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?

Perfectly correct. The last expression above isn't, though. – Bernard Mar 01 '15 at 13:21 — Bernard, Mar 01 '15 at 13:21
@Bernard what expression is not correct? – dimaastronom Mar 01 '15 at 13:40 — dimaastronom, Mar 01 '15 at 13:40

score 2 · Answer 1 · answered Mar 01 '15 at 13:22

2

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.

answered Mar 01 '15 at 13:22

kryomaxim

2,882

Thx, I ll try it. – dimaastronom Mar 01 '15 at 13:58

score 2 · Answer 2 · answered Mar 01 '15 at 13:22

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}

answered Mar 01 '15 at 13:22

Olivier Oloa

120,989

score 1 · Answer 3 · answered Mar 01 '15 at 20:10

1

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.

answered Mar 01 '15 at 20:10

Lucian

48,334
2
83
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Integrate $\int x^2e^{-x^2}dx$

3 Answers3