0
$\begingroup$

How do I evaluate the following integral, the answer according to Wolfram Alpha is $2$, but I keep on getting $0$ after using integration by parts.$$\frac12\int_{-\infty}^\infty x^2e^{-|x|}\ dx$$

$\endgroup$
  • 1
    $\begingroup$ Split the integral for the region $-\infty<x<0\ $ and $\ 0<x<\infty$. $\endgroup$ – Tunk-Fey Jun 10 '14 at 11:13
  • 1
    $\begingroup$ It equals $\int_{0}^{\infty}x^{2}e^{-x}dx=\Gamma\left(3\right)=2$ because of symmetry. $\endgroup$ – drhab Jun 10 '14 at 11:14
  • $\begingroup$ @Tunk-Fey That's probably not the asker's issue, since he/she is getting 0 as a result. $\endgroup$ – Ataraxia Jun 10 '14 at 11:15
2
$\begingroup$

$$\frac12\int_{-\infty}^\infty x^2e^{-|x|}dx = \frac12\int_{-\infty}^0x^2e^{x}dx +\frac12\int_{0}^\infty x^2e^{-x}dx \\ = \int_{0}^\infty x^2e^{-x}dx = [x^2(-e^{-x})]_{0}^\infty + 2\int_{0}^\infty xe^{-x} dx \\ =0+ 2[x(-e^{-x})]_{0}^\infty + 2\int_{0}^\infty e^{-x}dx = 0+0+2[(-e^{-x})]_{0}^\infty =2 $$

$\endgroup$
2
$\begingroup$

Hint: By parity, $$ \frac12\int_{-\infty}^\infty x^2e^{-\vert x \rvert}dx = \int_{0}^\infty x^2e^{- x }dx $$ Then, use integration by parts, now that there is no absolute values to cause trouble.

It cannot be zero, as your function (which happens to be the second raw moment of a Laplace distribution, incidentally) is always positive (except at $0$).

$\endgroup$
2
$\begingroup$

Hint : $$ \int_{-\infty}^\infty x^2e^{-|x|}\ dx=\int_{-\infty}^0 x^2e^{x}\ dx+\int_{0}^\infty x^2e^{-x}\ dx=2\int_{0}^\infty x^2e^{-x}\ dx. $$

$\endgroup$

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.