Can't understand how the integral in the Riemann zeta function converges I am struggling to understand how the integral in the  Riemann zeta function converges at $\Re(z)<1$ by using convergence rules. As you probably know the integral is:
$$\int_0^\infty \frac{t^{z-1}}{\exp(t)-1}\, dt.$$
 A: This is the integral 3.411.1 in Gradshteyn and Ryzhik, with the assignments $\mu = 1$, $\nu = z$.  It (only) converges for $\Re z > 1 $ because zeta has a simple pole at $z=1$.  (This is the only pole of zeta in $\mathbb{C}$.)
Why do you believe this integral converges for any/some $\Re z \leq 1$?  Note that the Riemann zeta function is defined as the analytic continuation of the above integral (or of the Dirichlet series) around the pole at $z = 1$ to the rest of the complex plane.  The sum and integral don't converge on that continuation, but there are various operators that transform these into new integrals and series which converge in various parts of the continuation.
A: First, let's show that the integral from $1$ to infinity converges.  We have for $\sigma > 1$,
$$\int_1^r \frac{t^{\sigma-1}}{e^t-1} dt \leq r^{\sigma-1} \int_1^r\frac{dt}{e^t-1}$$
Now by using $u$-substitution and partial fraction (or a term by term geometric series integration), an antiderivative of $(e^t-1)^{-1}$ is $\log(1-e^{-t})$, so 
$$r^{\sigma-1} \int_1^r\frac{dt}{e^t-1} = r^{\sigma-1}\log(1-e^{-r}) + C $$
for some constant $C$.  If you can show that the limit $\lim\limits_{r \to \infty} r^{\sigma-1}\log(1-e^{-r})$ exists, then the integral from $1$ to infinity will converge.  In fact, you can show that this limit is zero, since $\log(1-e^{-r})$ goes to zero way faster than $r^{y}$ goes to infinity when $y > 0$.
Next, you have to show that the integral from $0$ to $1$ converges.  You can write
$$\int_0^1 \frac{t^{\sigma-1}}{e^t-1} dt = - \int_1^{\infty} \frac{t^{-\sigma-1}}{e^{-t}-1} dt$$
and probably do something similar.
