I was reading the scattering section of Griffiths QM, the part where he derives the integral form of the Schrödinger equation, where he evaluates the following integral for the Green's function of the Helmholtz equation: $$ \frac1{4\pi^2 r}\int_{-\infty}^\infty \frac{s\sin(sr)}{k^2 - s^2}\mathrm ds. $$ He breaks it up into two pieces, $$ G(\mathbf r) = \frac{i}{8\pi^2 r}\left\{ \int_{-\infty}^\infty \frac{se^{isr}}{(s-k)(s+k)}\mathrm ds - \int_{-\infty}^\infty \frac{se^{-isr}}{(s-k)(s+k)}\mathrm ds \right\} \equiv \frac{i}{8\pi^2 r}(I_1 - I_2), $$ and introduces Cauchy's integration formula, which states that $$ \oint\frac{f(z)}{z - z_0} \mathrm dz = 2\pi i f(z_0) $$ if $z_0$ is inside the integral and zero otherwise. He then says to skirt the poles by going over the one at $-k$ and under the one at $+k$. Then, close $I_1$'s contour by drawing a large semi-circle with positive imaginary part, and $I_2$'s contour with a large semi-circle of negative imaginary part, such that their contributions vanish as the radius goes to infinity, obtaining $$ G(\mathbf r) = -\frac{e^{ikr}}{4\pi r}. $$
He invites the reader to try different conventions for skirting the pole, stating that this gives a different but equally valid Green's function. So I tried going over both poles instead of under one and over the other, and I got that $I_1$ was zero, because now its contour contains no poles.
My idea to reconcile this is that $I_2$ would now contain two poles, and therefore (by residue theorem? I study physics and don't know much complex analysis) give the same answer as if we had gone with the original skirting convention.
Question 1: Is the residue theorem idea correct? If so,
Question 2: Given that $I_1$ and $I_2$ are separate integrals, why am I not able to skirt the poles in different ways for each? Say I go over over for $I_1$ and under under for $I_2$. Then I would have zero for both integrals, which does not seem like a valid Green's function.