Definite Integral from 0 to Half Pi of Logarithm of Sine x/Proof 1
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Theorem
- $\ds \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$
Proof
By Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Lemma, we have:
- $\ds \int_0^\pi \map \ln {\sin x} \rd x = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x$
We also have:
| \(\ds \int_0^{\pi/2} \map \ln {\sin x} \rd x\) | \(=\) | \(\ds \int_0^{\pi/2} \map \ln {\map \sin {\frac \pi 2 - x} } \rd x\) | Integral between Limits is Independent of Direction | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^{\pi/2} \map \ln {\cos x} \rd x\) | Sine of Complement equals Cosine |
giving:
| \(\ds 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x\) | \(=\) | \(\ds \int_0^{\pi/2} \map \ln {\sin x \cos x} \rd x\) | Sum of Logarithms | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^{\pi/2} \map \ln {\frac 1 2 \sin 2 x} \rd x\) | Double Angle Formula for Sine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac \pi 2 \map \ln {\frac 1 2} + \int_0^{\pi/2} \map \ln {\sin 2 x} \rd x\) | Primitive of Constant, Sum of Logarithms | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac \pi 2 \map \ln {\frac 1 2} + \frac 1 2 \int_0^\pi \map \ln {\sin u} \rd u\) | substituting $u = 2 x$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac \pi 2 \ln 2 + \int_0^{\pi/2} \map \ln {\sin u} \rd u\) | Logarithm of Reciprocal |
Therefore:
- $\ds \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$
$\blacksquare$