Integral of exponential and modified Bessel function of second kind

Question

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.

Answer 1

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)))

Answer 2

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 $\mu = 1$), $$ \int_0^\infty e^{-\alpha x} K_\nu(\beta \sqrt{x}) \, d x = \alpha^{-\frac{1}{2}} \beta^{-1} \Gamma\Bigl( 1 + \frac{\nu}{2} \Bigr) \Gamma\Bigl( 1 - \frac{\nu}{2} \Bigr) \exp\Bigl( \frac{\beta^2}{8 \alpha} \Bigr) W_{-\frac{1}{2}, \frac{1}{2} \nu} \Bigl( \frac{\beta^2}{4 \alpha} \Bigr). $$