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$\displaystyle\int^{2\pi}_{0} e^{ia \cos{\theta}}d\theta$ where $a$ is some constant. Can it be solved with some substitution? I tried it by expanding the exponential series but that was not proper result. I want analytic results.

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2 Answers

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This is a very challenging integral. I don't feel the result is very intuitive, but I learned how to solve it in a PDE's class years ago. First, Euler's identity:

$$\int^{2\pi}_{0} e^{ia \cos{\theta}}d\theta = \int^{2\pi}_{0} \cos(a \cos{\theta})d\theta+\int^{2\pi}_{0}i\sin(a \cos{\theta})d\theta $$ First, let's solve $\int^{2\pi}_{0} \cos(a \cos{\theta})d\theta$. To do so, introduce the Taylor Series definition to the outermost cosine argument. $$\int^{2\pi}_{0} \cos(a \cos{\theta})= \int^{2\pi}_{0} \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(a \cos{\theta})^{2n}d\theta \\ =\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}a^{2n}\int^{2\pi}_{0}\cos^{2n}(\theta)d\theta$$ Now you will need to know how to integrate $\int_0^{2\pi} \cos^{2n}(\theta)d\theta$. I will tell you for the sake of solving this problem that $$\int_0^{2\pi}\cos^{2n}(\theta)d\theta = \frac{(2n)!}{2^{2n}(n!)^2}2 \pi$$ but that is a result you should verify and prove yourself. Now let's plug in our result for the integral of $\cos^{2n}(\theta)$ and proceed. $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}a^{2n}\int^{2\pi}_{0}\cos^{2n}(\theta)d\theta = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}a^{2n} \left(\frac{(2n)!}{2^{2n}(n!)^2}2 \pi \right) \\ = \sum_{n=0}^\infty (-1)^n \frac{(2n)!}{(2n)!}\cdot \frac{a^{2n}}{2^{2n}} \frac{2\pi}{(n!)^2} \\ = 2\pi\sum_{n=0}^\infty (-1)^n\left(\frac{a}{2}\right)^{2n} \frac{1}{(n!)^2}$$ As other users have pointed out, the sum $\sum_{n=0}^\infty (-1)^n\left(\frac{a}{2}\right)^{2n+\alpha} \frac{1}{n!\Gamma (n+\alpha +1)}$ is very famous, and known as a Bessel function of the first kind, denoted $$\sum_{n=0}^\infty (-1)^n\left(\frac{a}{2}\right)^{2n+\alpha} \frac{1}{n!\Gamma (n+\alpha +1)} = J_\alpha (a)$$ We have the case of $\alpha=0$ in our sum, because as you can see $$\sum_{n=0}^\infty (-1)^n\left(\frac{a}{2}\right)^{2n+\alpha} \frac{1}{n!\Gamma (n+\alpha +1)}=\sum_{n=0}^\infty (-1)^n\left(\frac{a}{2}\right)^{2n} \frac{1}{n!\Gamma (n +1)} \\ = \sum_{n=0}^\infty (-1)^n\left(\frac{a}{2}\right)^{2n} \frac{1}{n!n!} \\ =J_0(a)$$ which is an identical match to our sum. We can conclude $$\int^{2\pi}_{0} \cos(a \cos{\theta})d\theta = 2\pi J_0(a)$$ Can you repeat similar steps to solve $\int^{2\pi}_{0}i\sin(a \cos{\theta})d\theta $?

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Use Euler's formula and the fact that $\displaystyle\int_0^\tfrac\pi2\cos(a~\cos x)~dx=\int_0^\tfrac\pi2\cos(a~\sin x)~dx=\frac\pi2~J_0(a)$

and $\displaystyle\int_0^\tfrac\pi2\sin(a~\sin x)~dx=\int_0^\tfrac\pi2\sin(a~\cos x)~dx=\frac\pi2~H_0(a)$. Then, employing the parity of

each such function, which are evident from their series expansions, you will ultimately arrive

at the desired result.

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