What is the integral $\int_{-\infty}^{+\infty} x^{-2} e^{-x^2}\,dx$?

Question

In my research, I reached to the $$ \int_{-\infty}^{+\infty}\!\! \frac{1}{\sqrt{2 \pi}x^2}e^{-x^2/2}\, dx. $$

I have tried polar change of systems, integral by parts and change of variables but I did not reach to any solution. I plotted the integrand and I see that it is very near to Dirac delta function. However, I did not found is it equal to it or not?

Answer 1

The integral does not converge, in fact near $x=0$ the function behaves like $1/x^2$ and it's well known that $$\int_0^1\frac1{x^p}\mathrm dx <+\infty \iff p<1$$

Answer 2

Noticing that the integrand is positive and $$\int _{-\infty}^{\infty} x^{-2}e^{-ax^2} dx =2 \int _{0}^{\infty} x^{-2}e^{-ax^2} dx = 2\left[\int _{0}^{1} x^{-2}e^{-ax^2} dx + \int _{1}^{\infty} x^{-2}e^{-ax^2} dx \right]> 2\int _{0}^{1} x^{-2}e^{-ax^2} dx. $$

For any $a>0,$ $$\int _{0}^{1} x^{-2}e^{-ax^2} dx > \int _{0}^{1} x^{-2}e^{-a} dx >+\infty,$$ Therefore $\int _{-\infty}^{\infty} x^{-2}e^{-ax^2} dx$ is divergent for any $a>0$ and hence $\int _{-\infty}^{\infty} x^{-2}e^{-x^2} dx$ is divergent too.