I'm trying to calculate
$$
I = \int_0^\infty dq\ \frac{e^{-qm^2 – R^2/(4q)}}{q^{d/2}}, \quad R, m \geq 0,\ d \geq 1
\tag1$$
To do it, I apply the change of variables,
$$
q = \alpha^2, \quad dq = 2\alpha d\alpha
\tag2$$
This renders,
$$
I = \int_0^\infty d\alpha\ \alpha^{1 – d}e^{-\alpha^2m^2 – R^2/(4\alpha^2)}
\tag3$$
I tryied to solve it and I looked for it in books and even Wolfram-Alpha, but I don't get any solutions. Any suggestions to solve Eq. (3)?
EDITION
Knowing the solution for $d = 0, d = 3$, would it be correct to use them and take derivatives respect to $R^2$ to generate the solutions to the rest $d$'s? For that pair of $d$-values, the solutions depend on $\sqrt{R^2} = R$, so would it be right to use
$$
\frac{\partial}{\partial(R^2)} = \frac{\partial R}{\partial(R^2)}\frac{\partial}{\partial R} = \Big(\frac{\partial R^2}{\partial R}\Big)^{-1}\frac{\partial}{\partial R} = \frac{1}{2R}\frac{\partial}{\partial R}
$$
or not? Probably it is a naive question, but I would like to have a second opinion. Actually, this is the same as integrating those results respect to $m^2$.
Best Answer
It is known that the modified Bessel function of the second kind can be written in an integral form
$$ K_{\nu}(z) = \frac{z^{\nu}}{2^{\nu+1}}\int_{0}^{\infty}\frac{e^{-t-z^{2}/4t}}{t^{\nu+1}}\,\mathrm{d}t. $$
This post has some more information on its derivation. Identifying $\nu = d/2-1$ and $z = Rm$, one can write the integral as
$$ \int_{0}^{\infty}\frac{e^{-m^{2}q-R^{2}/4q}}{q^{d/2}}\,\mathrm{d}q = 2\left(\frac{2m}{R}\right)^{d/2-1}K_{d/2-1}(Rm). $$