Integral of exponential of trigonometric functions

Question

I'm looking for a proof of the result

$$ J_0 \left( x \sqrt{a^2 + b^2}\right) = \frac{1}{2\pi} \int_0^{2\pi} e^{ix \, (a \cos \theta \, + \, b \sin \theta)} \, \text{d} \theta $$

$J_0$ is the Bessel function of order $0$.

Answer 1

We have that $$a\cos\theta + b\sin\theta = \sqrt{a^2+b^2}\cos(\theta-\psi),\qquad \psi=\arctan\frac{b}{a}\tag{1}$$ hence $$\frac{1}{2\pi}\int_{0}^{2\pi}e^{ix(a\cos\theta+b\sin\theta)}\,d\theta = \frac{1}{2\pi}\int_{0}^{2\pi}e^{ix\sqrt{a^2+b^2}\cos\theta}\,d\theta \tag{2}$$ and the claim follows by expanding $e^{ix\sqrt{a^2+b^2}z}$ as a Taylor series centered at $z=0$ and performing termwise integration, since: $$ \int_{0}^{2\pi}\cos^{2n}(\theta)\,d\theta = \frac{2\pi}{4^n}\binom{2n}{n}.\tag{3}$$