analytic function integral question Suppose $\phi$ is analytic in $\mathbb{C}$ and $g$ is continuous on a closed interval $[a,b]$ in $\mathbb{R}$.  Let $$f(z)=\int_a^bg(t)\phi(zt)dt.$$  Prove first that, in $\mathbb{C}\,$, $f$ is continuous, and then prove that $f$ is analytic. 
 A: *

*We show continuity. Let $x_0\in\Bbb C$ fixed, and $\gamma$ a closed curve such that for each $z$ which satisfies $|z-z_0|\leq 1$ and each $t\in [a,b]$, $|tz-z'|\geq 1$ for all $z'$ in the support of the curve $\gamma$. We use Cauchy integral formula, which gives, for each $z'$ in the interior of the curve 
$$\phi(z')=\frac 1{2\pi i}\int_{\gamma}\frac{\phi(\xi)}{\xi-z'}d\xi.$$
Applying it to $z'=tz$ and $z'=tz_0$, we get
\begin{align}
2\pi i(f(z)-f(z_0))&=\int_a^bg(t)\int_{\gamma}\phi(\xi)\left(\frac 1{\xi-tz}-\frac 1{\xi-tz_0}\right)d\xi dt\\
&=\int_a^b\int_{\gamma}g(t)\phi(\xi)\frac{\xi-tz_0-(\xi-tz))}{(\xi-tz)(\xi-tz_0)}d\xi dt\\
&=\int_a^b\int_{\gamma}g(t)\phi(\xi)t(z-z_0)\frac 1{(\xi-tz)(\xi-tz_0)}d\xi dt.
\end{align}
We deduce that 
$$|f(z)-f(z_0)|\leq \sup_{t\in [a,b]}|tg(t)|\sup_{\xi\in\gamma}|\phi(\xi)|\frac 1{2\pi}|z-z_0|,$$
which gives continuity. 

*Now we show that $f$ is holomorphic. We fix $z\in\Bbb C$, and we shall show that $$\lim_{h\to 0}\frac{f(z+h)-f(z)}h-\int_a^btg(t)\phi'(tz)=0.$$
We use again Cauchy's integral formula, applied to the same curve $\gamma$. We consider $|h|$ small enough. Write $A(h):=2\pi i\left(\frac{f(z+h)-f(z)}h-\int_a^btg(t)\phi'(tz)\right)$. Then 
$$A(h)=\int_a^bg(t)\left(\frac{\phi((z+h)t)-\phi(zt)}h-t\phi'(zt)\right)dt
$$
and 
\begin{align}
\frac{\phi((z+h)t)-\phi(zt)}h-t\phi'(zt)&=\int_{\gamma}\phi(\xi)\left(\frac 1{h(\xi-(z+h)t)}-\frac 1{(\xi-zt)h}-\frac t{(\xi-zt)^2}\right)d\xi\\
&=\int_{\gamma}\phi(\xi)\left(\frac{\xi-zt-\xi+zt+th}{h(\xi-(z+h)t)(\xi-zt)}-\frac t{(\xi-zt)^2}\right)d\xi\\
&=\int_{\gamma}\phi(\xi)\left(\frac{t}{(\xi-(z+h)t)(\xi-zt)}-\frac t{(\xi-zt)^2}\right)d\xi\\
&=t\int_{\gamma}\phi(\xi)\frac{\xi-zt-(\xi-(z+h)t)}{(\xi-(z+h)t)(\xi-zt)^2}d\xi\\
&=t\int_{\gamma}\phi(\xi)\frac{\xi-zt-\xi+zt+ht)}{(\xi-(z+h)t)(\xi-zt)^2}d\xi\\
&=ht^2\int_{\gamma}\phi(\xi)\frac 1{(\xi-(z+h)t)(\xi-zt)^2}d\xi.
\end{align}
We get that 
$$|A(h)|\leq h\sup_{t\in [a,b]}|t^2g(t)|\sup_{\xi\in\gamma}|\phi(\xi)|.$$

A: Continuity.
Only continuity of $\phi$ is needed. Let $z_0\in\mathbb{C}$. We show that $f$ is continuous at $z_0$. The function $\phi(t\,z)$ is continuous as a function of $(t,z)$, and hence uniformly continuous, on $[a,b]\times\{|z-z_0|\le1\}$. Given $\epsilon>0$ there is a $\delta>0$ such that
$$
|\phi(t\,z)-\phi(t\,z_0)|\le\frac{\epsilon}{\int_a^b|g(t)|\,dt},\quad t\in[a,b],\quad |z-z_0|\le\delta.
$$
Then
$$
|f(z)-f(z_0)|\le\int_a^b|g(t)|\,|\phi(z\,t)-\phi(t\,z_0)|\,dt\le\epsilon.
$$
Analiticity.
We apply Morera's theorem. Given any closed piecewise $C^1$ curve $\gamma$,
$$
\int_\gamma f(z)\,dz=\int_a^bg(t)\Bigl(\int_\gamma\phi(t\,z)\,dz\Bigr)\,dt=0.
$$
The change in the order of integration is justified by the continuity of $g(t)\,\phi(t\,z)$ on $[a,b]\times\gamma$, and tha analiticity of $\phi$ implies that $\int_\gamma\phi(t\,z)\,dz=0$ for all $t\in[a,b]$.
