\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}dx and \int_{-infty}^\infty x^2e^{-\frac{1}{2}x^2}dx how i compute these integrals via

Sam Longoria

Sam Longoria

Answered question


e12x2dx and x2e12x2dx
how i compute these integrals via Gauss Integral?

Answer & Explanation

Jonathan Burroughs

Jonathan Burroughs

Beginner2022-01-04Added 37 answers

Make the substitution x=2t in both, and use integration by parts in the second, by writing
Stuart Rountree

Stuart Rountree

Beginner2022-01-05Added 29 answers

An interesting alternate way to evaluate the second integral x2e12x2dx using the results form the first one is to utilize a technique called ''differentiation under the integral sign''.
From a substitution and our first integral's value in hand, we can conclude that a>0R
Now we differentiate both sides with respect to a (under the integral sign on the left), and we obtain
Factoring out the 1 from both sides and plugging in a=12 gives us our answer:


Expert2022-01-11Added 613 answers

The one on the left is what I usually think of when I hear the term ''Gaussian integral''. But some people, maybe especially physicists, omit the fraction 1/2. Including the 1/2 makes sense because then the variance of the probability distribution that you get when you normalize the function to be a probability density is 1.
If you know that
then you can write x=2w and dx=2dw. As w goes from to , so does x. So
For e12x2dx begin by observing that it's an even function integrated over an interval that is symmetric about 0, so it's equal to
Then substitute u=12x2, du=x dx=2u, so du2u=dx. The integral becomes

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