I am working on the following problem.
Consider a continuous function $\phi:[-1,1]\to \mathbb{C}$ on $[-1,1]$. Let $$g(z):=\int_{-1}^{1}\frac{\phi(t)}{t-z}\:dt.$$
I am interested in finding the coefficients $a_n$ of the Taylor series $\sum_{n=0}^{\infty}a_n(x-i)^n$ of $g$ centered at $z=i$. I know that $a_n=\frac{g^{(n)}(a)}{n!}$, so the problem essentially boils down to computing $g^{(n)}(z)$, for each $n$.
This may sound trivial, but I have trouble computing $g^{(n)}(z)$, for $n>0$. If I can compute the first derivative, then may be I can figure out the rest. Any help will be appreciated. Thanks in advance.
Following @AnneBauval's idea:
\begin{align*} g(z) &= \int_{-1}^1 \dfrac{\phi(t)}{(t-i)-(z-i)}dt \\ \\ &= \int_{-1}^1 \dfrac{\phi(t)}{(t-i)}\sum_{n=0}^\infty\left(\dfrac{z-i}{t-i}\right)^ndt \\ \\ &= \sum_{n=0}^\infty \left( \int_{-1}^1 \dfrac{\phi(t)}{(t-i)^{n+1}}dt \right)(z-i)^n \end{align*}
for $|z-i|<1$.