Definite Integral from 0 to Half Pi of Logarithm of Sine x/Proof 2
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Theorem
- $\ds \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$
Proof
By Lemma, we have:
- $\ds \int_0^\pi \map \ln {\sin x} \rd x = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x$
By Product of Sines of Fractions of Pi, we have:
- $\ds \prod_{k \mathop = 1}^{n - 1} \map \sin {\frac {k \pi} n} = \frac n {2^{n - 1} }$
Therefore, we have:
\(\ds \sum_{k \mathop = 1}^{n - 1} \map \ln {\map \sin {\frac {k \pi} n} }\) | \(=\) | \(\ds \map \ln {\prod_{k \mathop = 1}^{n - 1} \map \sin {\frac {k \pi} n} }\) | Sum of Logarithms: Natural Logarithm | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \ln {\frac n {2^{n - 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \ln n - \paren {n - 1} \ln 2\) | Difference of Logarithms |
We have:
\(\ds \int_0^\pi \map \ln {\sin x} \rd x\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac \pi n \sum_{k \mathop = 1}^{n - 1} \map \ln {\map \sin {\frac {\pi k} n} }\) | Definition of Riemann Integral | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\frac \pi n \ln n - \pi \paren {1 - \frac 1 n} \ln 2}\) |
We can show the first term to vanish:
\(\ds \lim_{n \mathop \to \infty} \frac \pi n \ln n\) | \(=\) | \(\ds \pi \lim_{u \mathop \to \infty} u e^{-u}\) | letting $n = e^u$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Limit to Infinity of $x^n e^{-a x}$ |
So:
\(\ds \lim_{n \mathop \to \infty} \paren {\frac \pi n \ln n - \pi \paren {1 - \frac 1 n} \ln 2}\) | \(=\) | \(\ds -\pi \ln 2 \lim_{n \mathop \to \infty} \paren {1 - \frac 1 n}\) | Combined Sum Rule for Limits of Real Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds -\pi \ln 2\) |
giving:
- $\ds \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$
$\blacksquare$
Sources
- Jack D'Aurizio (https://math.stackexchange.com/users/44121/jack-daurizio), Evaluate $\ds \int_0^{\pi/2} \map \ln {\cos x} \rd x$, URL (version: 2015-08-19): https://math.stackexchange.com/q/690644