Calculating $\int_\Bbb{R}x^2e^{-x^2} dx$ using $\int_\Bbb{R}x^2e^{-tx^2} dx = -\frac{d}{dt}\int_\Bbb{R}e^{-tx^2} dx$

Question

How can I calculate $$ \int_\mathbb{R} x^2e^{-x^2} dx $$ with the help of this Identity $$ \int_\mathbb{R} x^2e^{-tx^2} dx = -\frac{d}{dt}\int_\mathbb{R} e^{-tx^2} dx $$

Answer 1

Let us consider the function $$ f(x,t)=e^{-t x^{2}} $$ which is defined for $x \in \mathbb{R}\setminus \{0\}$ and $t \in \mathbb{R}$. Obviously, $f$ is differentiable in $t$ : $$ f^{\prime}(x,t)=\frac{\partial}{\partial t} f(x, t)=-x^2e^{-tx^2} $$ The derivative $f^{\prime}(x, t)$ obviously has a majorant for all $t \in \mathbb{R}$. $$ g(x)=0, $$ which is integrable on $\mathbb{R}$, since $$ \int_{\mathbb{R}} 0 \hspace{1mm} d x=\int_{-\infty}^{\infty} 0 \hspace{1mm} d x=0 . $$ The function $x \mapsto f(x, t)$ is also integrable on $\mathbb{R}$ since for $t\geq 1$. $$ \int_{-\infty}^{\infty} e^{-t x^{2}} d x\leq \int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{{\pi}}. $$ Thus the function satisfies $$ F(t)=\int_{\mathbb{R}} e^{-t x^{2}} d x $$ all conditions and we obtain for all $t \geq 1$ $$ F^{\prime}(t)=\int_{\mathbb{R}} -x^2e^{-tx^2} d x=\int_{-\infty}^{\infty} -x^2e^{-tx^2} d x. $$ Thus the integral and the derivative are interchangeable and we obtain $$ \int_\mathbb{R} x^2e^{-tx^2} dx =\int_\mathbb{R} -\frac{d}{dt} e^{-tx^2} dx=-\frac{d}{dt} \int_\mathbb{R} e^{-tx^2} dx $$ With the substitution $$ u=\sqrt{t}x \rightarrow \frac{du}{dx}=\sqrt{t} \rightarrow dx=\frac{1}{\sqrt{t}}du $$ we have $$ \frac{1}{\sqrt{t}} \int_\mathbb{R} e^{-u^2}du $$ and we know that the gaussian integral is $\sqrt{\pi}$ so that $$ \int_\mathbb{R} x^2e^{-tx^2} dx =-\frac{d}{dt} \frac{1}{\sqrt{t}} \sqrt{\pi} = \dfrac{\sqrt{{\pi}}}{2\sqrt{t^3}} $$ and for $t=1$ we finally get $$ \int_\mathbb{R} x^2e^{-tx^2} dx= \dfrac{\sqrt{{\pi}}}{2}. $$