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Starting with $\int_{0}^{\pi}{e^{2\cos(x)}\, dx}$ we have:

$$\begin{align} \int_{0}^{\pi}{e^{2\cos(x)}\, dx} &= \int_{0}^{\pi}{\sum_{k=0}^{\infty}{\frac{[2\cos(x)]^k}{k!}}\, dx}\\ &= \sum_{k=0}^{\infty}{\frac{2^k}{k!}\int_{0}^{\pi}{\cos^k(x)\, dx}} \end{align}$$

I think the sum/integral interchange is valid. Continuing with the integral we see that for $k=2n+1, n\in\mathbb{N}$ $$\int_{0}^{\pi}{\cos^{2n+1}(x)\, dx}=\int_{-\pi}^{\pi}{\cos^{2n+1}(x)\, dx}=0$$

That leaves the $k=2n$ case, let’s start by defining:

$$I_{2n}:=\int_{0}^{\pi}{\cos^{2n}(x)\, dx}$$

Continuing:

$$\int_{0}^{\pi}{\cos^{2n}(x)\, dx}=\int_{0}^{\pi}{\cos^{2n-1}(x)\cos(x)\, dx}$$

(IBP, first terms go to zero and the integral is left over)

$$\begin{align}\int_{0}^{\pi}{\cos^{2n-1}(x)\cos(x)\, dx}&=(2n-1)\int_{0}^{\pi}{\cos^{2n}(x)+(2n-3)\sin^2(x)\cos^{2n-2}(x)\, dx}\\&=(2n-1)\int_{0}^{\pi}{\cos^{2n}(x)+(2n-3)(1-\cos^2(x))\cos^{2n-2}(x)}\\&=(2n-1)\int_{0}^{\pi}{\cos^{2n}(x)+(2n-3)[\cos^{2n-2}(x)-\cos^{2n}(x)]\, dx}\\&=(2n-1)\int_{0}^{\pi}{(4-2n)\cos^{2n}(x)+(2n-3)\cos^{2n-2}(x)\, dx}\\&=(2n-1)(4-2n)\int_{0}^{\pi}{\cos^{2n}(x)\, dx}+(2n-1)(2n-3)\int_{0}^{\pi}{\cos^{2n-2}(x)\, dx}\\&=(2n-1)(4-2n)I_{2n}+(2n-1)(2n-3)I_{2n-2}\\&=-2(2n-1)(n-2)+(2n-1)(2n-3)I_{2n-2}\end{align}$$

In the end we have, assuming everything is correct here:

$$\begin{align}I_{2n} &= -2(2n-1)(n-1)I_{2n}+(2n-1)(2n-3)I_{2n-2}\\\implies I_{2n}&=\frac{(2n-1)(2n-3)}{4n^2-6n+3}I_{2n-2}\end{align}$$

But this doesn’t work for $I_2$ for example, so something must be wrong here.

AnthonyML
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1 Answers1

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The easiest way to find $I_{2n} = \int_0^\pi \cos(x)^{2n} \; dx$ is using complex exponentials: $\cos(x) = (\exp(ix) + \exp(-ix))/2$. Expand out using the binomial theorem and note that $\int_0^\pi \exp(ikx)\; dx = 0$ for all even $k$ except $0$. You should get $$ \int_0^\pi \cos^{2n}(x) = {2n \choose n}2^{-2n} \pi$$ and thus $$ \int_0^\pi \exp(t \cos(x))\; dx = \sum_{n=0}^\infty \pi \frac{(t/2)^{2n}}{(n!)^2} = \pi I_0(t) $$ where $I_0$ is a modified Bessel function.

Robert Israel
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