Doubts while understanding why the Laplace transform is holomorphic? I am trying to understand why the Laplace transform is holomorphic. Let's be more precise, let $\phi(w):\mathbb{R}^+\to\mathbb{C}$ be continuous. Let $A>0$ and $c\in\mathbb{R}$ such that
$$
|\phi(w)|\leq Ae^{cw}
$$
Consider now the Laplace transform of $\phi$ that I will call $\Phi$
$$
\Phi(z)=\int_0^{\infty}e^{-zw}\phi(w)\,dw
$$
We can easily show that the absolute value of the integral above is smaller than infinity in the domain $B_c=\{z\in\mathbb{C}:Re(z)>c\}$. Apparently you can prove that $\Phi(z)$ is holomorphic in B but I don't understand the steps given, which are the following. Because $|e^{-zw}|=e^{-wRe(z)}\leq e^{-c_1w}$ for $z\in B_d=\{z\in\mathbb{C}:Re(z)>d\}$ we can find $F:\mathbb{R}^+\to\mathbb{R}^+$ integrable and independent of $z$ such that $|e^{-zw}\phi(w)|\leq F(w)$ and deduce that $\Phi$ is holomorphic on $B_c$. I don't understand the last sentence. First, I would like more details on why we can fin an $F$ with the above properties, and why this guarantees the holomorphicity of $\Phi$.
 A: Morera's theorem. Let be $\gamma\subset B_c$ a closed curve. Using Fubini (the line integral is an ordinary integral in some interval $[a,b]$) and the holomorphicity of $z\mapsto e^{-zw}\phi(w)$:
$$\int_{\gamma}\Phi(z)\,dz = \int_{\gamma}\int_0^{\infty}e^{-zw}\phi(w)\,dwdz = \int_0^{\infty}\int_{\gamma}e^{-zw}\phi(w)\,dzdw =  \int_0^{\infty}0\,dw = 0.$$
A: As you explained before, you have $$|e^{-zw}|=e^{-wRe(z)}\leq e^{-c_1w}.$$ Hence  $$|e^{-zw}\phi(w)|\leq e^{-c_1w} |\phi(w)| \leq Ae^{-c_1w}e^{cw}.$$ The $F$ you want is therefore given by $$F(w) = A e^{(c-c_1)w}.$$ It is integrable since $c_1 >c$ by construction. Now, why is $\Phi$ holomorphic. By definition, $\Phi$ is holomorphic if $$\frac{\partial \Phi}{\partial \overline{z}} =0.$$ Now we have to check the hypothesis to permute derivative and integral. Actually it will be essentially the same as what we did previously with $F$. So you can write $$\frac{\partial \Phi}{\partial \overline{z}} = \int_0^{\infty}\frac{\partial}{\partial \overline{z}}(e^{-zw}\phi(w))\,dw.$$ And of course this is equal to $0$ since $$\frac{\partial}{\partial \overline{z}}(e^{-zw})=0$$ by holomorphicity of the exponential. 
