Showing a complex function is analytic

Question

Define $H(z) = \int^{1}_{0} \frac{h(t)}{t-z} dt$ where $h(t)$ is some complex valued, continuous function on $[0,1]$. Show $H(z)$ is analytic on $\mathbb{C}/[0,1]$

My attempt:

$H'(z) = \lim_{\Delta z \to 0} \frac{H(z + \Delta z) - H(z)}{\Delta z}$ Equivalently, $$\lim_{\Delta z \to 0} \frac{1}{\Delta z} \cdot \big( \int^{1}_{0} \frac{h(t)}{t - z - \Delta z} dt - \int^{1}_{0} \frac{h(t)}{t - z} dt \big)$$

Adding the integrals and making a common denominator we get

$$ \lim_{\Delta z \to 0} \frac{1}{\Delta z} \int^{1}_{0} \frac{h(t)(t-z) - h(t)(t-z-\Delta z)}{t^2 - 2tz + z^2 + \Delta z t} dt$$ which is equal to

$$\lim_{\Delta z \to 0} \frac{1}{\Delta z} \int^{1}_{0} \frac{\Delta z}{t^2 - 2tz + z^2 + \Delta z t} dt$$

and

$$\lim_{\Delta z \to 0} \int^{1}_{0} \frac{1}{t^2 - 2tz + z^2 + \Delta z t } dt$$

Taking the limit under the integral, we see that

$$H'(z) = \int^{1}_{0} \frac{1}{(t-z)^2}dt = \frac{1}{z-t}_{|_{0}^{1}} = \frac{1}{z^2 - z}$$

My big problem with my attempt is that I'm not sure if integrals necessarily work with any complex $z$ and if taking the limit under the integral is allowed. Can anyone give me insight?

Answer 1

Your argument suggests that $H^\prime (z)=\int_0^1 \frac{h(t)}{(t-z)^2}dt$. So we prove that $$\lim_{\Delta z \to 0} \frac{H(z + \Delta z) - H(z)}{\Delta z}=\int_0^1 \frac{h(t)}{(t-z)^2}dt .$$ Now \begin{align} &\lim_{\Delta z \to 0} \frac{H(z + \Delta z) - H(z)}{\Delta z}-\int_0^1 \frac{h(t)}{(t-z)^2}dt\\ &=\lim_{\Delta z \to 0} \frac{1}{\Delta z}\left( \int^{1}_{0} \frac{h(t)}{t - z - \Delta z} dt - \int^{1}_{0} \frac{h(t)}{t - z} dt \right)-\int_0^1 \frac{h(t)}{(t-z)^2}dt\\ &=\lim_{\Delta z \to 0} \frac{1}{\Delta z} \int^{1}_{0} \frac{h(t)\Delta z}{(t-z-\Delta z)(t-z)} dt-\int_0^1 \frac{h(t)}{(t-z)^2}dt\\ &=\lim_{\Delta z \to 0} \int^{1}_{0} \frac{h(t)}{(t-z-\Delta z)(t-z)} dt-\int_0^1 \frac{h(t)}{(t-z)^2}dt\\ &=\lim_{\Delta z \to 0} \int^{1}_{0} \frac{h(t)\Delta z}{(t-z-\Delta z)(t-z)^2} dt. \end{align} Let $M=\max\limits_{0\le t\le 1}|h(t)|,$ $\delta=\operatorname{dist}(z, [0,1])>0$ for $z\not \in [0,1].$ If we take $ \Delta z$ so small, then $|t-z-\Delta z|\ge \delta/2$ for all $t\in [0,1].$ Therefore we have \begin{align} &\left|\int^{1}_{0} \frac{h(t)\Delta z}{(t-z-\Delta z)(t-z)^2} dt\right|\le |\Delta z|\int^{1}_{0} \frac{|h(t)|}{|t-z-\Delta z||t-z|^2} dt\\ &\le |\Delta z|\int^{1}_{0} \frac{M}{(\delta/2)\delta^2} dt\\ &=\frac{2M|\Delta z|}{\delta^3}\to 0\quad (|\Delta z|\to 0). \end{align} Thus we have $$ \lim_{\Delta z \to 0} \frac{H(z + \Delta z) - H(z)}{\Delta z}-\int_0^1 \frac{h(t)}{(t-z)^2}dt =0$$ and we know that $H(z)$ is analytic on $\mathbb{C}/[0,1].$

Answer 2

How about Morera's Theorem... Write $U = \mathbb C \setminus [0,1]$.

If $t \in [0,1]$, then $$ h_t(z) = \frac{h(t)}{t-z} $$ is analytic in $U$. Let $D \subset U$ be a closed disk. Then $$ \int_{\partial D} h_t(z)\;dz = 0\tag{1} $$ Therefore $$ \int_{\partial D} H(z)\;dz = \int_{\partial D}\left(\int_0^1h_t(z)\;dt\right)dz =\int_0^1\left(\int_{\partial D} h_t(z)\;dz\right)dt = \int_0^1 0\;dt = 0 . $$ This is true for all closed disks $D \subset U$, therefore by Morera's Theorem $H(z)$ is analytic in $U$.