Let $f:\Bbb{R} \to \Bbb{C}$ a continuous function and assume that $\int_{-\infty}^{+\infty}\frac{|f(t)|}{|t|+1}dt< +\infty$
Prove that $F(z)=\int_{-\infty}^{+\infty}\frac{f(t)}{t-z}dt$ is holomorphic at every $z \in \Bbb{C} \setminus \Bbb{R}$
From a theorem in complex analysis i belive that the derivative of this function is $G(z)=\int_{-\infty}^{+\infty}\frac{f(t)}{(t-z)^2}dt$
Let $z_0 \in \Bbb{C} \setminus \Bbb{R}$ ,so exists $>0$ such that $D(z_0,r)\subseteq \Bbb{C} \setminus \Bbb{R}$
If we take a smaller disk with center $z_0$ and radius $\frac{r}{2}$ we have that $|z-t| \geq \frac{r}{2}$ and $|t-w| \geq \frac{r}{2},\forall w \in D(z_0,\frac{r}{2}),\forall t \in \Bbb{R}$
So doing some algebraic manipulations i arrived at the inequality $$|\frac{F(w)-F(z_0)}{w-z_0}-G(z_0)| \leq \int_{-\infty}^{+\infty}\frac{|f(t)|}{|t-z|}dt\frac{|w-z_0|}{\delta^2}$$
But i could not proceed any further and use my hypothesis.
Can someone help with this?
Thank you in advance.
Since $z_{0}\notin{\bf{R}}$, then $|t-z_{0}|\geq|\text{Im}(z_{0})|>0$ for all $t\in(-\infty,\infty)$.
Consider now $w\in B_{\delta}(z_{0})$ for a small $\delta>0$. We let $\delta<|\text{Im}(z_{0})|$.
We have $|t-w|\geq|t-z_{0}|-|z_{0}-w|\geq|t-z_{0}|-\delta\geq|\text{Im}(z_{0})|-\delta>0$ for any $t\in(-\infty,\infty)$.
the function $\dfrac{|f(t)|}{|t-z_{0}|}$ is continuous on the compact set $\{|t|\leq 2|z_{0}|+2\delta+1\}$, and hence $\displaystyle\int_{|t|\leq 2|z_{0}|+\delta+1}\dfrac{|f(t)|}{|t-z_{0}|}dt<\infty$.
Now we want to find that a dominated integrable function of \begin{align*} \left|\dfrac{f(t)}{(t-w)(t-z_{0})}\right| \end{align*} which the dominated integrable function must be independent of $w$, if we succeed to do that, then the rest is just by Lebesgue Dominated Convergence Theorem.
On $\{|t|\leq 2|z_{0}|+2\delta+1\}$, we have \begin{align*} \left|\dfrac{f(t)}{(t-w)(t-z_{0})}\right|\leq\dfrac{1}{|\text{Im}(z_{0})|-\delta}\dfrac{|f(t)|}{|t-z_{0}|}, \end{align*} and it is already shown that the latter is integrable on the compact set.
On $\{|t|>2|z_{0}|+2\delta+1\}$, one sees that \begin{align*} |t|-|z_{0}|-\delta>\dfrac{1}{2}(|t|+1), \end{align*} so \begin{align*} |t-w||t-z_{0}|&\geq|\text{Im}(z_{0})|(|t-z_{0}|-|w-z_{0}|)\\ &\geq|\text{Im}(z_{0})|(|t|-|z_{0}|-\delta)\\ &\geq\dfrac{|\text{Im}(z_{0})|}{2}(|t|+1), \end{align*} but then \begin{align*} \left|\dfrac{f(t)}{(t-w)(t-z_{0})}\right|\leq\dfrac{2}{|\text{Im}(z_{0})|}\dfrac{|f(t)|}{|t|+1} \end{align*} which is integrable by assumption. So a dominated integrable function can be taken as \begin{align*} \dfrac{1}{|\text{Im}(z_{0})|-\delta}\dfrac{|f(t)|}{|t-z_{0}|}\chi_{|t|\leq 2|z_{0}|+2\delta+1}+\dfrac{2}{|\text{Im}(z_{0})|}\dfrac{|f(t)|}{|t|+1}\chi_{|t|>2|z_{0}|+2\delta+1}, \end{align*} we are done.