What is $\int_{\gamma}\int_{0}^{1} \frac{g(t) \, dt}{t-z}\,dz$ in terms of $I=\int_{[0,1]}g$ for a closed piecewise smooth curve $\gamma$. Q. Let $g:[0,1] \rightarrow \mathbb{C}$ be continuous, and let $V=\mathbb{C} \setminus[0,1] .$ Define $f: V \rightarrow \mathbb{C}$ by
$$
f(z)=\int_0^1 \frac{g(t) \, dt}{t-z}
$$
Let $\gamma$ be a closed piecewise smooth curve in $\mathrm{V}$ and set $I=\int_{[0,1]} g$, what is $\int_\gamma f$ in terms of $I$?
I couldn't proceed further since $V$ is not simply connected.
Is there any role for $$f'(z)=\int_0^1 \frac{g(t) \, dt}{(t-z)^2} \text{?}$$
 A: $$
\int\limits_\gamma \int_0^1 \frac{g(t)}{t-z} \, dt \,dz
$$
Since the curve along which $z$ moves is continuous, and is the image of the compact set $[0,1],$ and is disjoint from the set $[0,1],$ there is some positive number $\eta$ such that $|t-z|>\eta$ for every $t\in[0,1]$ and every $z$ on the curve. In other words, the denominator $t-z$ does not approach $0$ anywhere, so the function $(t,z)\mapsto 1/(t-z)$ is bounded. And $g$ is a continuous function on the compact set $[0,1],$ so it is also bounded. Therefore the order of integration can be changed with impunity, getting
$$
\int_0^1 \int\limits_\gamma \frac{g(t)}{t-z} \, dz \,dt.
$$
As $z$ moves along the curve $\gamma,$ $t$ does not change, so the above is equal to
$$
\int_0^1 \left( g(t) \int\limits_\gamma \frac{dz}{t-z} \right) \, dt
$$
Now recall that
\begin{align}
& \int\limits_\gamma \frac{dz}{t-z} \\[8pt]
= {} & \big( {-2}\pi i \times(\text{the number of times $\gamma$ winds around $t$})  \big)
\end{align}
So the integral you seek is
$$
-2\pi i\times \int_0^1 g(t)\, dt \times (\text{the winding number}).
$$
A: Note that for $|z|>1+\epsilon >1$ one can develop $f$ in Laurent series:
$$f(z)=-\sum_{k \ge 0} \left( \int_0^1g(t)t^k\,dt \right)z^{-k-1}$$
by the usual trick of writing $\frac{g(t)}{t-z}=\frac{-g(t)}{z(1-t/z)}$ expanding in geometric series which is absolutely convergent etc
But now $\int_\gamma f\,dz$ depends only on the homotopy class (in $V$) of $\gamma$ by the usual Cauchy theorem, so we have essentially two cases, if $\gamma$ is contractible, the integral is zero and if it is not, we can deform it into $|z|=2$ say since $\pi_1(V)=\mathbb Z$ and assuming the positive orientation say, we get that:
\begin{align}
& \int_\gamma f \, dz=\int_{|z|=2} f \, dz \\[8pt]
= {} & \int_{|z|=2} \left(-\sum_{k \ge 0} \left(\int_0^1g(t)t^k \, dt\right)z^{-k-1}\right) \, dz \\[8pt]
= {} & {-2}\pi i\int_0^1g(t) \, dt
\end{align}
since only the term $k=0$ with power $z^{-1}$ gives a nonzero integral as all the other powers have antiderivatives
