I have a question regarding Riemann's first proof of the functional equation that was given in his paper on the Riemann zeta function. I am an undergraduate working on a fairly short undergraduate report on the zeta function and as such the proof of the functional equation that I give, while going into the contour integration and residue theorem, etc, also leaves out certain technicalities that would otherwise bog it down.
I am aware that the functional equation attained via this method can be used to extend the zeta function to the whole complex plane (apart from 1). However, as far as I have been able to tell, the proof of the functional equation assumes that $\operatorname{Re}(s)>1$ (this is used for instance in the proof that $\Gamma(s) \zeta(s) = \int_{0}^{\infty} \frac{u^{s-1}}{e^{u}-1} du$ for $\operatorname{Re}(s)>1$, which is later incorporated in the proof of the functional equation). As a result, the functional equation $\zeta(s)=2(2\pi)^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$ appears to my mind only to additionally define the zeta function for $\operatorname{Re}(s)<0$, which leaves the strip $0\leq\operatorname{Re}(s)\leq1$ "uncontinued".
I know that this contour proof $is$ used to extend the zeta function to the whole complex plane, so my question therefore is: where exactly during the proof can we say that the zeta function can be extended analytically to $\operatorname{Re}(s)\leq1$ rather than just $\operatorname{Re}(s)<0$, and how can I - bearing in mind I am aiming for a rather rough undergraduate-level explanation - explain that although we have started by assuming $\operatorname{Re}(s)>1$, the functional equation we obtain is actually correct for the strip $0\leq\operatorname{Re}(s)\leq1$ as well as the continued region $\operatorname{Re}(s)<0$?
The integral
$$I(s) = \int_\gamma \frac{(-z)^{s-1}}{e^z-1}\,dz,$$
where $\gamma$ is a contour enveloping the positive real axis, exists for all $s \in \mathbb{C}$ and defines an entire function. For $\operatorname{Re} s > 1$, we have
$$2 \sin \pi s\,\Gamma(s)\zeta(s) = iI(s)$$
from the known $\Gamma(s)\zeta(s) = \int_0^\infty \frac{u^{s-1}}{e^u-1}\,du$, and that defines $\zeta(s)$ via
$$\zeta(s) = \frac{i}{2\sin \pi s\,\Gamma(s)}I(s)$$
as a meromorphic function on the entire plane. The functional equation is then derived from this, after the continuation to an entire meromorphic function has been established.