Definite Integral from 0 to Half Pi of Logarithm of Sine x by Cosine of 2nx

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Theorem

For $n \in \N_{>0}$:

$\ds \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \rd x = -\frac \pi {4 n}$


Proof

First we have:

\(\ds \lim_{x \mathop \to 0} \map \ln {\sin x} \sin 2 n x\) \(=\) \(\ds \lim_{x \mathop \to 0} \frac {\map \ln {\sin x} } {\csc 2 n x}\) Definition of Cosecant
\(\ds \) \(=\) \(\ds \lim_{x \mathop \to 0} \frac {\cot x} {- 2 n \cot 2 n x \csc 2 n x}\) L'Hôpital's Rule: Corollary $2$
\(\ds \) \(=\) \(\ds \lim_{x \mathop \to 0} \frac {\cos x} {- 2 n \cos 2 n x} \frac {\sin^2 2 n x} {\sin x}\) Definition of Cosecant, Definition of Cotangent
\(\ds \lim_{x \mathop \to 0} \frac {\sin^2 2 n x} {\sin x}\) \(=\) \(\ds \lim_{x \mathop \to 0} \frac {4 n \sin 2 n x \cos 2 n x} {\cos x}\) L'Hôpital's Rule
\(\ds \) \(=\) \(\ds 0\)

hence $\ds \lim_{x \mathop \to 0} \map \ln {\sin x} \sin 2 n x = 0$.


Thus:

\(\ds \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \rd x\) \(=\) \(\ds \frac 1 {2 n} \int_0^{\pi/2} \map \ln {\sin x} \map \rd {\sin 2 n x}\) Primitive of $\cos a x$
\(\ds \) \(=\) \(\ds \frac 1 {2 n} \paren {\bigintlimits {\map \ln {\sin x} \sin 2 n x} 0 {\pi/2} - \int_0^{\pi/2} \sin n x \map \rd {\map \ln {\sin x} } }\) Integration by Parts
\(\ds \) \(=\) \(\ds -\frac 1 {2 n} \int_0^{\pi/2} \sin n x \map \rd {\map \ln {\sin x} }\) From above
\(\ds \) \(=\) \(\ds -\frac 1 {2 n} \int_0^{\pi/2} \sin 2 n x \frac {\cos x} {\sin x} \rd x\) Primitive of Cotangent Function
\(\ds \) \(=\) \(\ds -\frac 1 {2 n} \int_0^{\pi/2} \frac {\sin \paren {2 n + 1} x + \sin \paren {2 n - 1} x} {2 \sin x} \rd x\) Werner Formula for Sine by Cosine
\(\ds \) \(=\) \(\ds -\frac 1 {4 n} \int_0^{\pi} \frac {\sin \paren {\paren {2 n + 1} u/2} + \sin \paren {\paren {2 n - 1} u/2} } {2 \sin \paren {u/2} } \rd u\) substituting $u = 2 x$
\(\ds \) \(=\) \(\ds -\frac 1 {4 n} \int_0^{\pi} \paren {\frac 1 2 + \sum_{k \mathop = 1}^n \map \cos {k u} + \frac 1 2 + \sum_{k \mathop = 1}^{n - 1} \map \cos {k u} } \rd u\) Lagrange's Cosine Identity
\(\ds \) \(=\) \(\ds -\frac 1 {4 n} \int_0^{\pi} 1 \rd u\) All integrals involving $\cos k u$ evaluate to $0$
\(\ds \) \(=\) \(\ds -\frac \pi {4 n}\) Definite Integral of Constant

$\blacksquare$