dx= \int_{0}^{\frac{\pi}{2}} \ln(\cos (x)) \ dx$. This could be another starting point just crossed my mind. Oh, this is even faster! – OFFSHARING Sep 12 '12 at 20:59

For what is worth, the general case of what Sasha did is

If $$I=\int_0^\pi xf(\sin x )dx$$


$$I=\frac \pi2\int_0^\pi f(\sin x )dx$$

and the argument is completely analogous.

share|cite|improve this answer
By the variable change $u=\pi-x$ – OFFSHARING Sep 12 '12 at 19:32
@Chris'ssister Yes, that is what Sasha did. – Pedro Tamaroff Sep 12 '12 at 19:34
It's helpful this formula. I'm glad to see it here. (+1) – OFFSHARING Sep 12 '12 at 19:35
Thank you very much @Chris's sister. ;-) – Nancy Rutkowskie Sep 13 '12 at 8:55
@Nancy R: welcome! :D – OFFSHARING Sep 13 '12 at 8:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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