Integral of the Laurent series with Bessel function coefficients

95 Views Asked by At

I'm trying to integrate a function of the following form, where $A$ and $B$ are both positive:

$$\int_0^{\infty}\exp\left(\frac{A}{2}\left(\frac{B}{1+x^2}-\frac{1+x^2}{B}\right)\right)dx.$$

My first inclination was to put it into Laurent series form, which I eventually discovered was

$$\int_0^{\infty} \sum_{k=-\infty}^{\infty} J_k(A)\left(\frac{B} {1+x^2}\right)^kdx.$$

Where $J_n(x)$ is the $n$th order Bessel function of the first kind.

Integrating the series directly was useless.

Then, I thought I could try out a complex representation. I know this function is even, and with some basic analysis of the Laurent series shows that if I integrate around a semi circle of radius $R$ pointed up from the $x$-axis, the integral around the contour should be:

$$\oint_C\exp\left(\frac{A}{2}\left(\frac{B}{1+x^2}-\frac{1+x^2}{B}\right)\right)dx = J_1(A)B\pi.$$

I figured I could take the limit as $R$ increased without bound and get my answer, but swiftly realized that the circular contour does not seem to die off to zero as $R$ increases without bound.

My question is either: what is the value of the following integral

$$\lim_{R \to \infty}Ri\int_0^\pi\exp\left(\frac{A}{2}\left(\frac{B}{1+R^2e^{2\theta i}}-\frac{1+R^2e^{2\theta i}}{B}\right)\right)e^{\theta i}d\theta.$$

Or

How else can I evaluate the initial integral? A numerical solution is fine, so long as it covers $A$ and $B$ being both large and close to zero.