How do you find the integral of #int 2x e^ (-x^2)dx# from negative infinity to infinity?

Question

Elijah Abram · Accepted Answer

See the explanation section, below.

First notice that the integrand is continuous on #(-oo,oo)# so we only nee to evaluate two integrals. Choose a number in #(-oo,oo)# to separate the integral. #0# is usually easy to work with. So #int_-oo^oo 2x e^ (-x^2)dx = int_-oo^0 2x e^ (-x^2)dx +int_0^oo 2x e^ (-x^2)dx # (Provided that both integrals exist.) #int_-oo^0 2x e^ (-x^2)dx = lim_(ararr-oo) int_a^0 2x e^ (-x^2)dx # # = lim_(ararr-oo) [- e^ (-x^2)]_a^0 # # = lim_(ararr-oo) [-1- (- e^ (-a^2))] = -1 # And #int_0^oo 2x e^ (-x^2)dx = lim_(brarroo) int_0^b 2x e^ (-x^2)dx # # = lim_(brarroo) [- e^ (-x^2)]_0^b # # = lim_(brarroo) [(- e^ (-b^2))-(-1)] = 1 # Both limits DO exists, so, #int_-oo^oo 2x e^ (-x^2)dx = int_-oo^0 2x e^ (-x^2)dx +int_0^oo 2x e^ (-x^2)dx # # = -1+1=0#

Aurora Alvarado · Answer

undefined To find the integral of $ \int_{-\infty}^{\infty} 2x e^{-x^2} \, dx $, you can use the method of integration by parts. Let $ u = x $ and $ dv = 2e^{-x^2} \, dx $. Then, $ du = dx $ and $ v = -e^{-x^2} $.

Using the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substitute the values:

$$ \int_{-\infty}^{\infty} 2x e^{-x^2} \, dx = -xe^{-x^2} \Bigg|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} (-e^{-x^2}) \, dx $$

As $ x $ approaches $ -\infty $ and $ \infty $, $ e^{-x^2} $ approaches 0, so the first term becomes 0:

$$ = 0 - (- \int_{-\infty}^{\infty} e^{-x^2} \, dx) $$

This integral is a known result and equals $ \sqrt{\pi} $, so the final answer is:

$$ = \sqrt{\pi} $$