1
$\begingroup$

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

$\endgroup$

closed as off-topic by JonMark Perry, Sahiba Arora, Namaste, Antonios-Alexandros Robotis, Nosrati Aug 17 '17 at 18:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JonMark Perry, Sahiba Arora, Namaste, Antonios-Alexandros Robotis, Nosrati
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Sorry Michael. Just formatted it. Please help me in this question $\endgroup$ – Sitanshu Aug 16 '17 at 20:57
  • $\begingroup$ Note: I reformatted your equation. Please check to make sure I didn't introduce any errors. $\endgroup$ – lulu Aug 16 '17 at 20:59
  • $\begingroup$ I tried doing integration by parts but then how to calculate $$\int_0^\frac{π}2 cot x sin2nx dx $$ $\endgroup$ – Sitanshu Aug 16 '17 at 21:01
  • $\begingroup$ Thanks Lulu. That's right :) $\endgroup$ – Sitanshu Aug 16 '17 at 21:04
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Yeah, that's a more simple way to answer this. Thanks, as Fourier series was not in my syllabus , so this answer perfectly fits. :) $\endgroup$ – Sitanshu Aug 17 '17 at 2:18
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Great job!! :)) $\endgroup$ – Tolaso Aug 16 '17 at 22:10
  • 1
    $\begingroup$ Bravo Olivier. Thanks a ton for solving this. :) :) $\endgroup$ – Sitanshu Aug 16 '17 at 23:37

Not the answer you're looking for? Browse other questions tagged or ask your own question.