$$\int_{-\infty}^\infty e^{-x²-\frac{1}{x²}}dx$$ if I do the substitution $u=x²+\frac{1}{x²}$ then as $x\to\infty$, $u\to \infty$ and as $x\to -\infty$, $u\to\infty$ since there is x squared. What do I do? I know the function is even, and I can split the integral in $2\int^\infty_0$ but that seems to get the same bounds as well. What do I do?
Please do not tell me a solution/method, I still want to try this integral on my own without hints or anything
If your domain is $]-\infty, \infty[$, the substitution $u=x^2 + \frac{1}{x^2}$ isn't injective, hence, it gives inconsistency when computing the integral. You can solve this problem by splitting the integral.
$J=\int_{-\infty}^{\infty}e^{-x^2-\frac{1}{x^2}}dx.$ since $e^{-x^2-\frac{1}{x^2}}$ is an even function
$J=2\int_0^{\infty}e^{-x^2-\frac{1}{x^2}}dx$
substituting $x=\frac{1}{u}$ :we find $J=2\int_0^{\infty}\frac{1}{u^2}e^{-u^2-\frac{1}{u^2}}du$
from $(1) $ and $(2)$ we have :
$2J=2\int_0^{\infty }(1+\frac{1}{x^2})e^{-x^2-\frac{1}{x^2}}dx\implies J=\int_0^{\infty}(1+\frac{1}{x^2})e^{-((x-\frac{1}{x})^2+2)}dx=e^{-2}\int_0^{\infty}(1+\frac{1}{x^2})e^{-(x-\frac{1}{x^2})^2}dx$
substituting $v=x-\frac{1}{x} \implies dv =(1+\frac{1}{x^2})$ we find : $J=e^{-2}\int_{-\infty}^{\infty}e^{-v}dv=\sqrt{\pi}e^{-2}$
