integration - Integral $int_0^inftyfrac{exp(ialphacos u)-J_0(alpha)}{1+beta u}mathrm{d}u$

I was studying the motion of a particle in a certain magnetic field and one of the quantities that arose was given by the titular integral
$$F(\alpha,\beta)=\int_0^\infty\frac{\exp(i\alpha\cos u)-J_0(\alpha)}{1+\beta u}\mathrm{d}u$$
Here $\alpha\in\mathbb{R}$, $\beta>0$ and $J_0$ is a Bessel function.

The integral does exist. I found the following theorem in a book on real analysis:
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuous on $[a,b]$ while $g:\mathbb{R}\rightarrow\mathbb{R}$ is nonnegative, decreasing and continuously differentiable on $[a,b]$. Then $\exists\xi\in[a,b]$:
$$\int_a^bf(x)g(x)\mathrm{d}x=g(a)\int_a^\xi f(x)\mathrm{d}x$$
Taking into account the fact that $\exp(i\alpha\cos u)-J_0(\alpha)$ is periodic with period $2\pi$ and that its integral over one period is $0$, we can apply the above theorem seperately to the real and imaginary parts of the original integral to find that for all $a,b>0$:
$$\left\vert\int_a^b\frac{\exp(i\alpha\cos u)-J_0(\alpha)}{1+\beta u}\mathrm{d}u\right\vert<\frac{6\pi}{1+\beta a}$$
This implies convergence.

However, I can't find any way to express $F$ in terms of other functions, be it ordinary or special. This wouldn't be too much of a problem if the integrand weren't oscillatory, making numerical evaluation difficult. Therefore, my question is twofold:

• is there a closed form for $F$, perhaps in terms of special functions,


• if not, are there at least any efficient methods for numerical computation of $F$?

Answer

You can also directly apply Dirichlet's test (the integral of the numerator is zero when taken over a period, therefore uniformly bounded on all finite intervals).

We have
$$\cos^{2 k} u = 2^{-2 k} \binom {2 k} k + 2^{1 - 2 k} \sum_{1 \leq j \leq k} \binom {2 k} {k - j} \cos 2 j u, \\ \sum_{k \geq 0} \frac {(i \alpha \cos u)^{2 k}} {(2 k)!} = J_0(\alpha) + \sum_{j \geq 1} \sum_{k \geq j} 2^{1 - 2 k} \binom {2 k} {k - j} \frac {(i \alpha)^{2 k}} {(2 k)!} \cos 2 j u = \\ J_0(\alpha) + 2 \sum_{j \geq 1} (-1)^j J_{2 j}(\alpha) \cos 2 j u.$$
Doing the same calculation for odd indices and integrating termwise, we obtain
$$I(\beta, j) = \int_0^\infty \frac {\cos j u} {1 + \beta u} du = \frac {\left( \frac \pi 2 - \operatorname{Si}(\omega) \right) \sin \omega - \operatorname{Ci}(\omega) \cos \omega} \beta, \quad \omega = \frac j \beta, \\ F(\alpha, \beta) = 2 \sum_{j \geq 1} i^j J_j(\alpha) I(\beta, j).$$

$J_j(\alpha)$ decays superexponentially for a fixed $\alpha$ (roughly speaking, it oscillates when $j < \alpha$ and then decays rapidly).