I am trying to numerically integrate
$$\int_0^a f(x) \int_{\sqrt{x}}^\infty \frac{\exp(-u^2)}{\sqrt{u^2-x}}du dx$$ where a is some positive real number and f(x) is some well behaved function. The inner integral is singular at 0 so the outer integral must be done using a method that can handle a singular end point. The inner integrand is also singular at the lower end point so it also needs to be handled in a similar way.
I have seen an integral like this done in a paper that claimed a "romberg method adapted to a singular endpoint" was used but I have not been able to find much on this method. I was curious if someone could explain how to handle this numerically. I was able to find this paper but it is not easy to see how to apply it to this problem.
An additional comment the original question: As was pointed out below the singularity at the lower bound of the u integral at $\sqrt{x}$ is removable (see post below). While this does simplify my task, the question on how to numerically deal with a singular endpoint remains since the lower limit for the x integral is a singular endpoint.
In this case you can eliminate the singularity for the inner integral at $u = \sqrt{x}$ with a variable transformation. First let $s = u^2$, then
$$\int_{\sqrt{x}}^{\infty} \frac{\exp{(-u^2)}}{\sqrt{u^2-x}}du = \int_{x}^{\infty} \frac{\exp{(-s)}}{2\sqrt{s}\sqrt{s-x}}ds.$$
Then letting $t^2 = s-x$, we get
$$\int_{\sqrt{x}}^{\infty} \frac{\exp{(-u^2)}}{\sqrt{u^2-x}}du = \int_{0}^{\infty} \frac{\exp{[-(x+t^2)]}}{\sqrt{x+t^2}}dt.$$
That may help.
In general, if $f \in C([0,1])$, then for $n \geq 2$ the change of variable $t^n=x$ transforms an improper into a proper integral:
$$\int_{0}^{1} x^{-\frac1{n}}f(x)dx=n\int_{0}^{1} t^{n-2}f(t^n)dt$$
Furthermore, the integrand decays rapidly as $u \rightarrow \infty,$ so the semi-infinite domain should not present much difficulty. There are many approaches for infinite domains, in general, including Gauss-Laguerre and Gauss -Hermite quadrature.