# Difficult integral .

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?....

damgo
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.

progtommlp
-a dx exp x^2

Tom Mattson
Staff Emeritus
Gold Member
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.

progtommlp
-ax^2 exp(arctan(x))

Tom Mattson
Staff Emeritus
Gold Member
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.

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

Tom Mattson
Staff Emeritus
Gold Member
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.

Tom Mattson
Staff Emeritus
Gold Member
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.

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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Gamma integration.