Do these interval converge or diverge?

Question

{Int from 0 to infinity} lnx/(x^(1/2)) dx

{Int from 1 to infinity} x^3/(x^(5)+2) dx (using the comparison test)

how can this problem be solved?

thanks.

Answer 1

1) For x > e, ln(x) >1 and, therefore,

lnx/(x^(1/2) > 1/(x^1/2) = 1/sqrt(x) > 0 (1)

We know that

Int (1 to oo) 1/sqrt(x) dx = 2 sqrt(x) (1 to oo) = oo, because sqrt(x) --> oo as x --> oo. In virtue of (1) it follows {Int from 0 to infinity} lnx/(x^(1/2)) dx diverges to oo.

2) For x >1,

0 < x^3/(x^(5)+2) < x^3/(x^5) = 1/x^2 (2)

Since Int (1 to oo) 1/x^2 = [- 1/x] (1 to o) = 1 , (2) shows that {Int from 1 to infinity} x^3/(x^(5)+2) dx converges.

Answer 2

Have you been taught "L'Hosptial's Rule" yet? It looks like this is what you need. If you don't know it yet, then you can always look it up online. It has to do with derivatives. You take the first derivative of the top and the first derivative of the bottom, and perform the division again. If this leads you to a definite answer, then this is your divergence or convergence.