My goal is to evaluate
$I = \int_{-\infty}^\infty (1+bx^2)^{-k/2} e^{-ax^2} dx$
where $a,b>0$ and $k\in\mathbb{N}$. My first thought was to use the fact that we are integrating an even function on a symmetric interval and then substitute $y=x^2$, giving
$I = \int_0^\infty y^{-1/2}(1+by)^{-k/2}e^{-ay}dy$
Mathematica tells me that this evaluates to
$I=\frac{\Gamma \left(\frac{1-k}{2}\right) \left(\frac{b}{a}\right)^{-\frac{k}{2}} ~ _1F_1\left(\frac{k}{2};\frac{k+1}{2};\frac{a}{b}\right)}{\sqrt{a}}+\frac{\sqrt{\pi } \Gamma \left(\frac{k-1}{2}\right) ~_1F_1\left(\frac{1}{2};\frac{3-k}{2};\frac{a}{b}\right)}{\sqrt{b} \Gamma \left(\frac{k}{2}\right)}$
But I haven't been able to figure out what this follows from. Identity 16.5.3 from here can be used to find a solution in terms of$~_2F_0$, but I would prefer to not have a result involving an asymptotic series.
Are there any known results that would help me to understand the computer calculation?
For any future readers, this is actually pretty straightforward. The integral
$=\int_0^\infty y^{-1/2}(1+by)^{-k/2}e^{-ay}dy$
is known to be an integral representation of the Tricomi confluent hypergeometric function (see 13.4.4 here).
Then we can use the fact that
$U(a,b,z) = \frac{\Gamma(1-b)}{\Gamma(a+1-b)}~_1F_1(a,b,z) + \frac{\Gamma(b-1)}{\Gamma(a)}z^{1-b}~_1F_1(a+1-b,2-b,z)$
to get the desired result.