I am interested numerically calculating the following integral (it's rather unstable): $$ U(p,q)= \frac{2}{\pi} \frac{1}{p q} \int_0^{\infty} \sin(pr) \sin(qr) V(r) dr $$ $V(r)$ is some function with the behavior: $$ \lim_{r\to 0^+} V(r)=\infty\;\;\text{and}\;\;\lim_{r\to \infty} V(r)=0 $$ And of course it has to go to zero fast enough for the integral to converge, for example $$V(r)= \frac{\mathrm{e}^{-r}}{r}$$
In fact the above $V(r)$ has a nice analytic solution for testing: $$ \frac{1}{pq}\int_0^{\infty } \frac{e^{-r}}{r} \sin (p r) \sin (q r) \, dr=\frac{1}{4pq} \left[\ln \left((p+q)^2+1\right)-\ln \left((p-q)^2+1\right)\right] $$
In general $U(p,q)$ is unstable. One thing I have tried to help this for large $p$ and $q$ is to WLOG let $p>q$ and $b=\frac{q}{p}$ then under the variable transformation $x=pr$ we have: $$ U(p,q)= \frac{2}{\pi} \frac{1}{p^{2}q} \int_0^{\infty}\sin{(x)}\sin{(bx)} V\left( \frac{x}{p} \right) dx $$
This helps with numeric stability, but not enough! I am still getting unstable results for pretty much every method.It was suggested in passing to me that I use a fast Fourier transform, but I'm not sure if that is actually useful.
What techniques can I use to help make this integral more numerically stable?
If you can compute Fourier transform of $V(|r|)$ then $U(p,q)$ can be rather simply expressed in terms of it: $$ \int_0^\infty \sin pr \sin qr V(r) dr = \int_0^\infty \frac{1}{2}\left[ \cos (p-q) r - \cos (p+q) r \right] V(r) dr = \\ = \frac{1}{2}\int_0^\infty \cos ((p-q) r) V(r) dr - \frac{1}{2}\int_0^\infty \cos ((p+q) r) V(r) dr = \\ = \frac{1}{4} \int_{-\infty}^\infty \cos ((p-q) r) V(|r|) dr - \frac{1}{4} \int_{-\infty}^\infty \cos ((p+q) r) V(|r|) dr. $$ And thus $$ U(p, q) = \frac{\hat V(p-q) - \hat V(p+q)}{2\pi p q} $$ where $$ \hat V(s) = \int_{-\infty}^\infty V(r) e^{isr} dr. $$