# Abstract Nonsense

## Another integral

Here is a semi-interesting integral

Problem: Compute $\displaystyle J=\int_0^{\infty}e^{-(x^2+\tfrac{1}{x^2})}dx$

Solution:Clearly

$\displaystyle J=\int_0^1 e^{-(x^2+\tfrac{1}{x^2})}dx+\int_1^{\infty}e^{-(x^2+\tfrac{1}{x^2})}dx$

Let $x=\frac{1}{z}\implies dx=\frac{-dz}{z^2}$ in the first to see that

$\displaystyle J=\int_0^{\infty}\frac{1}{x^2}e^{-(x^2+\tfrac{1}{x^2})}+\int_1^{\infty}e^{-(x^2+\tfrac{1}{x^2})}dx$

Which upon combination gives

$\displaystyle J=\int_1^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x^2+\tfrac{1}{x^2})}dx$

Note though that

$\displaystyle x^2+\frac{1}{x^2}=x^2-2x\frac{1}{x}+\frac{1}{x^2}-2=\left(x-\frac{1}{x}\right)^2-2$

So, letting $\displaystyle z=x-\frac{1}{x}\implies dz=1+\frac{1}{x^2}dx$ we see that

$\displaystyle J=\int_0^{\infty}e^{-z^2-2}dz=\frac{\sqrt{\pi}}{2e^2}$

$\blacksquare$