# Using Cauchy-Goursat Theorem to prove $g(z)=0$

I am wondering about complex-valued integrals of the following form: Let $C$ be the circle $|z|=s$, described in the positive sense, and let $f(s)$ be holomorphic everywhere. Given a function $g(z)$:

$$g(z) = \int_C \frac{f(s)}{s-z} \ ds$$

where $s$ denotes points on $C$, and $z$ is any point interior to $C$. So long as $z$ is an interior point to $C$, an integral of this form is trivial to evaluate using the Cauchy integral formula. However, my concern has to do with evaluating $g(z)$ such that $|z| > s$. My thoughts on how to proceed are summed up as follows:

• Clearly, the integrand $\frac{f(s)}{s-z}$ is holomorphic everywhere so long as $|z| \neq s$; that is, the points on $C$ are singularities.
• By the Cauchy-Goursat Theorem, we should be able to show that if the integrand is holomorphic on $C$ and within the disk enclosed by $C$, then the integral vanishes on $C$.
• We would then apply the principle of path deformation to show that along any arbitrary path $C_*$ exterior to $C$, $g(z)=0$, and since we have the liberty of picking any $C_*$ exterior to $C$, it follows that $g(z)=0$ $\forall z$ s.t. $|z|>s$.

My confusion is how one goes about satisfying the conditions of the Cauchy-Goursat Theorem when clearly the integrand is not even defined on $C$. In other words, I am not sure how to deal with the presence of the singularity based on the above statement of the Cauchy-Goursat Theorem (from Brown and Churchill).

• If $|z|>s$, then $z$ lies outside of the contour $C$. In particular, the integrand is holomorphic throughout the interior of $C$ and so by Cauchy-Goursat, the integral evaluates to 0. No path deformation is needed. – MightyTyGuy Apr 22 '17 at 7:14
• Do we not have to require that the integrand is holomorphic throughout the interior of $C$ and on $C$ itself? – David A. Lee Apr 22 '17 at 7:17