$\int_0^\infty t^{-3/2} \exp(a/t - bt) dt$ type integral How would one go about evaluating the following integral?
$$ \int_{0}^{\infty} \frac{Q}{\sqrt{4\pi t^3}} \exp\Big(-\frac{(Q - \alpha t)^2}{4t} - t(U^2 - \alpha U) \Big) dt$$
We have that $Q, \alpha > 0$, while $U$ is complex. Mathematica actually evaluates the whole thing to simply $e^{Q(\alpha - U)}$, under the assumption that $\Re(2U-\alpha)^2 > 0$.
My first attempt was to define $f(Q)$ as the integral above and show that it satisfies the differential equation of the exponential function; after differentiating the integrand I don't really see a way to recombine the terms though. I get something akin to
$$f'(Q) = \frac{1}{Q} f(Q) + \frac{\alpha}{2} f(Q) - \int_{0}^{\infty} \frac{Q^2}{4\sqrt{\pi t^5}} \exp\Big(-\frac{(Q - \alpha t)^2}{4t} - t(U^2 - \alpha U) \Big)  $$
which is nowhere near $(\alpha - U) f(Q)$.
I've also looked through various integral tables without success, I only found a reference that integrals of products of rational and exponential functions could be resolved as sums of the $\mathrm{Ei}$ function, but do not see how in this case.
 A: Assume that $(2U - \alpha)^{2}$ is positive.
A more convenient way to express the integral is $$ I =\frac{Q}{2 \sqrt{\pi}} \, \exp \left(\frac{\alpha Q}{2} \right) \int_{0}^{\infty}\exp \left(- \frac{\left(2U- \alpha\right)^{2}}{4}t- \frac{Q^{2}}{4t} \right) \, \frac{\mathrm dt}{t^{3/2}}. $$
Making the substitution $w = \frac{\left(2U- \alpha\right)^{2}}{4}t$, we get $$I = \frac{Q}{2 \sqrt{\pi}} \, \exp \left(\frac{\alpha Q}{2} \right) \frac{4}{(2U- \alpha)^{2}} \left( \frac{(2U - \alpha)^{2}}{4} \right)^{3/2}\int_{0}^{\infty} \exp \left(-w - \left(\frac{Q(2U- \alpha)}{2} \right)^{2}\frac{1}{4w}\right) \, \mathrm d w. $$
Then relating $I$ to a common integral representation of the modified Bessel function of the second kind, we get
$$\begin{align}I &= \frac{Q}{2 \sqrt{\pi}} \, \exp \left(\frac{\alpha Q}{2} \right) \frac{2U - \alpha}{2} \, 2 \left(\frac{Q(2U- \alpha)}{4} \right)^{-1/2} K_{1/2} \left(\frac{Q(2U- \alpha)}{2} \right) \\ &= \frac{Q}{2 \sqrt{\pi}} \, \exp \left(\frac{\alpha Q}{2} \right)(2U - \alpha)  \left(\frac{Q(2U- \alpha)}{4} \right)^{-1/2} \sqrt{\frac{\pi}{2}} \, e^{-Q(U - \alpha/2)} \left( \frac{Q(2U- \alpha)}{2} \right)^{-1/2}  \\ & = e^{Q(\alpha - U)}, \end{align} $$
where I used the fact that $K_{1/2}(z) = \sqrt{\frac{\pi}{2}} \frac{e^{-z}}{\sqrt{z}}$.
To argue that the result holds for $\Re(2U -\alpha)^{2} >0$, we can use the fact that the Laplace transform is analytic in the region of absolute convergence, and then appeal to the identity theorem.
