Integral $\int\cos (2nx)\,\log \sin(x)\; dx $

Question

Please help me out wrt this question - $$\int_0^\frac{\pi}2 \cos (2nx)\; \log \sin(x)\; dx = -\frac {\pi}{4n}$$

Here $n>1$.

I tried doing integration by parts but then how to calculate

$$\int_0^\frac{\pi}2 \cot (x) \,\sin (2nx)\, dx $$ The question is given under Improper Integrals.

Thank you in advance. I really need to sort this out

Answer 1

A different approach.

By integrating by parts, one has $$ \begin{align} \int_0^{\Large \frac{\pi}2} \cos (2nx) \log \sin (x)\; dx&=\left[ \frac{\sin (2nx)}{2n}\cdot \log \sin (x)\right]_0^{\Large \frac{\pi}2} -\frac{1}{2n}\int_0^{\Large \frac{\pi}2} \sin (2nx)\: \frac{\cos (x)}{\sin (x)}\; dx \\&=\color{red}{0}-\frac{1}{2n}\int_0^{\Large \frac{\pi}2} \sin (2nx)\: \frac{\cos (x)}{\sin (x)}\; dx. \tag1 \end{align} $$Let $$ u_n:=\int_0^{\Large \frac{\pi}2} \sin (2nx)\: \frac{\cos (x)}{\sin (x)}\; dx,\quad n\ge1. $$ One may observe that, for $n\ge1$, $$ \begin{align} u_{n+1}-u_n&=\int_0^{\Large \frac{\pi}2} \left[\frac{}{}\sin (2nx+2x)-\sin(2nx)\right]\cdot \frac{\cos (x)}{\sin (x)}\; dx \\&=\int_0^{\Large \frac{\pi}2} \left[2\frac{}{}\sin (x)\cdot \cos(2nx+x)\right]\cdot \frac{\cos (x)}{\sin (x)}\; dx\quad \left({\small{\color{blue}{\sin p-\sin q=2 \sin \frac{p-q}{2}\cdot \cos \frac{p+q}{2}}}}\right) \\&=\int_0^{\Large \frac{\pi}2} 2\cdot\cos(2nx+x)\cdot \cos (x)\; dx \\&=\int_0^{\Large \frac{\pi}2} \left[\frac{}{}\cos(2nx+2x)+\cos(2nx)\right] dx\qquad \quad \left({\small{\color{blue}{2\cos a \cos b= \cos (a+b)+ \cos (a-b)}}}\right) \\\\&=\color{red}{0} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left({\small{\color{blue}{\sin(m\cdot \pi)=0,\, m=0,1,2,\cdots }}}\right) \end{align} $$ giving $$ u_{n+1}=u_n=\cdots=u_1=2\int_0^{\Large \frac{\pi}2} \cos^2 (x)\; dx=\frac \pi2, \tag2 $$ then inserting $(2)$ in $(1)$ yields

$$ \int_0^{\Large \frac{\pi}2} \cos (2nx) \log \sin (x)\; dx=-\frac {\pi}{4n},\qquad n\ge1 $$

as wanted.

Answer 2

Here is an approach.

Hint. One may observe that, for $x \in \left(0,\frac \pi2\right)$, $$ \begin{align} \log \left(\cos x \right)&=\text{Re}\log \left(\frac{e^{ix}+e^{-ix}}{2}\right) \\\\&=\text{Re}\left(ix+\log \left(1+e^{-2ix}\right)-\log 2\right) \\\\&=-\log 2+\text{Re}\left(\log \left(1+e^{-2ix}\right)\right) \\\\&=-\log 2+\text{Re}\sum_{n=1}^\infty \frac{(-1)^{n+1} }n e^{-2nix} \\\\&=-\log 2+\text{Re}\sum_{n=1}^\infty \frac{(-1)^{n+1}}n \left(\cos (2nx)-i\sin(2nx)\right) \\\\&=-\log 2+\sum_{n=1}^\infty \frac{(-1)^{n+1} }n \:\cos (2nx) \end{align} $$ then by the uniqueness of the coefficent of a Fourier series one deduce $$ \frac 2\pi\int_0^{\Large \frac{\pi}2} \cos (2nx) \log \cos (x)\; dx = \frac {(-1)^{n+1}}{2n},\qquad n\ge1. $$ By the change of variable $u=\dfrac \pi2-x\,,$ $x=\dfrac \pi2-u\,,$ $du=-dx$, giving $$ \begin{align} &\cos (x) = \sin (u), \\ &\cos (2nx) = \cos \left(2n \cdot \frac \pi2- 2n \cdot u \right)=\cos (n \pi-2nu)=(-1)^n \cos (2nu) \end{align} $$ one gets

$$ \int_0^{\Large \frac{\pi}2} \cos (2nu) \log \sin (u)\; du =(-1)^n\int_0^{\Large \frac{\pi}2} \cos (2nx) \log \cos(x)\; dx=-\frac {\pi}{4n},\qquad n\ge1, $$

as announced.