proof $f$ is analytic in open set $D$

Question

I'm struggling to prove that $f(z)=\int^{1}_{0}{\frac{1}{1-zt^2}dt}$ is analytic in open set: $D=\left\{z\in\mathbb{C}|\left|{z}\right|<1\right\}$ I don't know how to prove that, I though using Morera's theorem but I don't see how.

Answer 1

We apply Morera's theorem. We need to show that:

1. $f$ defines a continuous function on $D$;
2. $\oint_{\Delta} f(z)\,\mathrm dz=0$ for any triangle $\Delta\subset D$.

Proof of 1. This easily follows from the dominated convergence theorem. (We have $$\left\lvert\frac1{1-zt^2}\right\rvert\le\frac{1+|z|}{(1-\Re z)^2}$$ for all $z\in D$ and $t\in[0,1]$.)

Proof of 2. Let $\Delta:=[a,b,c]\subset D$  be a triangle. Then $$ \oint_\Delta f(z)\,\mathrm dz=\int_{[a,b]}f(z)\,\mathrm dz+\int_{[b,c]}f(z)\,\mathrm dz+\int_{[c,a]}f(z)\,\mathrm dz.$$ We have \begin{align*}\int_{[a,b]}f(z)\,\mathrm dz&=\int_0^1(b-a)f\bigl(a+u(b-a)\bigr)\,\mathrm du\\[.4em] &=\int_0^1\left(\int_0^1\frac{b-a}{1-\bigl(a+u(b-a)\bigr)t^2}\,\mathrm dt\right)\mathrm du \\[.4em] &=\int_0^1\left(\int_0^1\frac{b-a}{1-\bigl(a+u(b-a)\bigr)t^2}\,\mathrm du\right)\mathrm dt&\color{red}{/!\backslash}\\[.4em] &=\int_0^1\left(\int_{[a,b]}\frac{\mathrm dz}{1-zt^2}\right)\mathrm dt, \end{align*} and similarly on $[b,c]$ and $[c,a]$. We deduce that \begin{align*} \oint_\Delta f(z)\,\mathrm dz&=\int_0^1\left(\int_{[a,b]}\frac{\mathrm dz}{1-zt^2}+\int_{[b,c]}\frac{\mathrm dz}{1-zt^2}+\int_{[c,a]}\frac{\mathrm dz}{1-zt^2}\right)\mathrm dt\\[.4em] &=\int_0^1\left(\oint_\Delta\frac{\mathrm dz}{1-zt^2}\right)\mathrm dt, \end{align*} and each inner integral there is $0$ because $z\mapsto\frac1{1-zt^2}$ is holomorphic on $D$ (for any fixed $t\in[0,1]$).

$\color{red}{/!\backslash}$ One should justify that one can change the order of integration here.