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 Post subject: Logrithmic IntegralPosted: Sat May 12, 2018 7:35 am

Joined: Thu Nov 12, 2015 5:26 pm
Posts: 102
$$\int^{\pi}_{0}x^2\ln(\sin x)dx$$

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 Post subject: Re: Logrithmic IntegralPosted: Sat May 12, 2018 1:43 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 457
Location: Ioannina, Greece
Using that the Fourier series of $\log(\sin{x})$ on $(0,\pi)$ is(*) \begin{align}
\log(\sin{x})=-\log2-\mathop{\sum}\limits_{n=1}^{\infty}\frac{\cos(2nx)}{n}
\end{align} then \begin{align*}
\int_{0}^{\pi}x^2\log(\sin{x})\,dx&\stackrel{(1)}{=}\int_{0}^{\pi}\Big(-x^2\log2-\mathop{\sum}\limits_{n=1}^{\infty}\frac{x^2\cos(2nx)}{n}\Big)\,dx\\
&=-\log2\int_{0}^{\pi}x^2\,dx-\int_{0}^{\pi}\mathop{\sum}\limits_{n=1}^{\infty}\frac{x^2\cos(2nx)}{n}\,dx\\
&=-\frac{\pi^3\log2}{3}-\mathop{\sum}\limits_{n=1}^{\infty}\Big(\frac{1}{n}\int_{0}^{\pi}x^2\cos(2nx)\,dx\Big)\\
&=-\frac{\pi^3\log2}{3}-\mathop{\sum}\limits_{n=1}^{\infty}\Big(\frac{1}{n}\frac{\pi}{2n^2}\Big)\\
&=-\frac{\pi^3\log2}{3}-\frac{\pi}{2}\mathop{\sum}\limits_{n=1}^{\infty}\frac{1}{n^3}\\
&=-\frac{\pi^3\log2}{3}-\frac{\pi}{2}\,\zeta(3)\,.\end{align*}

(*) which is not easy to prove! So, I hope that someone else will give a simpler solution.

_________________
Grigorios Kostakos

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 Post subject: Re: Logrithmic IntegralPosted: Sat May 12, 2018 4:31 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 157
Location: Melbourne, Australia
The Fourier series is the way to go here. Recall that

$$\cos(2kx) = \dfrac{e^{i2kx} + e^{-i2kx}}{2}$$

Hence,

\begin{align*}
\sum_{k=1}^{\infty} \dfrac{\cos(2kx)}k
&= \sum_{k=1}^{\infty} \dfrac{e^{i2kx} + e^{-i2kx}}{2k}
\\&= \dfrac12 \big(-\log (1-e^{i2x} )-\log (1-e^{-i2x} ) \big)
\\&= - \dfrac12 \log \big(2 - 2\cos(2x) \big)
\\&= - \dfrac12 \log\big(4 \sin^2(x)\big)
\\&= - \log 2 - \log\big(\sin(x)\big)
\end{align*}

and the result follows. The rest is history.

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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 Post subject: Re: Logrithmic IntegralPosted: Thu May 24, 2018 6:44 am

Joined: Thu Nov 12, 2015 5:26 pm
Posts: 102
Thanks kostakos and Riemann

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