I´m working on transforming an improper integral to an integral with limit 0 and 1. I know I can use the following identities, but they just work for limits from 0 to infinity.
Here are the indentities:
$$\theta=\int_0^\infty g(x)\,\mathrm dx,$$ we could apply the substitution $y=1/(x+1),\mathrm dy=-\mathrm dx/(x+1)^2=-y^2\mathrm dx,$ to obtain the identity $$\theta=\int_0^1h(y)\,\mathrm dy,$$ where $$h(y)=\dfrac{g\left(\tfrac1y-1\right)}{y^2}.$$
How can I apply them to the integral below?
$$\theta=\int_{-\infty}^\infty e^{-x^2}\mathrm dx=1.77245385090551602\ldots$$
Since the integrand $e^{-x^2}$ is even, we can write $\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\,dx = 2\int_{0}^{\infty}e^{-x^2}\,dx$, and then apply the transformation to get $\displaystyle\int_{0}^{1}\dfrac{2}{y^2}e^{-\left(\frac{1}{y}-1\right)^2}\,dy$. Of course, there are better ways to evaluate the Gaussian Integral.