Can you prove this result for any $m,n\in\mathbb{Z}^+$?

[sbhatnagar]

Prove that:

\[ \int_{0}^{\pi/2} \cos(nx) \cos^n(x) dx =\frac{\pi}{2^{n+1}}\]

\[ \int_{0}^{\pi} \frac{1-\cos(nx)}{1-\cos(x)} dx =n\pi \]

where \( n \in \mathbb{N} \). You can use induction, contour integration or any other method you like.

 

[PaulRS]

\[ \int_{0}^{\pi/2} \cos(nx) \cos^n(x) dx =\frac{\pi}{2^{n+1}}\]

First note that \[ \displaystyle\int_{0}^{\pi/2} \cos(nx) \cos^n(x) dx = \Re \{ \int_0^{\pi/2} e^{i\cdot n\cdot x} \cos^n(x) dx \}\]

Now since $\cos(x) = \frac{e^{i\cdot x} + e^{-i\cdot x}}{2}$ we find: \[ \displaystyle\int_0^{\pi/2} e^{i\cdot n\cdot x} \cos^n(x) dx = \int_0^{\pi/2} e^{i\cdot n\cdot x} \left(\frac{e^{i\cdot x} + e^{-i\cdot x}}{2} \right)^n dx = \frac{1}{2^n}\cdot \int_0^{\pi/2} e^{i\cdot n\cdot x} \left({e^{i\cdot x} + e^{-i\cdot x}}\right)^n dx \]

By the binomial theorem: \[\int_0^{\pi/2} e^{i\cdot n\cdot x} \left({e^{i\cdot x} + e^{-i\cdot x}}\right)^n dx = \int_0^{\pi/2} e^{i\cdot n\cdot x} \left( \sum_{k=0}^n { \binom{n}{k} \cdot e^{i\cdot k \cdot x} \cdot e^{- i\cdot (n-k)\cdot x}}\right) dx = \sum_{k=0}^n { \binom{n}{k} \cdot \int_0^{\pi/2} e^{i\cdot n\cdot x} \cdot e^{i\cdot k \cdot x} \cdot e^{- i\cdot (n-k)\cdot x} dx} = \sum_{k=0}^n { \binom{n}{k} \cdot \int_0^{\pi/2} e^{i\cdot 2k \cdot x} dx} \]

Finally we note that $\int_0^{\pi/2} e^0 dx = \frac{\pi}{2}$ and $\int_0^{\pi/2} e^{i\cdot 2k \cdot x} dx = \frac{e^{i\cdot \pi \cdot k} - 1}{i\cdot 2k} $ for $k> 0$. The latter expression is purely imaginary, thus:
\[\Re\{\int_0^{\pi/2} e^{i\cdot n\cdot x} \left({e^{i\cdot x} + e^{-i\cdot x}}\right)^n dx\} = \Re\{\sum_{k=0}^n { \binom{n}{k} \cdot \int_0^{\pi/2} e^{i\cdot 2k \cdot x} dx} \} = \frac{\pi}{2} \]
Hence:
\[ \displaystyle\int_{0}^{\pi/2} \cos(nx) \cos^n(x) dx = \Re \{ \int_0^{\pi/2} e^{i\cdot n\cdot x} \cos^n(x) dx \} = \Re\{\frac{1}{2^n}\cdot \int_0^{\pi/2} e^{i\cdot n\cdot x} \left({e^{i\cdot x} + e^{-i\cdot x}}\right)^n dx \} = \frac{1}{2^n}\Re\{\int_0^{\pi/2} e^{i\cdot n\cdot x} \left({e^{i\cdot x} + e^{-i\cdot x}}\right)^n dx\} = \frac{\pi}{2^{n+1}} \square\]

 

[Sherlock1]

\[ \int_{0}^{\pi} \frac{1-\cos(nx)}{1-\cos(x)} dx =n\pi \]

$\displaystyle n+2\sum_{k=1}^{n-1}(n-k)\cos{kx} = \frac{1-\cos{nx}}{1-\cos{x}}$, thus $\displaystyle I = \int_{0}^{\pi}\frac{1-\cos{nx}}{1-\cos{x}}\;{dx} = \int_{0}^{\pi}\left(n+2\sum_{k=1}^{n-1}(n-k)\cos{kx}\right)\;{dx} = n\pi+2\sum_{k=1}^{n-1}(n-k)\int_{0}^{\pi}\cos{kx}\;{dx} = n\pi. $

 

[sbhatnagar]

$\displaystyle n+2\sum_{k=1}^{n-1}(n-k)\cos{kx} = \frac{1-\cos{nx}}{1-\cos{x}}$, thus $\displaystyle I = \int_{0}^{\pi}\frac{1-\cos{nx}}{1-\cos{x}}\;{dx} = \int_{0}^{\pi}\left(n+2\sum_{k=1}^{n-1}(n-k)\cos{kx}\right)\;{dx} = n\pi+2\sum_{k=1}^{n-1}(n-k)\int_{0}^{\pi}\cos{kx}\;{dx} = n\pi. $

You have chosen the best method so far. Here are a few more ways to tackle the problem.

Approach 1

Let \( \displaystyle I_n=\int_{0}^{\pi}\frac{1-\cos{nx}}{1-\cos{x}}\;{dx} \).

\( \displaystyle I_{n+1}=\int_{0}^{\pi}\frac{1-\cos{(n+1)x}}{1-\cos{x}}\;{dx} \).

\( \displaystyle I_{n+1}-I_n=\int_{0}^{\pi}\frac{1-\cos{nx}}{1-\cos{x}}\;{dx} -\int_{0}^{\pi}\frac{1-\cos{nx}}{1-\cos{x}}\;{dx}= \int_{0}^{\pi}\frac{\sin{ \left\{ \frac{(2n+1)x}{2}\right \}}}{\sin{\left( \frac{x}{2}\right)}}dx\).

Let \( \displaystyle J_n= I_{n+1}-I_n= \int_{0}^{\pi}\frac{\sin{ \left\{ \frac{(2n+1)x}{2}\right \}}}{\sin{\left( \frac{x}{2}\right)}}dx\)

\( \displaystyle J_{n+1}= \int_{0}^{\pi}\frac{\sin{ \left\{ \frac{(2n+3)x}{2}\right \}}}{\sin{\left( \frac{x}{2}\right)}}dx\)

\( \displaystyle J_{n+1}-J_n= \int_{0}^{\pi}\frac{\sin{ \left\{ \frac{(2n+3)x}{2}\right \}-\sin\left\{ \frac{(2n+1)x}{2}\right \}}}{\sin{\left( \frac{x}{2}\right)}}dx = 2\int_{0}^{\pi} \cos{\left\{ \frac{(2n+2)x}{2}\right\} }dx = 0\)

This means \( \displaystyle J_{n+1}=J_n=J_{n-1}=\cdots=J_1=\pi\)

So \( \displaystyle J_{n}=\pi \ \Rightarrow I_{n+1}-I_n=\pi \)

So \( I_n = \pi+(n-1)\pi = n\pi \)

Approach 2

\( \displaystyle I_n=\int_{0}^{\pi}\frac{1-\cos{nx}}{1-\cos{x}}\;{dx} \)

Substitute \( z=e^{ix} \).

\( \displaystyle I_n=\int_{0}^{\pi}\frac{1-\cos{nx}}{1-\cos{x}}\;{dx} = \frac{1}{2}\oint_{C} \frac{1-\frac{z^n +z^{-n}}{2}}{1-\frac{z +z^{-1}}{2}}\frac{dz}{iz}=\frac{1}{2}\oint_{C}\frac{(z^n -1)^2}{iz^n (z-1)^2}dz\)

where C is the unit circle.

Let \( \displaystyle f(z)=\frac{(z^n -1)^2}{iz^n (z-1)^2} \)

\( f(z) \) has pole at \( z=0 \).

\(\displaystyle \text{Res}_{z=0}f(z)=\frac{n}{i} \)

So \(\displaystyle I_n= \pi i\displaystyle \text{Res}_{z=0}f(z)=\frac{n\pi i}{i} =n\pi \)

 

[Sherlock1]

Very nice! (Yes)

You have chosen the best method so far.

It took me ages to come up with that! (Sleepy)

 

[PaulRS]

\[ \int_{0}^{\pi} \frac{1-\cos(nx)}{1-\cos(x)} dx =n\pi \]

Here is a related result you guys might be interested in trying out: \[ \displaystyle\int_0^\pi \left(\frac{\sin(n\cdot x)}{\sin(x)}\right)^m = \pi \cdot [x^{m\cdot (n-1)}] \left\{ \left( \sum_{k=0}^{n-1} x^{2\cdot k} \right)^m\right\}\]
for $m,n\in\mathbb{Z}^+$

Notation: $[x^a]\{p(x)\}$ is the coefficient of $x^a$ in $p(x)$.

(Wink)

So for example, for $m=2$ in particular we get: $\displaystyle\int_0^\pi \left(\frac{\sin(n\cdot x)}{\sin(x)}\right)^2 = \pi \cdot [x^{2\cdot (n-1)}] \left\{ \left( \sum_{k=0}^{n-1} x^{2\cdot k} \right)^m\right\}= \pi \cdot [x^{n-1}] \left\{ \left( \sum_{k=0}^{n-1} x^k \right)^2\right\} = \pi \cdot n$ because for each $k\in \{0,1,..,n-1\}$ there is exactly one $j\in \{0,1,...,n-1\}$ such that $k+j = n-1$.