I encountered this integral in my work, and it would be really convenient if it had a closed form in terms of any known special functions (which Mathematica could handle):
$$J(\alpha,\beta)=\int_0^\infty \frac{e^{-x^2} I_0 \left(\beta x \right) d x}{\sqrt{ \alpha^2+x^2}}$$
It's easy to see that:
$$J(\alpha,0)=\frac12 e^{\alpha^2/2} K_0 \left( \frac{\alpha^2}{2} \right)$$
Using series expansion of the Bessel function and a little help from Mathematica, I was able to get a series expression:
$$J(\alpha,\beta)=\frac12 e^{\alpha^2/2} K_0 \left( \frac{\alpha^2}{2} \right)+\frac{\sqrt{\pi}}{16} \beta^2 \sum_{k=0}^\infty \frac{(2k+1)!}{k!^3 (k+1)^2} U \left(\frac{1}{2},-k,\alpha^2 \right) \frac{\beta^{2k}}{4^{2k}}$$
Where $U$ is the confluent hypergeometric function, which in this case (according to Mathematica) is a linear combination of Bessel functions $K_0$ and $K_1$ though I wasn't able to get the general expression for the coefficients yet.
This series is good for small values of $\beta$, but for large $\beta$ I need too many terms, and the computation time is slower than I'd like.
I know that it's simple to get a quadrature formula for the integral, but still, a closed form would be better.
The integral can be thought as a Hankel transform with imaginary argument, but I wasn't able to find it in the tables of Hankel transforms either.
Clarification re: R. Burton's comment:
The final expression has the form:
$$\alpha e^{-\beta^2/2} J(\alpha,\beta)$$
Which should make the expression finite for every choice of variables.
Now that I think about it, I should probably use asymptotics for the Bessel function for large $\beta$:
$$I_0 \left(\beta x \right) \asymp \frac{1}{\sqrt{2 \pi \beta x}}e^{\beta x}$$
Then I get:
$$J(\alpha,\beta) \asymp \frac{1}{\sqrt{2 \pi \beta}} \int_0^\infty \frac{e^{-x^2+\beta x} d x}{\sqrt{x}\sqrt{ \alpha^2+x^2}}, \quad \beta \to \infty$$
Which I still don't know the closed form for.
A more simple integral which does have a closed form:
$$\int_0^\infty e^{-x^2} I_0 \left(\beta x \right) d x= \frac{\sqrt{\pi}}{2} e^{\beta^2/8} I_0 \left(\frac{\beta^2}{8} \right)$$