Given $f \in L^1$, prove that $$F(z) = \frac{1}{2\pi i} \int^\infty_{-\infty} \frac{f(t)}{t-z}\,dt$$ is an analytic function and $$F'(z)=\frac{1}{2\pi i} \int^\infty_{-\infty} \frac{f(t)}{(t-z)^2}\,dt.$$
My attempt: It is sufficient to prove its analytic by showing that the derivative of $F(z)$ exists. (Any flaws with this argument?)
\begin{align} F'(z)&=\lim_{\Delta z_n\to 0} \frac{F(z+\Delta z_n) - F(z)}{\Delta z_n}\\ &=\lim_{\Delta z_n\to 0} \frac{1}{\Delta z_n} \frac{1}{2\pi i} \left[\int^\infty_{-\infty} \left( \frac{f(t)}{t-(z+\Delta z_n)}-\frac{f(t)}{t-z} \right) dt \right]\\ \mbox{(a bit algebra)} &= \frac{1}{2\pi i} \lim_{\Delta z_n\to 0} \int^\infty_{-\infty} \frac{f(t)}{(t-z)(t-z-\Delta z_n)} \, dt \\ \end{align}
I came so close to justify the exchange of the limit with the integral (Hence, DCT). I somehow cannot find a appropriate $g\in L^1$ to bound the integrand.
If $z$ is not real, you can always bound the denominator away from zero, and so you can use $f(t)/(t-z+\delta)^2$ for an appropriate (and small enough) $\delta>0$.
But when $t$ is real I see problems: for instance let $f$ be any $L^1$ function that is constant on some interval around $2$. Then $$ \int_{-\infty}^\infty\frac{f(t)}{t-2} $$ doesn't exists (it blows up at $2$).
Are you really supposed to show that $F$ is analytic on all of the complex plane?