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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Logarithmic Functions: $15.103$