# Example on Analytic continuation

Given $$f_1(z)=\int_0^\infty t^{z-1}e^{-t}dt$$ where $$\operatorname{Re}(z)>0$$ $$f_2(z)=\sum_{n=0}^\infty\frac{(-1)^n}{(n+z)(n!)}+\int_1^\infty t^{z-1}e^{-t}dt$$ except for the values $$z=0,-1,\ldots$$.

How to prove that $$f_2$$ is an analytic continuation of $$f_1$$?

I have been simply trying to represent function $$f_2$$ as a Taylor series of the $$f_1$$'s to show that $$f_1$$ is equal to $$f_2$$ in the intersecting domains. But dont know how to do this.

• I think it may be simpler to work with the integral of $t^{z-1}e^{-t}$ from $0$ to $1$ since that's the bit that changes between the two forms. I'm still working on it but based on my experience I suspect the way to go is to replace $e^{-t}$ with its Taylor expansion and then interchange the sum and the integral. Apr 12, 2021 at 9:38

Hints: $$\int_1^\infty t^{z-1}e^{-t}dt$$ is an entire function. When $$\Re z >0$$ we can write $$\int_0^1 t^{z-1}e^{-t}dt$$ as $$\sum \frac {(-1)^{n}} {(n+z)n!}$$ by expanding $$e^{-t}$$ in its Taylor series and integrating term by term. But now we can see that this series converges for any $$z$$ which is not of the form $$-n$$ with $$n \in\{0,1,2...\}$$. The sum is analytic in $$\mathbb C \setminus \{0,-1,-2,...\}$$ because the series converges uniformly on compact subsets of this region. Thus we have found an analytic continuation of $$\int_0^\infty t^{z-1}e^{-t}dt$$ to $$\mathbb C \setminus \{0,-1,-2,...\}$$.