Integrate x.e^(-x^2) from - infinity to + infinity

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- Sep 26th 2005, 07:56 AMbobby77Integration
Integrate x.e^(-x^2) from - infinity to + infinity

- Sep 26th 2005, 10:55 AMEne Dene
$\displaystyle \int_{-\infty}^{\infty} x\cdot e^{-x^2}dx $ ?

If that is so, then the integral is 0, becoase integral of odd function from -infinity to +infinity is 0. - Sep 26th 2005, 11:25 AMticbol
Here is one way.

INT.(-inf.-->+inf.)[x e^(-x^2)]dx

Let u = e^(-x^2)

So du = e^(-x^2)*(-2x dx)

INT.(-inf.-->+inf.)[x e^(-x^2)]dx

= INT.(-inf.-->+inf.)[e^(-x^2)](x dx)

= INT.(-inf.-->+inf.)[e^(-x^2)](-2x dx)*(-1/2)

= (-1/2)INT.(-inf.-->+inf.)[e^(-x^2)](-2x dx)

That is (-1/2)INT.[du]

which is (-1/2)(u) +C

so,

= (-1/2)[e^(-x^2)](-inf.-->+inf.)

= (-1/2)[e^(-(+inf.)^2) -e^(-(-inf.)^2)]

= (-1/2)[e^(-inf.) -e^(-inf)]

= 0 ---------answer.