Definite Integral from 0 to Pi of x by Logarithm of Sine x
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Theorem
- $\ds \int_0^\pi x \map \ln {\sin x} \rd x = -\frac {\pi^2} 2 \ln 2$
Proof
\(\ds \int_0^\pi x \map \ln {\sin x} \rd x\) | \(=\) | \(\ds \int_0^\pi \paren {\pi - x} \map \ln {\map \sin {\pi - x} } \rd x\) | Integral between Limits is Independent of Direction | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \int_0^\pi \map \ln {\sin x} - \int_0^\pi x \map \ln {\sin x} \rd x\) | Sine of Supplementary Angle, Linear Combination of Definite Integrals |
So:
\(\ds 2 \int_0^\pi x \map \ln {\sin x} \rd x\) | \(=\) | \(\ds \pi \int_0^\pi \map \ln {\sin x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \pi \int_0^{\pi/2} \map \ln {\sin x} \rd x\) | Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Lemma | |||||||||||
\(\ds \) | \(=\) | \(\ds -\pi^2 \ln 2\) | Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$ |
giving:
- $\ds \int_0^\pi x \map \ln {\sin x} \rd x = -\frac {\pi^2} 2 \ln 2$
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Logarithmic Functions: $15.104$