It diverges.

First we split the integral into two pieces:

`int_0^oo = int_0^1 + int_1^oo`

Consider the first integral, `int_0^1`

If `0<=x<=1` , then `1 <= e^x <= e`

So `1 >= 1/(e^x) >= 1/e`

Thus:

`int_0^1 1/x 1/(e^x) dx >= int_0^1 1/x 1/e dx = 1/e int_0^1 1/x dx`

`int 1/x dx = "ln" |x|`

`lim_(a->o) int_a^1 1/x dx =`

`lim_(a->0) ["ln" |x| ] |_a^1`

`=lim_(a->0) (0-"ln" a)`

`=oo`

Thus:

`int_0^1 1/x dx` diverges, so `1/e int_0^1 1/x dx` diverges

Thus `int_0^1 1/(xe^x) dx >=1/e int_0^1 1/x dx ` diverges

And we don't even have to figure out the second integral, `int_1^oo`