showing a function defined from an integral is entire

Solution 1:

Let $z$ fixed. We have $$e^{zt}=\sum_{n=0}^{+\infty}\frac{t^n}{n!}z^n,$$ and this series is normally convergent on $[0,1]$. Consequently, we can switch the series and the integral, to get $$\int_0^1f(z)e^{zt}dt=\int_0^1f(t)\sum_{n=0}^{+\infty}\frac{t^n}{n!}z^ndt=\sum_{n=0}^{+\infty}\frac 1{n!}\left(\int_0^1f(t)t^ndt\right)z^n,$$ and $f$ is actually a power series (of infinite radius of convergence, hence entire.

Solution 2:

Related problem 1, related problem 2. You can use Morera's theorem which states that a continuous, complex-valued function ƒ defined on a connected open set D in the complex plane that satisfies $$ \oint_{\gamma} f(z)\, dz = 0 $$ for every closed piecewise C1 curve $\gamma$ in $D$ must be holomorphic on $D$. Applying this theorem to your case, we have

$$ \oint_{\gamma} f(z) dz = \oint \int_{0}^{1} f(t){\rm e}^{zt}dt dz = \int_{0}^{1}f(t)\oint_{\gamma} {\rm e}^{zt} dz \,dt = \int_{0}^{1} f(t) (0) dt = 0$$

The inner integral equals $0$, since ${\rm e}^{zt} $ is analytic and hence by Cauchy theorem the integral is zeros. The interchanging of the integrals is justified by the uniform convergence of the $\int_{0}^{1} f(t) {\rm e }^{zt} $ in $z$.

Solution 3:

You can differentiate under the integral sign with respect to $z$ in such situations... so not only is it entire but its derivative is given by $\int_0^1 tf(t)e^{zt}\,dt$. To prove it, you apply the dominated convergence theorem to the difference quotients ${F(z + h) - F(z) \over h} = \int_0^1 f(t)e^{zt} {e^{ht} - 1 \over h}\,dt$ as $h \rightarrow 0$.