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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.

mick
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Nancy Rutkowskie
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3 Answers3

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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.

Community
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Sasha
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    @Nancy R: for a fast, simple solution you may also use the variable change $x = 2u$ at the last integral from Sasha's work. – user 1591719 Sep 12 '12 at 19:39
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    @Chris'ssister This is just simple and beautiful, simply beautiful! You ought to write it up. – Sasha Sep 12 '12 at 19:49
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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 $\displaystyle -\frac{\pi^2}{2} \ln(2).$

user 1591719
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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! – user 1591719 Sep 12 '12 at 20:59
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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.

Pedro
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