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.

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\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.

