$\int_{0}^{\infty}\tau^{-p/2}e^{-1/(2\tau)}\text{ d}\tau$, $p > 2$

Question

Consider, for $p > 2$:

$$\int_{0}^{\infty}\tau^{-p/2}e^{-1/(2\tau)}\text{ d}\tau\text{.}$$ How is this calculated? Apparently the solution to the qualifying exam that I have says this is a "straightfoward calculation."

Here's what I've tried:

1) $$\int_{0}^{\infty}x^{n}e^{-ax}\text{ d}x = \dfrac{\Gamma(n+1)}{a^{n+1}}$$ won't work, since this requires that $n > -1$.

2) Using the definition of the Gamma function doesn't work here - substitute $u = \dfrac{1}{2\tau}$, so $\text{d}u = \dfrac{-1}{2\tau^2}\text{ d}\tau$, and $\text{d}\tau = -2\tau^2\text{ d}u$ and the integral above results in $$\int_{0}^{\infty}-2\tau^{-p/2+2}e^{-u}\text{ d}u\text{.}$$ For the Gamma function to work, $-p/2 + 2 > 0$, which means that $p/2 < 2$, or $p < 4$. Not quite what I'm looking for.

This comes up in probability as being proportional to the expected value of the inverse $\chi^2$ distribution. The Wikipedia indicates to me that $$\dfrac{1}{\Gamma\left(p/2\right)2^{p/2}}\int_{0}^{\infty}\tau^{-p/2}e^{-1/(2\tau)}\text{ d}\tau = \dfrac{1}{p-2}$$ so $$\int_{0}^{\infty}\tau^{-p/2}e^{-1/(2\tau)}\text{ d}\tau = \dfrac{\Gamma\left(p/2\right)2^{p/2}}{p-2}\text{.}$$

Answer 1

Let $\Gamma (x)=\int_0^\infty e^{-t}t^{x-1}\,dx$ for $x>0$. Then, for $p>2$ we have

$$\begin{align} \int_0^\infty \tau^{-p/2}e^{-1/(2\tau)}\,d\tau&=2^{p/2-1}\int_0^\infty e^{-t}t^{p/2-2}\,dt\\\\ &=2^{p/2-1}\Gamma(p/2-1) \tag 1 \end{align}$$

for $p/2-2>-1\implies p>2$ as was assumed.

Using the functional relationship $\Gamma(1+x)=x\Gamma(x)$, we can write $\Gamma(1-p/2)=\frac{\Gamma(p/2)}{p/2-1}=2\frac{\Gamma(p/2)}{p-2}$. Hence, $(1)$ becomes

$$\int_0^\infty \tau^{-p/2}e^{-1/(2\tau)}\,d\tau=\frac{2^{p/2}\Gamma(p/2)}{p-2}$$

for $p>2$ as was to be shown!