Integral of exponential and modified Bessel function of second kind I want to do the following
\begin{equation}
\int_{0}^{\infty}{\rm e}^{-\alpha x}\,
\operatorname{K}_{\nu}\left(\beta\,\sqrt{\, x\,}\,\right){\rm d}x.
\end{equation}
As per G&R ${\bf 6.611.3}$
\begin{equation}
\int_{0}^{\infty}{\rm e}^{-\alpha x}\,
\operatorname{K}_{\nu\,}\left(\beta\, x\right)\,{\rm d}x
=
\frac{\pi\sin\left(\nu\,\theta\right)}{\beta\sin\left(\nu \,\pi\right)\sin\left(\theta\right)}
\end{equation}
and $\displaystyle\cos\left(\theta\right)=\frac{\alpha}{\beta}$.
How do I do the modifications? Please guide.
 A: Two approaches, based on different G&R integrals (they lead to the same result, but you get the illusion of choice!).
First, 6.614.3 says
$$
\int_0^\infty e^{-\alpha x} K_{2 \nu}(2 \sqrt{\beta x}) \, d x = \frac{e^{\frac{1}{2} \frac{\beta}{\alpha}}}{2 \sqrt{\alpha \beta}} \Gamma(\nu + 1) \Gamma(1 - \nu) W_{-\frac{1}{2}, \nu} \Bigl( \frac{\beta}{\alpha} \Bigr) 
$$
for $\mathrm{Re}(\alpha) > 0$ and $|\mathrm{Re}(\nu)| < 1$.
Next (as suggested by Gary in comments), 6.631.3 says
$$
\int_0^\infty x^\mu e^{-\alpha x^2} K_\nu(\beta x) \, d x = \frac{1}{2} \alpha^{-\frac{1}{2} \mu} \beta^{-1} \Gamma\Bigl( \frac{1 + \nu + \mu}{2} \Bigr) \Gamma\Bigl( \frac{1 - \nu + \mu}{2} \Bigr) \exp\Bigl( \frac{\beta^2}{8 \alpha} \Bigr) W_{-\frac{1}{2} \mu, \frac{1}{2} \nu} \Bigl( \frac{\beta^2}{4 \alpha} \Bigr)
$$
for $\mathrm{Re}(\mu) > | \mathrm{Re}(\nu)| - 1$.
In both of these, $W_{\lambda, \kappa}(z)$ is one of the Whittaker functions.
In view of the first, we have (replace $\beta$ with $\beta^2 / 4$, $\nu$ with $\nu/2$):
$$ 
\int_0^\infty e^{-\alpha x} K_\nu(\beta \sqrt{x}) \, d x = \frac{e^{\frac{1}{8} \frac{\beta^2}{\alpha}}}{\sqrt{\alpha} \beta} \Gamma\Bigl( \frac{\nu}{2} + 1 \Bigr) \Gamma\Bigl( 1 - \frac{\nu}{2} \Bigr) W_{-\frac{1}{2}, \frac{\nu}{2}}\Bigl( \frac{\beta^2}{4 \alpha} \Bigr). 
$$
In view of the second, first (again as per Gary's comment) perform the change of variables $t^2 = x$, $2 t \, d t = d x$, so that
$$
\int_0^\infty e^{-\alpha x} K_\nu(\beta \sqrt{x}) \, d x = 2 \int_0^\infty t e^{-\alpha t^2} K_\nu( \beta t) \, d t
$$
and so by 6.631.3 (idenfity $\mu = 1$),
$$
\int_0^\infty e^{-\alpha x} K_\nu(\beta \sqrt{x}) \, d x = \alpha^{-\frac{1}{2}} \beta^{-1} \Gamma\Bigl( 1 + \frac{\nu}{2} \Bigr) \Gamma\Bigl( 1 - \frac{\nu}{2} \Bigr) \exp\Bigl( \frac{\beta^2}{8 \alpha} \Bigr) W_{-\frac{1}{2}, \frac{1}{2} \nu} \Bigl( \frac{\beta^2}{4 \alpha} \Bigr). 
$$
A: Using only Mathematica 13.1:
$$\int_0^{\infty } \exp (-\alpha  x) K_v\left(\beta  \sqrt{x}\right) \, dx=-\frac{e^{\frac{\beta ^2}{8 \alpha }} \sqrt{\pi } \beta  \left(K_{\frac{1}{2} (-1+v)}\left(\frac{\beta ^2}{8
   \alpha }\right)-K_{\frac{1+v}{2}}\left(\frac{\beta ^2}{8 \alpha }\right)\right) \csc \left(\frac{\pi  v}{2}\right)}{8 \alpha ^{3/2}}$$
If:   $-2<\Re(v)<2\land \Re(\alpha )\geq 0\land \Re(\beta )>0$
MMA code:
Integrate[ Exp[-\[Alpha]*x] BesselK[v, \[Beta] Sqrt[x]], {x, 0,  Infinity}] == -(( E^(\[Beta]^2/(8 \[Alpha])) Sqrt[\[Pi]] \[Beta] (BesselK[1/2 (-1 + v), \[Beta]^2/( 8 \[Alpha])] -  BesselK[(1 + v)/2, \[Beta]^2/(8 \[Alpha])]) Csc[(\[Pi] v)/2])/( 8 \[Alpha]^(3/2)))
