Given that $\phi:[0,\infty)\rightarrow \mathbb{C}$ is a bounded continuous function, prove that $$H(z)=\int^{\infty}_0\frac{\phi(t)}{t^2+z}dt$$ is a holomorphic function on $\mathbb{C}-\{z\in\mathbb{R},z\le0\}.$
It's my first time to come across such a question and I lacked a clue. I just thought that this formula was very similar to the Cauchy integral formula, but the denominator was raised to the power of 2. Can you give me some hints?
One strategy is to show that $H$ admits a power series about any point in $\mathbb{C}\setminus\{(-\infty,0]\{0\}$ with the help of a little integration theory (dominated convergence).
Here is a sketch of a solution. I let the OP deal with the details should she/he is interested.
One can consider the map $$F(z)=\int^{\infty}_0\frac{\phi(t)}{t^2-z}dt$$ to make things look more symmetric. The OP's map is $H(z)=F(-z)$.
The map $F$ is analytic (and hence holomorphic) on $D=\{z=x+iy: x>0\}$. To wit, choose $M>0$ so that $|\phi(t)|\leq M$ for all $t$. Fix $a=\alpha+i\beta\in D$. Notice that $\rho=\inf\{|a-t^2|:t\in\mathbb{R}\}>0$.
Let $B(a;r)$ be an open ball with $r<\rho$. Notice that $$\frac{\phi(t)}{t^2-z}=\frac{\phi(t)}{(t^2-a)-(z-a)}=\frac{\phi(t)}{t^2-a}\frac{1}{1-\tfrac{z-a}{t^2-a}}=\frac{\phi(t)}{t^2-a}\sum^\infty_{n=0}\Big(\frac{z-a}{t^2-a}\Big)^n\tag{0}\label{zero}$$ The series converges absolutely and uniformly on $B(a;r)$ since $\Big|\frac{z-a}{t^2-a}\Big|\leq \frac{r}{\rho}<1$; furthermore, $$ \left|\frac{\phi(t)}{t^2-a}\sum^\infty_{n=0}\Big(\frac{z-a}{t^2-a}\Big)^n\right|\leq\frac{M}{\sqrt{(t^2-\alpha)^2+\beta^2}} \frac{\rho}{\rho-r}\in L_1(0,\infty) $$ The conclusion follows by an application of dominated convergence. In addition, one can obtain the expression: $$F(z)=\sum^\infty_{n=0}c_n(z-a)^n,\qquad z\in B(a;r)$$ where $c_n=\int^\infty\frac{\phi(t)}{(t^2-a)^{n+1}}\,dt$.
