Integrating integer powers of $\cos(\theta)$

Question

Consider the following integral for $n\in\mathbb{N}$:

$$I_n = \int_0^\pi\cos^n\theta\,d\theta \tag1$$

which, using integration by parts, one can show to be $I_n = 0$ for $n$ odd and to be equal to

$$I_{2m} = \frac{(2m-1)!!}{(2m)!!}\pi \tag2$$

for $n = 2m$ even.

However, I've tried to evaluate $(1)$ using the binomial theorem to write powers of cosine as a sum over powers of exponential functions:

\begin{align} I_n &= \frac{1}{2^n}\int_0^\pi\left(e^{i\theta}+e^{-i\theta}\right)^n\,d\theta\\ &=\frac{1}{2^n}\sum_{k=0}^n{n\choose k}\int_0^\pi e^{ik\theta}e^{-i(n-k)\theta}\,d\theta\\ &=\frac{1}{2^n}\sum_{k=0}^n{n\choose k}\int_0^\pi e^{i(2k-n)\theta}\,d\theta\\ &=\frac{1}{2^n}\sum_{k=0}^n{n\choose k}\frac{e^{i\pi(2k-n)}-1}{i(2k-n)} \end{align}

However, this result seems to contradict $(2)$ - note that if $n = 2m$ is even, then $2k-2m$ is an even integer and $e^{2\pi i(k-m)} = 0$ for any value of $k$, which would imply that $I_n$ is nonzero only for odd values of $n$.

Furthermore, if $n = 2m -1$, then we have $e^{i\pi(2k-2m+1)} = -1$ and we thus have

\begin{align} I_{2m-1} &= \frac{1}{2^{2m-1}}\sum_{k=0}^{2m-1}{{2m-1}\choose k}\frac{-2}{i(2k-2m-1)}\\ &=\frac{i}{2^{2m}}\sum_{k=0}^{2m-1}{{2m-1}\choose k}\frac{1}{2(k-m)-1} \tag3 \end{align}

which is purely imaginary, which is obviously wrong.

So what's the problem here? Am I somehow wrong in using the binomial theorem? Have I made a computational error that explains this odd result? Can this approach to computing $(1)$ be salvaged?

EDIT.

Taking into consideration that the sum is actually non-zero if the argument of the exponential function is itself zero, which occurs if, for $n$ = $2m$, we have $k = m$, this gives us:

\begin{align} I_{2m} &= \frac{1}{2^{2m}}{{2m}\choose m}\pi = \frac{(2m)!}{2^{2m+1}m!}\pi \end{align}

However, I don't see how this is equal to $(2)$.

I have also tried to show that $(3)$ is $0$, by symmetry, but I have no managed to show this yet either.

Answer 1

First, note that we have

$$\int_0^\pi e^{i(2k-n)x}\,dx=\begin{cases}\pi&,n=2k\\\\\frac{1-(-1)^n}{i(n-2k)}&,n\ne 2k\end{cases}$$

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Now, if $n$ is even, say $n=2m$, then we can write

$$\begin{align} \int_0^\pi \cos^{2m}(x)\,dx&=\frac{1}{2^{2m}}\sum_{k=0}^{2m}\binom{2m}{k}\pi \delta_{m,k}\\\\ &= \frac{\pi}{2^{2m}}\binom{2m}{m}\\\\ &=\frac{\pi}{2^{2m}} \frac{(2m)!}{(m!)^2}\tag1 \end{align}$$

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Now, let's take a look at $\frac{(2m-1)!!}{(2m)!!}$. We can write

$$\begin{align} \frac{(2m-1)!!}{(2m)!!}&=\frac{(2m-1)(2m-3)\cdots (3)(1)}{(2m)(2m-2)\cdots (2)(1)}\\\\ &=\left(\frac{(2m)(2m-2)\cdots (2)(1)}{(2m)(2m-2)\cdots (2)(1)}\right)\left(\frac{(2m-1)(2m-3)\cdots (3)(1)}{(2m)(2m-2)\cdots (2)(1)}\right)\\\\ &=\frac{(2m)!}{(2^m)(m!)(2^m)(m!)}\\\\ &=\frac{(2m)!}{2^{2m}(m!)^2}\\\\ &=\frac{1}{2^{2m}}\binom{2m}{m}\tag2 \end{align}$$

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Substituting the left-hand side of $(2)$ into $(1)$ yields

$$\int_0^\pi \cos^{2m}(x)\,dx=\pi \frac{(2m-1)!!}{(2m)!!}$$

as was to be shown!

Answer 2

There seems to be a sign error in your expansion of the odd-power integral. It should be $$ I_{2m-1} = \frac{1}{2^{2m-1}}\sum_{k=0}^{2m-1}{{2m-1}\choose k}\frac{-2}{i(2k-2m+1)}. $$

Now notice that $\binom{2m-1}{2m-1-k} = \binom{2m-1}{k}$ and that $2(2m-1-k)-2m+1 = -(2k-2m+1),$ and perhaps it will be clearer that the last $m$ terms of the sum exactly cancel the first $m$ terms, leaving a final sum equal to zero.