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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.

Daniele Tampieri
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Shaurya Anand
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  • 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. – Stephen Donovan Apr 12 '21 at 09:38

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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 $\mathbb C \setminus \{0,-1,-2,...\}$.

Kavi Rama Murthy
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    Minor footnote for the OP: the interchange of the sum and integral is justified by the Fubini-Tonelli theorem because the sum that we get is absolutely convergent. See here for details: https://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged – Stephen Donovan Apr 12 '21 at 09:53