Integral of $\int_0^{2\pi}{e^{({x\sin\theta)}}\ \text{d}\theta}$

Question

I would like to compute $$f(x)= \int_0^{2\pi}{e^{({x\sin\theta)}}\ \text{d}\theta}$$

I already know that it equals $2\pi\sum_{j\geq0}{(\frac{x^j}{2^jj!})^2}$. I'm already happy since it provides an efficient numerical solution to the problem. However I would like to know whether this fonction can be expressed simply with common functions.

Using series expansion, I found that $g(x)=\frac1{2\pi}f(2x)$ solves the second order linear differiential equation $g(0)=1$, $g'(0)=0$, $g''(x)+\frac{g'(x)}x-4g(x)+2=0$. I don't find it very helpful...

Answer 1

Personally, I suggest a Taylor series of the whole integrand function, namely:

$$e^{x\sin\theta} = \sum_{k = 0}^{+\infty} \frac{(x\sin\theta)^k}{k!}$$

then arranging the writing:

$$\sum_{k = 0}^{+\infty}\frac{x^k}{k!}\int_0^{2\pi} \sin^k\theta\ \text{d}\theta$$

The integral is not trivial, and it'n not easily calculable unless $k$ is some small number like $k = 2$ or $k = 4$ or so on.

Notice that for every odd $k$, the integrand is zero!

This will mean that the sum will automatically cover only the even values of $k$.

Anyway, the exact solution of such an integral does exist, and it's expressed in terms of Hypergeometric Functions, and I'll give you the immediate solution, so take it as it comes. Then I'll write you the solution of the bounded integral.

$$\int\sin^k\theta\ \text{d}\theta = -\ _2F_1\left[\frac{1}{2},\ \frac{1-k}{2},\ \frac{3}{2},\ \cos^2\theta\right]\cos\theta \sin^{1+k}\theta\left(\sin^2\theta\right)^{\frac{-1-k}{2}}$$

More on Hypergeometric Functions

https://en.wikipedia.org/wiki/Hypergeometric_function

Now, the definite integral is instead easier and its result comes in terms of the Euler Gamma Function

$$\int_0^{2\pi} \sin^k\theta\ \text{d}\theta = \frac{(1 + (-1)^k)\sqrt{\pi}\ \Gamma\left(\frac{1+k}{2}\right)}{\Gamma\left(1 + \frac{k}{2}\right)}$$

Notice that for odd values of $k$, the integrand is zero as wanted!

More on Euler Gamma

https://en.wikipedia.org/wiki/Gamma_function

So, the final result is given:

$$\mathcal{I(x)} = \int_0^{2\pi} e^{x\sin\theta} = \sum_{k = \text{even}} \frac{x^k}{k!}\frac{(1 + (-1)^k)\sqrt{\pi}\ \Gamma\left(\frac{1+k}{2}\right)}{\Gamma\left(1 + \frac{k}{2}\right)} = \sum_{k} \frac{x^k}{k!}\mathcal{G}(k)$$

Some Terms

We may proceed in finding the very first three terms, for example, namely for $k = 0,\ 2,\ 4$:

For $k = 0$, $\mathcal{G}(0) = 2\pi$ (this is trivial, and doable by hands too with the integral of $\sin^0\theta$)

For $k = 2$, $\mathcal{G}(2) = \pi$

For $k = 4$, $\mathcal{G}(4) = \frac{3\pi}{4}$

For $k = 6$, $\mathcal{G}(6) = \frac{5\pi}{8}$

So the first terms of the series are

$$\mathcal{I}(x) \approx 2\pi + \frac{x^2}{2} + \frac{3\pi}{96}x^4 + \frac{5\pi}{5670}x^6 + \cdots$$

The amazing thing of this series, is that if we simplify the fractions, and if we collect a $2\pi$ term, we get the result

$$\mathcal{I}(x) = 2\pi\left[1 + \frac{x^2}{4} + \frac{x^4}{64} + \frac{x^6}{2304} + \cdots \right]$$

Which is nothing but the Series expansion of the Bessel Function of the first kind!

Indeed, Mathematica says that the exact result of the integration is:

$$\int_0^{2\pi} e^{x\sin\theta}\ \text{d}\theta = 2\pi (\text{Bessel1}[0, x])$$

More on Bessel Function of the first kind

http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html

This is basically the explanation of your result, and it provides the best (and exact) solution you could get for such integral.