# How does $\int\limits_0^\infty{{x^{v-1}}{e^{-\frac{\beta}{x}-\gamma x}}dx}=2{({\frac{\beta}{\gamma}})^{\frac{v}{2}}}{K_v}({2\sqrt{\beta\gamma}})$?

I have been reading Gradshteyn and Ryzhik's Table of Integrals, Series, and Products 7th Edition and see the 9th integral in chapter 3 section 3.471.

$$\int\limits_0^\infty{{x^{v-1}}{e^{-\frac{\beta}{x}-\gamma x}}dx}=2{({\frac{\beta}{\gamma}})^{\frac{v}{2}}}{K_v}({2\sqrt{\beta\gamma}})$$

Where $$\operatorname{Real} \left( \beta \right) > 0,\operatorname{Real} \left( \gamma \right) > 0$$

The book does not provide any detail explanations on why the Bessel function just pop up in the answer so are there anyway to solve this integral explicitly step by step ?

Thank you for your enthusiasm !

• This identity is referenced in DLMF – K.defaoite Dec 20 '20 at 0:52

Recall the definition of $$K$$ (second equation after the the graphs https://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions:_I%CE%B1,_K%CE%B1 $$\begin{eqnarray*} K_{\alpha}(x)= \int_0^{\infty} e^{x \cosh t } \cosh( \alpha t) dt. \end{eqnarray*}$$ Now try the substitution $$\begin{eqnarray*} x= \sqrt{ \frac{\beta}{\gamma} } e^t \\ \frac{dx}{x} = dt \end{eqnarray*}$$
and your integral becomes $$\begin{eqnarray*} \int_{- \infty}^{\infty} \left( \sqrt{ \frac{\beta}{\gamma} } e^t\right)^{\nu} e^{- \sqrt{ \beta \gamma} (e^t+e^{-t})} dt \end{eqnarray*}$$ Now split the interval integral at zero and sub $$e^t+e^{-t}=2 \cosh(t)$$ $$\begin{eqnarray*} = \left( \sqrt{ \frac{\beta}{\gamma} } \right)^{\nu} \int_{0}^{\infty} \left( e^{ \nu t} + e^{-\nu t} \right) e^{- 2\sqrt{ \beta \gamma} \cosh (t) } dt \end{eqnarray*}$$ and the result follows.
• I am sorry but where does $\sqrt {\beta \gamma } \cosh \left( t \right)$ at the equal sign in the second last integral from negative infinity to positive infinity come from ?. I am sorry but in my major this function is not popularly known. – Tuong Nguyen Minh Dec 2 '20 at 6:37