Not the Normal Distribution

Consider a more general form of this integral, let $y>0$ be a real number and $\text{I}(y)=\int_{-\infty}^{\infty}e^{-x^2-(\frac{y}{x})^2} \ \text{d}x$ Take derivative with respect to $y$ $\text{I}'(y)=\int_{-\infty}^{\infty} -\frac{2y}{x^2}e^{-x^2-(\frac{y}{x})^2} \ \text{d}x$ Let $u=\frac{y}{x}$ in the $\text{I}'$ $\text{I}'(y)=\int_{\infty}^{-\infty}2e^{-(\frac{y}{u})^2-u^2} \ \text{d}u=-2\int_{-\infty}^{\infty }e^{-(\frac{y}{u})^2-u^2} \ \text{d}u=-2\text{I}(y)$ Therefore, $\text{I}'(y)+2\text{I}(y)=0$ , solve this simple differential equation to get $\text{I}(y)=a \exp(-2y)$ , Note that $a=\text{I}(0)=\int_{-\infty}^{\infty}e^{-x^2} \ \text{d}x=\sqrt{\pi}$ So $\text{I}(y)=\sqrt{\pi} \exp(-2y)$ and for your particular case of $y=1$ $\text{I}(1)=\sqrt{\pi} \exp(-2)$

Write $I=\int_{-\infty}^{\infty}e^{-(x^2+1/x^2)}dx$ . Then, note that, since the integrand is an even function, $I=2\int_{0}^{\infty}e^{-(x^2+1/x^2)}dx$ By a transformation $x\to 1/x$ , we see that the integral becomes, $I=2\int_{0}^\infty \frac{1}{x^2}e^{-(x^2+1/x^2)}dx\\\implies I=\int_0^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x^2+1/x^2)}dx\\\implies I=\int_{0}^{1}\left(1+\frac{1}{x^2}\right)e^{-(x^2+1/x^2)}dx+\int_{1}^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x^2+1/x^2)}dx$ Now, using the transformation $x-1/x\to x$ , we get $I=e^{-2}\int_{-\infty}^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{e^2}$ Thus the answer is $100(0.5+2)=\boxed{250}$ .

The infinities aren't so huge. We may take -26.616 to 26.616 instead. Limit to 0 is 0.

0.239875543936122+ for Sqrt (Pi)/ e^2 = 0.23987554393612289473607300327359...

100 (1/ 2 + 2) = 250