# Difficult integral .

eljose79
difficult integral.....

I have problems to calculate the integral

I(dxexp(-ax**2)), where the limits run from -infinite to infinite.
x**2+1

and a>0..can someone help?....

## Answers and Replies

Hint: the square of that integral when a=1 is equal to [inte][inte]exp(-x^2-y^2)dxdy. Try and solve that first. :)

If you can't figure it out, a web search for "Gaussian intregal" will reveal the clever trick.

-a dx exp x^2

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Originally posted by damgo
Hint: the square of that integral when a=1 is equal to [inte][inte]exp(-x^2-y^2)dxdy.

I think that elijose's integral is not Gaussian, but that he meant:

[integral][-inf..+inf](exp(-ax^2)*(x^2+1)^(-1)dx

which is more difficult.

Man, I have got to learn how to post math symbols in here.

-ax^2 exp(arctan(x))

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Originally posted by progtommlp
-ax^2 exp(arctan(x))

No.

You can take the derivative of your answer to see that it's not correct.

Oh, hmmm. Have you tried doing the contour integral over the upper half-circle?

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You could try that. You could also try the following program:

1. Recognize that the integrand is even, so double the integral and reduce the range of integration from 0 to infinity.

2. Make the change of variables: u^2=x^2+1. The resulting integrand can be split into two terms, one with integrand exp(-au^2)/u^2 and one with 1/u^2. Don't forget to transform the limits of integration.

3. The second integral is elementary, while the first can be done by treating it as a function of a. Differentiate with respect to a, and the u^2 from the exponent cancels with the u^2 in the denominator (there's the trick!), and you have a Gaussian.

If it doesn't work, don't shoot me--I just thought of it. You can always try the contour integral if it doesn't work my way.

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Originally posted by Tom
3. The second integral is elementary, while the first can be done by treating it as a function of a. Differentiate with respect to a, and the u^2 from the exponent cancels with the u^2 in the denominator (there's the trick!), and you have a Gaussian.

Oops, my mistake. You get a regular exponential, not a Gaussian. But hey, that's easier anyway.

bogdan
it has something to do with probabilities...something linked to Weibull distribution...or something...but I can't find it in a book in which I know it was...wait...a second...

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