Trying to show given integral is an analytic function I am trying to re-study complex analysis on my own. I came across this problem in a set of notes. I am given a complex-valued continuous function $h(t)$ on $[0,1]$. Then the function $H(z)$
is defined as follows:
$$H(z) = \int_0^1 \dfrac{h(t)}{t-z} dt $$
I am supposed to show that $H(z)$ is analytic on $\mathbb{C}-[0,1]$ and get its derivative.
I got the latter through first principles i.e. by simply evaluating the limit
$$ \dfrac{1}{\Delta z} [H(z +\Delta z) - H(z)]$$
However the hint says I should use uniform convergence of integrals to show analyticity.This I dont get. I do not wish to use Cauchy's or Morera's theorem or any of the related ilk. 
Am I supposed to approximate $h(t)$ which is continuous on a compact set by polynomials by invoking the Weierstrass theorem??Please help. 
 A: Let $z\in\mathbf C\setminus[0,1]$. You can evaluate the limit and prove existence by letting $(\Delta z_n)_{n\in\mathbf N}$ be any sequence in $\mathbf C\setminus\{0,z\}$ which converges to zero and considering the sequence with terms \begin{align*}\frac{H(z+\Delta z_n)-H(z)}{\Delta z_n}&=\int\limits_0^1\frac{1}{\Delta z_n}\left(\frac{h(t)}{t-z-\Delta z_n}-\frac{h(t)}{t-z}\right)\,\mathrm dt\\&=\int\limits_0^1 \frac{h(t)}{(t-z-\Delta z_n)(t-z)}\,\mathrm dt\end{align*}
Now the integrand clearly converges pointwise to $t\mapsto \frac{h(t)}{(t-z)^2}$ for $z\in \mathbf C\setminus[0,1]$. If you can show the integrand converges uniformly, you are allowed to interchange limit and integral, by the following theorem: If $(f_n)_{n\in\mathbf N}$ is a sequence of Riemann integrable functions on $[a,b]$ that converges uniformly to a limit $f$, then $f$ is Riemann integrable with integral $$\int\limits_a^b f=\lim\limits_{n\to\infty}\int\limits_a^bf_n$$
In this case we find
$$\lim\limits_{n\to\infty}\frac{H(z+\Delta z_n)-H(z)}{\Delta z_n}=\int\limits_0^1 \frac{h(t)}{(t-z)^2}\,\mathrm dt$$
As $(\Delta z_n)$ is an arbitrary sequence in $\mathbf C\setminus\{0,z\}$ which converges to $0$ we find that $H$ is holomorphic in $z$ with derivative 
$$H'(z)=\lim\limits_{\Delta z\to0}\frac{H(z+\Delta z)-H(z)}{\Delta z}=\int\limits_0^1\frac{h(t)}{(t-z)^2}\,\mathrm dt$$
