How do you evaluate the integral $int xe^(-x^2)dx$ from $-oo$ to $oo$ ?

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Answer 1 · 2016-09-23T18:23:59

$int_-∞^∞xe^(-x^2)=0$

Let $color(red)(u=e^(-x^2))$

$du=-2xe^(-x^2)dx$

$color(blue)(-(du)/2=xe^(-x^2)dx)$

So
$intxe^(-x^2)dx$

$intcolor(blue)(-(du)/2)$

$=-color(red)(u)/2+C$ where C is a constant

$=-color(red)(e^(-x^2))/2+C$

Therefore,

$int_-∞^∞xe^(-x^2)=(-e^(-(-∞)^2)/2)-(-e^(-∞^2)/2)$
$int_-∞^∞xe^(-x^2)=(-e^(-∞)/2)-(-e^(-∞)/2)$

Knowing that $e^(-∞)=0$

Then,

$int_-∞^∞xe^(-x^2)=0-0=0$

How do you evaluate the integral int xe^(-x^2)dxint xe^(-x^2)dx from -oo-oo to oooo?