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.