State whetever the given integral converges or diverges, and jusify your claim. int(1/(x*exp(x)), x = 0 .. infinity)

1 Answer

mlehuzzah's profile pic

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted on

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`


`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)`




`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`