How could I solve $\int_{-\infty}^{+\infty} x^2e^{-x^2}dx$ apply special function gamma

Question

I try solve the integral $$\int_{-\infty}^{+\infty} t^2e^{-t^2}dt$$ I do not know but I think that I should apply $gamma\ function$, which is $$ \Gamma (x)=\int_{0}^{\infty} t^{x-1}e^{-t}dt$$

Like the function $f(x)=x^2e^{-x^2}$ is continuous in $0$, I did $$\int_{-\infty}^{+\infty} t^2e^{-t^2}dt=\int_{-\infty}^{0} t^2e^{-t^2}dt+\int_{0}^{+\infty} t^2e^{-t^2}dt$$

but I do not know how I could go..., could someone give me one idea?

Answer 1

Hint: Let $t^2 = u$ then $du = 2tdt$ thus we get

$$2\int_{0}^{\infty}t e^{-t^2} t\,dt = \int_{0}^{\infty} u^{1/2}e^{-u}du = \color{red}{\Gamma\left(\frac{3}{2}\right)}$$

Answer 2

Integration by parts brings that integral to the usual gaussian integral: $$\int_{\mathbb{R}}t^2 e^{-t^2}\,dt = \int_{0}^{+\infty} t\cdot\left(2t e^{-t^2}\right)\,dt = \int_{0}^{+\infty}e^{-t^2}\,dt = \frac{\sqrt{\pi}}{2}.$$

Answer 3

Hint: Evaluate $I(a)=\displaystyle\int_{-\infty}^\infty e^{-ax^2}~dx,$ and then differentiate both sides with regard to a.