computation of $\int_{0}^{\infty}\frac{1}{\theta^{2n+1}}\exp\left(- \frac1{\theta^2}\sum_{i=1}^n x_i^2\right)d\theta$

Question

I encountered this integral in proving that a function is an approximation to the identity. But I don't know how to solve this integral. I would greatly appreciate any help. How can I calculate $$\int_{0}^{\infty} \displaystyle\frac{1}{\theta^{2n+1}}\exp\left(-\frac1{\theta^2}\displaystyle\sum_{i=1}^n x_i^2\right)d\theta\,?$$

Answer 1

Hint: Using a $u$-sub $x = \theta^{-1}$, we get \begin{align} \int^\infty_0 \frac{1}{\theta^{2n+1}} e^{-c/\theta^2}\ d\theta = \int^\infty_0 x^{2n-1}e^{-cx^2}\ dx. \end{align} where $c = \sum x_i^2$. Then use integration by parts.

Answer 2

The best I could come up with is as follows: $-z=\sum_{i=1}^n x_i^2,$ and $ v=1/\theta^2.$ This is a setup for a "$u-$substitution." You get the following: $$\frac{dv}{-2v\sqrt{v}}=d\theta,$$ and your integral becomes $$\int_0^\infty \frac{1}{\theta^{2n+1}}e^\frac{-z}{\theta^2}d\theta=\int_0^\infty v^n\sqrt{v}e^{-zv}\frac{dv}{-2v\sqrt{v}}$$ which yields

$$ -\frac{1}{2}\int_0^\infty v^{n-1}e^{-zv}dv.$$

To me this looks an awful lot like the Gamma function

$$ \Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx,$$ and this conversion might help you with your proof.