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I've come across an integral which I'm stumped on how the solution was reached. I want to know how it was derived so I can understand if it's possible to vary the limits of integration in the definite integral.

$\int_0^{2 \pi} \exp(j A \cos(\theta))\ d\theta = 2\pi J_0(A)$ (and for $\sin$ in place of $\cos$)

which can be extended to: $\int_0^{2 \pi} \exp(j A \cos(\theta))\exp(j B \sin(\theta))\ d\theta = 2\pi J_0(\sqrt{A^2 + B^2})$

I've seen these (or similar forms) listed in "Tables of Integrals, Series, and Products" by Gradshteyn and Ryzhik (who reference Bierens de Haan, D., Nouvelles tables d^integrales definies. Amsterdam, 1867), and Wolfram gives me the same answer.

I understand there's a use of the identity for Bessel function first kind and modified Bessel function first kind, but I'm not sure how the first integral is achieved. Any insight into how the first integral is derived would be much appreciated.

EDIT: How to solve integral $\int_0^{2\pi} e^{i(a\cos\phi + b\sin\phi)} \cos\phi\ d\phi$ shows me how to effectively get the second integral. Does anyone know how the first integral is derived? I'm curious to know if I can integrate from $0$ to $\frac{\pi}{3}$ in a situation where I have 6-fold rotational symmetry?

A rural reader
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Ian
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    The second integral follows directly from the first. To "solve" the first depends on how you define $J_n$; it can be defined by a series, integral, ODE, etc. – David Raveh Jun 20 '23 at 00:04
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    @DavidRaveh Yes, and I'd like to know how to derive the first integral so that I have the full picture of where all of those following steps come from. – Ian Jun 20 '23 at 00:41

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After some searching, I came across this helfpul website: http://www.math.tulane.edu/~vhm/Table.html, specifically http://www.math.tulane.edu/~vhm/formula_html/3.339.proof.pdf

Here, they expand the exponential function into its infinite series: $\int_0^\pi exp(z*Cos(\theta))d\theta = \pi\sum_{j=0}^{\infty}\frac{z^j}{j!}\int_0^\pi Cos^j(\theta)d\theta$

By the symmetry of Cosine between 0 and $\pi$, all odd j evaluate to $0$. For all even j = 2k, they use Wallis' formula: $\int_0^\pi Cos^{2k}(\theta)d\theta = \frac{(2k)!}{k!^2}\frac{\pi}{2^{2k}}$

Thus: $\int_0^\pi exp(z*Cos(\theta))d\theta = \pi\sum_{j=0}^{\infty}\frac{1}{k!^2}(\frac{z}{2})^{2k}$

Which gives us the power series for zeroth-order modified Bessel function, i.e., $\pi I_0(z)$

This answers my initial question, and addressing my edit, the integral result for $\frac{\pi}{3}$ as limit of integration doesn't allow for the same argument of cosine symmetry and requires hypergeometric function.

Ian
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