Query related to Eq. 3.471.9 of Book of Gradeshteyn (Integration tables series and products) Equation no. 3.471.9 of Integral series and products (By Gradeshteyn) is written below $$\int_0^{\infty}x^{v-1}e^{-\frac{\beta}{x}-\gamma x}dx=2\left(\frac{\beta}{\gamma}\right)^{\frac{v}{2}}K_{v}(2\sqrt{\beta \gamma})$$ although it is mentioned that $Re(\beta)>0$ and $Re(\gamma)>0$ there is nothing written about $v$. So my question related to the values of $v$. Is the above equation valid for all possible real values of $v$? And if it is not valid then how to solve the above integral for general real values of $v$. Many thanks in advance.
 A: The integral will converge as long as $\text{Re}(\beta) > 0$ and $\text{Re}(\gamma) > 0$, regardless of the value of $v$.  Both sides are analytic as functions of $v$ for fixed $\beta, \gamma$.  So the equation should work for all $v$.
A: To see that the integral is valid for all possible values of $\nu$, rewrite the integral as
$$\gamma^{-\nu} \int_0^{\infty} dx \, x^{\nu-1} \, e^{-x - \frac{\beta \gamma}{x}} $$
Let $u = x+\beta \gamma/x$, then 
$$x=\frac12 \left (u \pm \sqrt{u^2-4 \beta \gamma} \right ) $$
$$dx=\frac12 \left (1 \pm \frac{u}{\sqrt{u^2-4 \beta \gamma}} \right ) du $$
Then the integral may be rewritten as (see this answer)
$$ \gamma^{-\nu}\int_{\infty}^{2 \sqrt{\beta \gamma}} du \, \left (1 - \frac{u}{\sqrt{u^2-4 \beta \gamma}} \right ) \left (u - \sqrt{u^2-4 \beta \gamma} \right )^{\nu-1} e^{-u} \\+ \gamma^{-\nu}\int_{2 \sqrt{\beta \gamma}}^{\infty} du \, \left (1 + \frac{u}{\sqrt{u^2-4 \beta \gamma}} \right ) \left (u + \sqrt{u^2-4 \beta \gamma} \right )^{\nu-1} e^{-u} $$
which may be simplified to
$$ (2 \gamma)^{-\nu} \int_{2 \sqrt{\beta \gamma}}^{\infty} du \, \left (u^2-4 \beta \gamma \right )^{-1/2} \left [\left (u + \sqrt{u^2-4 \beta \gamma} \right )^{\nu} + \left (u - \sqrt{u^2-4 \beta \gamma} \right )^{\nu} \right ] e^{-u} $$
which is, after a rescaling,
$$\left (\frac{\beta}{\gamma} \right )^{\nu/2}  \int_1^{\infty} dv \, \left (v^2-1 \right )^{-1/2} \left [\left (v + \sqrt{v^2-1} \right )^{\nu} + \left (v + \sqrt{v^2-1} \right )^{-\nu} \right ] e^{-2 \sqrt{\beta \gamma} v} $$
which becomes, after subbing $v=\cosh{t}$, 

$$2 \left (\frac{\beta}{\gamma} \right )^{\nu/2} \int_0^{\infty} dt \, \cosh{(\nu t)} \, e^{-2 \sqrt{\beta \gamma} \cosh{t}} $$

The integral according to the DLMF (10.32.9) is indeed $K_{\nu} \left (2 \sqrt{\beta \gamma} \right ) $ and the result in G & R is verified for $\operatorname{Re}{\beta} \gt 0$ and $\operatorname{Re}{\gamma} \gt 0$.  
As far as $\nu$ goes, it appears that the result is valid for all possible real and complex values of $\nu$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}x^{\nu - 1}\expo{-\beta/x - \gamma x}\dd x & =
\int_{0}^{\infty}x^{\nu - 1}\exp\pars{-\root{\beta\gamma}\bracks{\root{\beta \over \gamma}{1 \over x} + \root{\gamma \over \beta}x}}\,\dd x
\end{align}

Set $\ds{x = \root{\beta \over \gamma}\expo{-\theta}}$:

\begin{align}
\int_{0}^{\infty}x^{\nu - 1}\expo{-\beta/x - \gamma x}\dd x & =
\int_{\infty}^{-\infty}\pars{\beta \over \gamma}^{\pars{\nu - 1}/2}
\expo{\pars{1 - \nu}\theta}
\expo{-2\root{\beta\gamma}\cosh\pars{\theta}}\bracks{-\pars{\beta \over \gamma}^{1/2}\expo{-\theta}}\dd\theta
\\[5mm] & =
\pars{\beta \over \gamma}^{\nu/2}\int_{-\infty}^{\infty}\expo{-\nu\theta}
\expo{-2\root{\beta\gamma}\cosh\pars{\theta}}\dd\theta
\\[5mm] & =
\pars{\beta \over \gamma}^{\nu/2}\int_{0}^{\infty}\pars{\expo{-\nu\theta} +
\expo{\nu\theta}}
\expo{-2\root{\beta\gamma}\cosh\pars{\theta}}\dd\theta
\\[5mm] & =
2\pars{\beta \over \gamma}^{\nu/2}\int_{0}^{\infty}\cosh\pars{\nu\theta}
\expo{-2\root{\beta\gamma}\cosh\pars{\theta}}\dd\theta
\\[5mm] &=
\bbx{2\pars{\beta \over \gamma}^{\nu/2}
\,\mrm{K}_{\nu}\pars{2\root{\beta\gamma}}} \\ &
\end{align}

where $\ds{\,\mrm{K}_{\nu}}$ is a Modified Bessel Function.

