I think this is a reverse product rule but I could not figure out how to reverse this.

$$\int_0^\infty \frac{\beta^\alpha}{\Gamma(\alpha)}{\theta^{-(\alpha+1)}}*e^{-\frac{\beta}{\theta}}d\theta$$

I pulled out the constants but then got stuck here: $$\frac{\beta^\alpha}{\Gamma(\alpha)}\int_0^\infty {\theta^{-(\alpha+1)}}*e^{-\frac{\beta}{\theta}}d\theta$$

$\alpha$ and $\beta$ are constants while $\Gamma$ is a function.

  • $\begingroup$ You want "integration by parts". $\endgroup$ – Michael Lugo Mar 10 '16 at 22:23
  • $\begingroup$ $\Gamma(\alpha)$ is also a constant So this is really just the interval $\int_0^\infty x^{a}e^{-b/x}dx$ modulo a constant. $\endgroup$ – Gregory Grant Mar 10 '16 at 22:24
  • $\begingroup$ yea gamma only depends on alpha, that's why i pulled it out. i'm going to try integration by parts $\endgroup$ – O.rka Mar 10 '16 at 22:35

The change of variable $\;u:=\dfrac {\beta}{\theta}\;$ should help much here giving $\;\theta:=\dfrac {\beta}u,\;d\theta=-\dfrac{\beta}{u^2}du\;$ and :

\begin{align} I(\alpha,\beta)&:=\dfrac{\beta^\alpha}{\Gamma(\alpha)}\int_0^{\infty} {\theta^{-(\alpha+1)}}\;e^{-\large{\frac{\beta}{\theta}}}d\theta\\ &=\dfrac{\beta^\alpha}{\Gamma(\alpha)}\int_{\infty}^0 \left(\frac {\beta}u\right)^{-(\alpha+1)}\;e^{-u}\frac {-\beta}{u^2}\,du\\ &=\dfrac{\beta^\alpha}{\Gamma(\alpha)}\int_0^{\infty} \beta\left(\frac{u}{\beta}\right)^{\alpha+1}\;e^{-u}\frac 1{u^2}\,du\\ &=\dfrac{1}{\Gamma(\alpha)}\int_0^{\infty} u^{\alpha-1}\;e^{-u}\,du\\ &=\dfrac{\Gamma(\alpha)}{\Gamma(\alpha)},\quad(*)\\ &=1 \end{align} $(*)$ from the definition of the $\Gamma$ function for $\,\alpha>0,\;\beta>0$.

  • $\begingroup$ woah...i never would have thought to do that. $\endgroup$ – O.rka Mar 11 '16 at 1:21
  • 1
    $\begingroup$ @O.rka: Well as indicated by Gregory Grant this had the form $\;\displaystyle \int_0^\infty x^{a}e^{-b/x}dx\;$ looking like a $\Gamma$ function except for the $\dfrac bx$ term so... Fine continuation, $\endgroup$ – Raymond Manzoni Mar 11 '16 at 8:43

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