# Verifing $\int_0^{\pi}x\ln(\sin x)\,dx=-\ln(2){\pi}^2/2$

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.

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Nice question (+1) –  Chris's sis Sep 12 '12 at 19:42

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.

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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. –  Chris's sis 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 $\displaystyle -\frac{\pi^2}{2} \ln(2).$

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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! –  Chris's sis Sep 12 '12 at 20:59

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

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

then

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

and the argument is completely analogous.

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By the variable change $u=\pi-x$ –  Chris's sis 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) –  Chris's sis 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 –  Chris's sis Sep 13 '12 at 8:56