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#

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