Question

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

Answer 1

`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`