Integrate $\int x^2e^{-x^2}dx$

Question

I need your advice in integrating $\int x^2e^{-x^2}dx$ by parts. I went this way, $$\int x^2e^{-x^2}dx=\int x*xe^{-x^2}dx=\frac{1}{2}\int x e^{-x^2}dx^2$$ I think I need to substitute $x^2=t$. Is that correct?

Answer 1

Yes, the Substitution $t=x^2$ is correct. Now use the Definition of the incomplete gamma function:

$\gamma(s,x) = \int_0^x t^{s-1}e^{-t} dt$. Here, $s$ is an arbitrary Exponent.

Answer 2

Integrating by parts gives \begin{align} \int x^2e^{-x^2}\, dx&=-\frac12\int x \times(-2 x)\:e^{-x^2}\, dx=-\frac12 xe^{-x^2} +\frac12\int e^{-x^2}\, dx \end{align}

Answer 3

Hint: Alternately, you could express $I(k)=\displaystyle\int e^{-kx^2}~dx$ in terms of the error function, and then differentiate both sides with regard to k.