Definite Integral from 0 to Half Pi of Square of Logarithm of Sine x

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Theorem

$\ds \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24}$


Proof

From Fourier Series for $\map \ln {\sin x}$ from $0$ to $\pi$:

$\ds \map \ln {\sin x} = -\ln 2 - \sum_{n \mathop = 1}^\infty \frac {\cos 2 n x} n$

Then, by Parseval's Theorem:

\(\ds \frac 2 \pi \int_0^\pi \paren {\map \ln {\sin x} }^2 \rd x\) \(=\) \(\ds 2 \paren {\ln 2}^2 + \sum_{n = 1}^\infty \frac 1 {n^2}\)
\(\ds \) \(=\) \(\ds 2 \paren {\ln 2}^2 + \frac {\pi^2} 6\) Basel Problem

We then have:

\(\ds \int_0^\pi \paren {\map \ln {\sin x} }^2 \rd x\) \(=\) \(\ds \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x + \int_{\pi/2}^\pi \paren {\map \ln {\sin x} }^2 \rd x\) Sum of Integrals on Adjacent Intervals for Integrable Functions
\(\ds \) \(=\) \(\ds \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x + \int_0^{\pi/2} \paren {\map \ln {\map \sin {\pi - x} } }^2 \rd x\)
\(\ds \) \(=\) \(\ds 2 \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x\) Sine of Supplementary Angle


So:

$\ds \frac 4 \pi \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = 2 \paren {\ln 2}^2 + \frac {\pi^2} 6$

Multiplying by $\dfrac \pi 4$:

$\ds \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24}$

$\blacksquare$


Sources

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