Let's consider the integral:
$$I(a,b)=\int_0^\infty \frac{1}{t}e^{-a^2/t^2-b t} \text{d}t,~~~~a,b>0$$
We can try to use the integral representation of the part of the function inside the integral:
$$ \frac{1}{t}e^{-a^2/t^2} \text{d}t=\frac{2}{\sqrt{ \pi}} \int_0^\infty \cos(2ax) e^{-t^2 x^2} \text{d}x$$
Then we will have:
$$I(a,b)= \frac{2}{\sqrt{ \pi}} \int_0^\infty \cos(2ax) \int_0^\infty e^{-x^2 t^2-bt}~\text{d}t~ \text{d}x$$
The inner integral can be done with the use of complementary error function:
$$x^2t^2+bt=x^2 \left(t^2+ \frac{b}{x^2} t+\frac{b^2}{4x^4} \right)-\frac{b^2}{4x^2}$$
$$\int_0^\infty e^{-x^2 t^2-bt}~\text{d}t=\frac{\sqrt{ \pi}}{2} \frac{1}{x} e^{b^2/(4 x^2)} \text{erfc}~\left(\frac{b}{2x} \right)$$
Finally, we obtain:
$$I(a,b)= \int_0^\infty \frac{1}{x} \cos(2ax) e^{b^2/(4 x^2)} \text{erfc}~\left(\frac{b}{2x} \right)~ \text{d}x$$
Is this correct? Are the two integrals equal?
My problem is that Mathematica can't numerically evaluate the last integral, it gives large values, while the first integral evaluates well and gives small values. (For example, if we take $a=1, b=1$).
Maybe the problem is that the function under the last integral oscillates too much. Or maybe I made a mistake somewhere.