How do you determine if the improper integral converges or diverges #int xe^-x dx # from 0 to infinity?

1 Answer
Steve M
Oct 30, 2016

# int_0^oo xe^-x dx# converges to #1#

Explanation:

First let is find # int xe^-x dx #

The formula for integration by parts is:
# intu(dv)/dxdx = uv - intv(du)/dxdx #

Essentially we would like to find one function that simplifies when differentiated, and one that simplifies when integrated (or is at least integrable).

In this case differentiating x simplifies whereas #e^-x# makes no difference.

Let # {(u=x, => ,(du)/dx=1),((dv)/dx=e^-x,=>,v =-e^-x ):}#

So IBP gives:

# int(x)(e^-x)dx = (x)(-e^-x)-int(-e^-x)(1)dx#
# :. intxe^-x dx = -xe^-x + inte^-xdx #
# :. intxe^-x dx = -xe^-x - e^-x (+C) #

This next part is not a vigorous proof, but it is adequate (unless you are studying for a Maths degree!)

And, so we have:
# int_0^oo xe^-x dx = [-xe^-x - e^-x]_0^oo #
# :. int_0^oo xe^-x dx = lim_(x->oo)(-xe^-x - e^-x) - lim_(x->0)(-xe^-x - e^-x)#

Now # lim_(x->oo)(xe^-x) = 0 # (as #e^-x -> 0 # faster than #x -> oo # as #x -> oo #

# :. int_0^oo xe^-x dx = (-0 - 0) - (0 - 1) = 1 #

Hence, # :. int_0^oo xe^-x dx# converges to #1#

Related questions
See all questions in Integral Test for Convergence of an Infinite Series
Impact of this question
9896 views around the world
You can reuse this answer
Creative Commons License