limit of exponential function

**Bud** · August 30th 2010, 03:40 AM

Hi there,

I found this line in a book which I would like to understand:

$K \int^{+\infty}_{-\infty}e^{-ax^{2}}dx = K \left(\frac{\pi}{a}\right)^{\frac{1}{2}}$

I tried to follow this step, but I did not succeed. I tried to calculated the limits after I determined the integral using l'hopital's rule:

$K \int^{+\infty}_{-\infty}e^{-ax^{2}}dx = \left[ -\frac{1}{2ax}e^{-ax^{2}} \right]^{+\inf}_{-\inf}$

Does somebody know how to handle this?
Thanks in advance

Cheers
Bud

**mr fantastic** · August 30th 2010, 03:57 AM

Originally Posted by Bud

Hi there,

I found this line in a book which I would like to understand:

$K \int^{+\infty}_{-\infty}e^{-ax^{2}}dx = K \left(\frac{\pi}{a}\right)^{\frac{1}{2}}$

I tried to follow this step, but I did not succeed. I tried to calculated the limits after I determined the integral using l'hopital's rule:

$K \int^{+\infty}_{-\infty}e^{-ax^{2}}dx = \left[ -\frac{1}{2ax}e^{-ax^{2}} \right]^{+\inf}_{-\inf}$ Mr F says: This is wrong. If you differentiate the anti-derivative you found, it is clear that you do not get the integrand. In fact, no anti-derivative exists in closed form using elementary functions.

Does somebody know how to handle this?
Thanks in advance

Cheers
Bud

Read this: Gaussian Integral -- from Wolfram MathWorld

**Amer** · August 30th 2010, 04:05 AM

this can't be done like that suppose

$\displaystyle \int_{-\infty}^{\infty} e^{-ax^2} dx = I$

then

$\displaystyle I^2 = \int_{\infty}^{\infty} e^{-ax^2} dx \int_{-\infty}^{\infty} e^{-ay^2} dy$

rite now

$\displaystyle I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-a(x^2+y^2)} dx dy$

transfer to polar

$r^2 = x^2 +y^2$

and the Jacobean is r

$\displaystyle I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-ar^2} r dr d\theta$

can u continue

**Bud** · August 30th 2010, 04:40 AM

Thanks you two. You proposed the same solution, i.e. this trick of combining two one-dimensional Gaussians and then changing to polar coordinates.

From there it was simple. Thanks!

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