I have the integral:
$$\int_{-\infty}^\infty e^{-x^2} \, dx$$
And I’d like to solve it using differentiation under the integral sign. I understand that I must convert $e^{-x^2}$ to $e^{-x^2}g(x,t)$, where $g(x,t)$ is just some term involving a new parameter $t$. I tried multiplying by $e^{-2xt}$, which doesn’t really seem very logical, however, I just do not know what to put. How do I find the appropriate term to make the whole process work?
let $u=x^2$, $du=2xdx$
$$\int_{-\infty}^\infty e^{-x^2} \, dx$$ $$\int_{-\infty}^\infty e^{-x^2}dx=\int_{0}^{\infty}u^{-\frac{1}{2}}e^{-u}du=\Gamma\left(\frac{1}{2}\right).$$
Here $\Gamma (s)$ is the Gamma function.
$$\Gamma(1-s)\Gamma(s)=\frac{\pi}{\sin\pi s}$$ $$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}.$$ $$\int_{-\infty}^\infty e^{-x^2} \, dx=\sqrt{\pi}$$
The easiest way to compute $\int_{-\infty}^{\infty}e^{-x²}dx$ is to sqaure it and to use Fubini's theorem and polar coordinates: $$\left(\int_{-\infty}^{\infty}e^{-x²}dx\right )^2=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}d\lambda=\int_0^{\infty}e^{-r^2}2\pi rdr=\pi$$ where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}^2$.
