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

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The discussion focuses on proving two integral results involving trigonometric functions for positive integers \( n \). The first result states that the integral of \( \cos(nx) \cos^n(x) \) from 0 to \( \pi/2 \) equals \( \frac{\pi}{2^{n+1}} \), which is derived using complex exponentials and the binomial theorem. The second result asserts that the integral of \( \frac{1-\cos(nx)}{1-\cos(x)} \) from 0 to \( \pi \) equals \( n\pi \), with various approaches discussed, including induction and contour integration. Both results are confirmed through detailed calculations and alternative methods, showcasing the versatility in tackling such problems. The thread concludes with a related integral result that invites further exploration.
sbhatnagar
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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.
 
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sbhatnagar said:
\[ \int_{0}^{\pi/2} \cos(nx) \cos^n(x) dx =\frac{\pi}{2^{n+1}}\]

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:
\[ \displaystyle\int_{0}^{\pi/2} \cos(nx) \cos^n(x) dx = \Re \{ \int_0^{\pi/2} e^{i\cdot n\cdot x} \cos^n(x) dx \} = \Re\{\frac{1}{2^n}\cdot \int_0^{\pi/2} e^{i\cdot n\cdot x} \left({e^{i\cdot x} + e^{-i\cdot x}}\right)^n dx \} = \frac{1}{2^n}\Re\{\int_0^{\pi/2} e^{i\cdot n\cdot x} \left({e^{i\cdot x} + e^{-i\cdot x}}\right)^n dx\} = \frac{\pi}{2^{n+1}} \square\]
 
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sbhatnagar said:
\[ \int_{0}^{\pi} \frac{1-\cos(nx)}{1-\cos(x)} dx =n\pi \]
$\displaystyle n+2\sum_{k=1}^{n-1}(n-k)\cos{kx} = \frac{1-\cos{nx}}{1-\cos{x}}$, 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. $
 
Sherlock said:
$\displaystyle n+2\sum_{k=1}^{n-1}(n-k)\cos{kx} = \frac{1-\cos{nx}}{1-\cos{x}}$, 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 \(\displaystyle I_n= \pi i\displaystyle \text{Res}_{z=0}f(z)=\frac{n\pi i}{i} =n\pi \)
 
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Very nice! (Yes)
sbhatnagar said:
You have chosen the best method so far.
It took me ages to come up with that! (Sleepy)
 
sbhatnagar said:
\[ \int_{0}^{\pi} \frac{1-\cos(nx)}{1-\cos(x)} dx =n\pi \]

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 $k+j = n-1$.
 
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