$\int_{0}^{\pi}x\ln \left ( \sin x \right )\, {\rm d}x$

Calculus (Integrals, Series)
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Tolaso J Kos
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$\int_{0}^{\pi}x\ln \left ( \sin x \right )\, {\rm d}x$

#1

Post by Tolaso J Kos »

Perhaps a well known integral, but anyway.

Evaluate the integral:

$$\int_{0}^{\pi}x\ln \left ( \sin x \right )\, {\rm d}x$$
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Papapetros Vaggelis
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Re: $\int_{0}^{\pi}x\ln \left ( \sin x \right )\, {\rm d}x$

#2

Post by Papapetros Vaggelis »

First of all , \(\displaystyle{I=\int_{0}^{\pi}x\,\ln\,\left(\sin\,x\right)\,\rm{dx}<\infty}\) because,

$$\begin{aligned}\left|\int_{0}^{\pi}x\,\ln\,\left(\sin\,x\right)\,\rm{dx}\right|&\leq \int_{0}^{\pi}x\,\left|\ln\,\left(\sin\,x\right)\right|\,\rm{dx}\leq \pi\,\int_{0}^{\pi}\left|\ln\,\left(\sin\,x\right)\right|\,\rm{dx}\\&=-\pi\,\int_{0}^{\pi}\ln\,\left(\sin\,x\right)\,dx\\&=\pi^2\,\ln\,2\end{aligned}$$

By substituting \(\displaystyle{x=\pi-t}\) , we have that \(\displaystyle{\rm{dx}=-\rm{dt}\,,t\in\left[0,\pi\right]}\) and

$$\begin{aligned} I&=-\int_{\pi}^{0}\left(\pi-t\right)\,\ln\,\left(\sin\,\left(\pi-t\right)\right)\,\rm{dt}\\&=\int_{0}^{\pi}\left(\pi-t\right)\,\ln\,\left(\sin\,t\right)\,\rm{dt}\\&=\pi\,\int_{0}^{\pi}\ln\,\left(\sin\,t\right)\,\rm{dt}-I\end{aligned}$$

so, \(\displaystyle{I=-\pi^2\,\ln\,2-I\iff 2\,I=-\pi^2\,\ln\,2\iff I=-\dfrac{\pi^2\,\ln\,2}{2}}\) .
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