# 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.

## 3 Answers

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 $v$.

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 $\nu$.


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.