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- Thread starter Mr Davis 97
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Stephen Tashi

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Try integration by parts.

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What I found was that ##\int_{-\infty}^{\infty}e^{-x^2} ~dx = 2\int_{-\infty}^{\infty}x^2 e^{-x^2} ~dx##, so by original integral is ##\sqrt{\pi} / 2##. To do this though I had to start from the original Gaussian integral and integrate by parts to get the integral that I have. Is that trick of integrating by parts the original Gaussian integral to get what I want what you had in mind?Try integration by parts.

- #4

Stephen Tashi

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Is that trick of integrating by parts the original Gaussian integral to get what I want what you had in mind?

No, I was thinking of starting with the original problem and using ##f(x) = x, g'(x) = x e^{-x^2} ##.

- #5

lurflurf

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for ##t+t^\ast>0\\

\int_{-\infty}^\infty \mathrm{f}(x^2)e ^{-t x^2} dx=\mathrm{f}\left(-\dfrac{d}{dt}\right)\int_{-\infty}^\infty e ^{-t x^2} dx%=\mathrm{f}\left(-\dfrac{d}{dt}\right)\sqrt{\frac{\pi}{t}}##

your case is f(x)=x

the case f(x)=x^n imvolves (2n-1)!!

that comes up all the time in

chemistry http://www.colby.edu/chemistry/PChem/notes/Integral.pdf

psychology http://www.umich.edu/~chem461/Gaussian Integrals.pdf

http://www.math.tamu.edu/~manshel/papers/Hermite-integrals.pdf

http://mathworld.wolfram.com/GaussianIntegral.html

\int_{-\infty}^\infty \mathrm{f}(x^2)e ^{-t x^2} dx=\mathrm{f}\left(-\dfrac{d}{dt}\right)\int_{-\infty}^\infty e ^{-t x^2} dx%=\mathrm{f}\left(-\dfrac{d}{dt}\right)\sqrt{\frac{\pi}{t}}##

your case is f(x)=x

the case f(x)=x^n imvolves (2n-1)!!

that comes up all the time in

chemistry http://www.colby.edu/chemistry/PChem/notes/Integral.pdf

psychology http://www.umich.edu/~chem461/Gaussian Integrals.pdf

http://www.math.tamu.edu/~manshel/papers/Hermite-integrals.pdf

http://mathworld.wolfram.com/GaussianIntegral.html

Last edited:

- #6

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$$ I(x) =\int_{-\infty}^{\infty}x^2e^{-x^2}dx=2\int_{0}^{\infty}x^2e^{-x^2}dx$$

Let ##u=x^2##, ##du=2xdx## and ##x=\sqrt {u}##

$$I(u)=\int_{0}^{\infty}u^{\frac {1} {2}}e^{-u}du = \Gamma(\frac {3} {2}) = \frac {1} {2} \sqrt {\pi}$$

Peace,

Fred

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