Why do we need to worry about removable singularities when using Residue Theorem?

So, the Residue Theorem says that, for analytic $f$ analytic except at isolated singularities, $S$ and for loop $\gamma$ not passing through any singularities that:$\DeclareMathOperator{\Res}{Res}$

$\int_{\gamma}{f(z)dz}=\sum_{s\in S}n(\gamma,s) \cdot \Res_{z=s}f(z)$

with $n(\gamma,s)$ being the winding number of $s$ with respect to $\gamma$.

Now, we know that if $S' \subset S$ is the set of removable singularities of $S$, then we can create a function $\hat{f}$ which is the analytic extension of $f$ i.e. its analytic everywhere that that $f$ is and on $S'$. Now, according to the Residue Theorem:

$\int_{\gamma}{\hat{f}(z)dz}=\sum_{s\in S - S'}n(\gamma,s) \cdot \Res_{z=s}\hat{f}(z)$

But, $f$ and $\hat{f}$ agree on everywhere except on $S'$ so that must mean that:

• for $s\in S-S'$ we get $\Res_{z=s}f(z)=\Res_{z=s}\hat{f}(z)$ since you can always create a curve around $s$ that doesn't with all other singularities on the exterior.
• $\int_{\gamma}{f(z)dz}=\int_{\gamma}{\hat{f}(z)\,dz}$ since they agree on $\gamma$

So, I conclude that:

$\int_{\gamma}{f(z)\,dz}=\sum_{s\in S - S'}n(\gamma,s) \cdot \Res_{z=s}\hat{f}(z)$

Which, to me, sounds like I can ignore removable singularities when using the Residue Theorem. Is this true? If it is, it seems weird to me that it isn't included in the presentation of the Residue Theorem.

The residue is zero at removable singularities so it doesn't matter if they are included or not. I guess that sometimes it is done just for convenience to avoid unnecessary extra conditions in statements.