# Help Solve Difficult Integral I(dxexp(-ax**2))

• eljose79
In summary, the conversation discusses a difficult integral involving the function exp(-ax^2). The limits of integration are from -infinity to infinity, and a must be greater than 0. A hint is given to try solving a simpler integral first, which involves the function exp(-x^2-y^2). Another member suggests using a contour integral or a program to solve it, while another mentions a possible connection to probabilities and the Weibull distribution.
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?...

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

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))

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?

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.

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...

Last edited:

Gamma integration.

## What is the purpose of solving a difficult integral?

Solving integrals is an important part of scientific research, as it helps us understand and analyze complex mathematical equations. It also allows us to model and predict real-world phenomena, making it a crucial tool in many scientific fields.

## What makes this integral difficult to solve?

The integral in question, I(dxexp(-ax**2)), involves an exponential function raised to a power, which can be challenging to integrate. Additionally, the variable x is squared, which adds another layer of complexity to the problem.

## What are the different methods for solving this type of integral?

There are several methods for solving difficult integrals, including substitution, integration by parts, and trigonometric substitution. Each method has its own advantages and is useful in different scenarios. It is important to choose the most appropriate method for each integral.

## What are the steps for solving this specific integral?

The first step in solving this integral is to apply the substitution method, where u = -ax**2 and du = -2ax dx. After substituting these values, we can use integration by parts to solve the resulting integral. Finally, we can use the inverse substitution to solve for the original variable, x.

## Is there a way to check if the solution is correct?

Yes, there are several ways to check if the solution to a difficult integral is correct. One way is to differentiate the solution and see if it matches the original function. Another way is to use numerical integration methods, such as Simpson's rule, to approximate the integral and compare it to the calculated solution.

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