# Show that function is holomorphic

Let $f$ : $[a, b] \rightarrow\mathbb C$ be continuous.

Show that the function $F : \mathbb C \rightarrow \mathbb C$ defined by $$F(z) = \int_a^b f(t)\exp(tz) dt$$ is holomorphic.

I am unsure of how to start, I think I should get some property of holomorphic functions like composition or integral of holomorphic functions is holomorphic but I'm not sure if any such thing exists.

You just have to observe that$$\int_a^bf(t)\exp(tz)\,\mathrm dt=\int_a^bf(t)\sum_{n=0}^\infty\frac{t^nz^n}{n!}\,\mathrm dt=\sum_{n=0}^\infty\left(\int_a^b\frac{f(t)t^n}{n!}\,\mathrm dt\right)z^n.$$

• So we can say that as it is analytic, it's holomorphic right. And why can we interchange summation and integral in this, there is some restriction on it I suppose? – john doe Nov 17 '17 at 11:36
• @johndoe We can interchange it because the convergence is uniform. – José Carlos Santos Nov 17 '17 at 11:37

We can use Morera's theorem.

Let $\gamma: [0,1] \to \mathbb C$ be a closed piecewise $C^1$ path. Then

$\displaystyle \int_\gamma F(z)\,dz =\int_0^1 F(\gamma(s))\gamma'(s)\,ds$

$\displaystyle =\int_0^1 \int_a^b f(t)\exp(t\gamma(s))\,dt\, \gamma'(s) \,ds$

$\displaystyle =\int_a^b \int_0^1 f(t)\exp(t\gamma(s))\gamma'(s) \,ds\, \,dt$

$\displaystyle =\int_a^b 0 \,dt = 0$

The switch in the order of integration is ok because the integrand is continuous and the domain is compact.

The last inner integral is $\displaystyle \int_\gamma f(t) \exp(tz)\,dz$ and so is zero by Cauchy's theorem, since the integrand is a holomorphic function for fixed $t$.