$\int_{-\pi/2}^{\pi/2} \cos(a \cos\theta) e^{im\theta} e^{-ib\sin\theta} \mathrm{d}\theta $ Integration

Question

I am struggling to find the integration of the expression below, $$\int_{-\pi/2}^{\pi/2} \cos(a \cos\theta) e^{im\theta} e^{-ib\sin\theta} \mathrm{d}\theta $$ where $a$ and $b$ are arbitrary constant and $m$ is an integer.

I have found the result for $m = 0$ which contains $J_0(\sqrt{a^2 + b^2})$ term. I think for this integration, it will involve the $m$-th order of Bessel function of the first kind.

Answer 1

You integral is $\frac{1}{2}\left(I(a,b)+I(-a,b)\right)$ where:

$$I(a,b)=\int_{-\pi/2}^{\pi/2}\exp(im\theta)\exp\left(ia\cos\theta-ib\sin\theta\right)\,d\theta$$ and by writing $a\cos\theta-b\sin\theta$ as $\rho\cos(\theta+\varphi)$, with $\rho=\sqrt{a^2+b^2}$ and $\varphi=\arctan\frac{b}{a}$,

$$ I(a,b) = e^{-im\varphi}\int_{\varphi-\pi/2}^{\varphi+\pi/2}\exp(im\theta)\exp(i\rho\cos(\theta))\,d\theta $$ where $$ e^{i\rho\cos\theta} = J_0(\rho)+2\sum_{n\geq 1}i^n J_n(\rho)\cos(n\theta) $$ gives:

$$ I(a,b) = e^{-im\varphi}\left[\pi\,J_0(\rho)+2\sum_{n\geq 1}i^n J_n(\rho)\int_{\varphi-\pi/2}^{\varphi+\pi/2}\exp(im\theta)\cos(n\theta)\,d\theta \right]$$

and your integral is represented as a fast-convergent series.