Fubini on path integrals? I've been given an exercise in more steps where I need to prove that the Riemann $\zeta$ function extends to a meromorphic function on all $\mathbb{C}$ with a single simple pole in $z=1$. To prove it I've been given the equality:  $sin(\pi s)\Gamma(s)\Gamma(1-s)=\pi$ which I now plug in in a formula which i derived before which leads to the following result:
$$\zeta(s)=-\frac{\Gamma(1-s)}{2\pi i}\int_\gamma\frac{(-z)^{(s-1)}}{e^z-1}$$
We know that $\zeta(s)$ is holomorphic for $\Re(s)>1$ and we want now to check the behaviour for $\Re(s)\le1$.
For $\Re(s)\le1$ we know that $\Gamma(1-s)$ is meromorphic with a simple pole at $s=1$. And now the idea was to check that the integral represents an holomorphic function, hence I wanted to use Fubini as shown below. Let $\gamma_s$ be a closed path in $\mathbb{C}$. Then:
$$\int_{\gamma_s} \underbrace{\int_\gamma \frac{(-z)^{s - 1}}{e^z - 1} dz}_{f(s)} ds \overbrace{=}^{Fubini} \int_\gamma \int_{\gamma_s} \underbrace{\frac{(-z)^{s - 1}}{e^z - 1}}_{\text{is holomorphic in s on $\mathbb{C}$}} ds dz \overbrace{=}^{\text{Cauchy Theorem}} \int_\gamma 0 dz = 0$$
Is this last step allowed a priori or do I need more assumptions to be able to swap paths? Thank you
 A: It's not a priori allowed to change the order of integration. You need to verify that some set of conditions that allows the change is met.
Here, the integrand
$$f(z,s) = \frac{(-z)^{s-1}}{e^z-1}$$
is continuous on $\operatorname{Tr} \gamma \times \operatorname{Tr} \gamma_s$, so for an application of Fubini's theorem the measurability condition is satisfied, and it suffices to see that
$$\int_{\gamma_s} \int_\gamma \lvert f(z,s)\rvert\,\lvert dz\rvert\:\lvert ds\rvert < \infty.$$
That is elementarily (but somewhat tediously, if one does it completely rigorously) obtained from the estimate
$$\lvert f(z,s)\rvert \leqslant e^{\pi\lvert t\rvert} \frac{\lvert z\rvert^{\sigma-1}}{\lvert e^z-1\rvert},$$
where $s = \sigma + it$.
Another way is to approximate the path $\gamma$ with parts $\gamma_N$ that are parameterised over a compact subinterval of $\gamma$'s parameter interval, and see that
$$I_N(s) = \int_{\gamma_N} f(z,s)\,dz \to \int_\gamma f(z,s)\,dz = I(s)$$
locally uniformly (also using the estimate of $\lvert f(z,s)\rvert$ above). Since the integral of $I_N(s)$ is over a compact interval, and the integrand is entire in $s$, $I_N(s)$ is known to be holomorphic by a general theorem.
Yet another way to see that $I(s)$ is holomorphic is to verify that differentiating under the integral sign is allowed.
