S(1 to e) dx/[x(lnx)^p] For which values does this converge?

Question

Can you use p-test on this?
I used u substitution x=lnx and got S(0 to 1) du/u^p. And I end up with 1/(1-p)
If p>1, does integral become u^(-p+1)/(1-p) ??
If p<1, seems like integral would be u^(p+1)/(p+1)
I'm confused about p test a bit....

Answer 1

Note that if we put u = ln(x), the du = dx/x. And we get

Int u^(-p) du. For p <>1 (different from 1) this actually yields (u^(1-p))/(1-p) + C = (ln(x))^(1-p)/(1-p) + C.

If p >1, then 1-p <0. And since ln(1) = 0 and ln(x) >0 for x >1, the integral diverges to oo.

If p <1, then 1-p >0 and (ln(x))^(1-p) -> 0 as x -> 1 +. Therefore, the integral converges to ln(e)/(1-p) = 1/(1-p).

Finally, if p =1, we get ln(ln(x) [from 1 to e] = oo because x -> 1+ => ln(x) -> 0 => ln(ln(x)) -> -oo. Since -oo is in the lower limit of integration, the integral is oo.