I have recently been trying to evaluate some integrals involving the modified Bessel function $K_0(x)$. The specific integrals are
$$L(x,u) = \int_0^{1} K_0\left( 2x \sqrt{r(1-r)} \right) \exp(2ixur) dx $$
where $x>0$ and $u \in \mathbb{R}$. I have found some (likely) answers in some special cases, namely $u=0$ and $u=1$. By expanding as power series and monkeying around with the terms, it appears that
$$L(x,0) = \frac{ \mathrm{Ei}(x)\exp(-x) - \mathrm{Ei}(-x)\exp(x) }{2x}$$
and
$$L(x,1) = \exp(ix)\left( \frac{\sin(x)}{x} - \mathrm{Ci}(x) \right),$$
where $\mathrm{Ei}$ is the exponential integral and $\mathrm{Ci}$ is the cosine integral. I say these are likely answers because they seem to work to arbitrary precision, but I haven't proved them rigorously.
It seems like it should be possible to get a closed form in terms of both $x$ and $u$, given the relationships that exist between exponential and trig integrals, but I haven't been able to figure it out.
Important note: for my purposes, it is sufficient to get an approximation to these integrals that converges faster than Mathematica's generic NIntegrate function. That said, a closed form would certainly be much more pleasing :)
One idea that seemed very nice for the $u=0$ case is to use Graf's formula, as in this answer. After some substitution, the integral looks like
$$L(x,0) = \frac{1}{2}\int_0^{\pi} K_0 \left( x \sin(\theta) \right) \sin(\theta) d\theta$$
and, using Graf, we get
$$L(x,0) = \sum_{m=-\infty}^{\infty}\frac{1}{1-4m^2} K_m\left(\frac{x}{2}\right) I_m\left( \frac{x}{2} \right)$$
but I can't see how to show this gives the exponential integral expression above. And in the case of $u\ne 0$ this substitution makes the exponential much worse to deal with.