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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.

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

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  • $\begingroup$ 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? $\endgroup$ – john doe Nov 17 '17 at 11:36
  • $\begingroup$ @johndoe We can interchange it because the convergence is uniform. $\endgroup$ – José Carlos Santos Nov 17 '17 at 11:37
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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 $t$.

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