Convergence of improper integrals with parameters

[hamsterman]

I'm having a lot of trouble with the subject. Here's one example I'd like explained.
F(t_1, t_2) = \int \limits_0^1 x^{t_1}\ln^{t_2}\frac{1}{x} dx
The book asks to find for what \vec{t} F converges. The answer is \vec{t}\in(-1; \infty)^2, but I don't see how to get that.

In general, what tools are there to test convergence? The only one I know is that (assuming that only f(b) is undefined) \int_a^b f converges if \lim \limits_{x \rightarrow b} \int_x^b f = 0.
Is it safe to assume that if g \sim f, x \rightarrow b and \int g converges then so does \int f?

My book shows some tests for uniform convergence, but that's not exactly the same thing, is it?

 

[sunjin09]

use the change of variable u=ln(1/x), it becomes much clearer.

 

[hamsterman]

So that's \int f = \int \limits_0^{\infty}e^{-(t_1+1)u}u^{t_2} \mathrm{d} u then.
It seems clear that \lim \limits_{u \rightarrow 0} f = u^{t_2} and f converges at infinity when e^{-(t_1+1)u} does.
However, when t₁ = 0, this integral is gamma function of t₂+1. It also converges on negative non integers. Why did I not find that? Also, do there exist other values of t₁ such that the integral converges for some t₂<-1 ?

 

[sunjin09]

So that's \int f = \int \limits_0^{\infty}e^{-(t_1+1)u}u^{t_2} \mathrm{d} u then.
It seems clear that \lim \limits_{u \rightarrow 0} f = u^{t_2} and f converges at infinity when e^{-(t_1+1)u} does.
However, when t₁ = 0, this integral is gamma function of t₂+1. It also converges on negative non integers. Why did I not find that? Also, do there exist other values of t₁ such that the integral converges for some t₂<-1 ?

According to this
http://en.wikipedia.org/wiki/Gamma_function
The integral representation of Gamma function is convergent only if t2+1>0. The full Gamma function is obtained by analytic continuation.

 

[hamsterman]

Thanks, that's good to know.