Is it possible to convert $\sigma = \int_0^\infty e^{-x^2} dx$ to an integral problem over $(0,1)$?

Question

Is it possible obtain a transformation to convert $\theta=\displaystyle\int_0^\infty e^{-x^2}\, dx$ to an integral problem over $(0,1)$?

Answer 1

Sure. Try, for example, with the change of variable

$\displaystyle y = \frac{2}{\pi} \arctan(x); \quad x = \tan\left( \frac{\pi y}{2} \right)$.

In this example, the $\arctan$ function is continuous, when $x \rightarrow \infty$, $y \rightarrow 1$, and when $x \rightarrow 0$, $y \rightarrow 0$.

You can express

$\displaystyle dx = \frac{\pi}{2} \sec^2\left( \frac{\pi y}{2} \right) \, dy$

Therefore, your integral would be

$\displaystyle \int_0^1 dy\,\frac{\pi}{2} \sec^2\left( \frac{\pi y}{2} \right) e^{-\tan^2\left( \frac{\pi y}{2} \right)}$.

Answer 2

Yes. This is the famous Gaussian integral, which, by a simple change in variable, becomes the $\Gamma$ function of argument $\dfrac12$. But this function also possesses a logarithmic form over the unit interval $(0,1)$. In general, we have

$$\int_0^\infty e^{-x^n}dx=\int_0^\infty\sqrt[n]x\cdot e^{-x}~dx=\int_0^1\sqrt[n]{\ln\frac1x}~dx=\Gamma\bigg(1+\frac1n\bigg).$$

$~\ln\dfrac1x$ can also be rewritten as $-\ln x$ or even $|\ln x|$, since it is negative on that interval.