1
$\begingroup$

Could anyone be able to solve the following integral $\int^{\infty}_{-\infty } |x| e^{-x²} dx$? I know that $\int^{\infty}_{-\infty } |x| e^{-x²} dx = 2 \int^{\infty}_{0 } x e^{-x²} dx$, but I don't know how to solve that. I let this change of variable $u=-x²$, but something was unclear. There is a sign error.

$\endgroup$
  • $\begingroup$ Hint: What is the derivative of $-\frac 12 e^{-x^2}$ ? $\endgroup$ – Fnacool Dec 2 '16 at 3:30
  • $\begingroup$ Just do $u=x^2$ as the substitution to avoid the sign error. $\endgroup$ – NickC Dec 2 '16 at 3:31
5
$\begingroup$

$$\int_{-\infty}^\infty|x|e^{-x^2}dx=2\int_0^\infty xe^{-x^2}dx$$ $$=\int_\infty^0 e^{-x^2}d(-x^2)$$ $$=e^{-x^2}\bigg|^0_\infty$$ $$=1$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.