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In what sets are these functions analytic? $$\begin{align} 1. \qquad & \int_0^z \exp(\zeta^2)\,d\zeta \\ 2.\qquad & \int_0^\infty t^{z-1}\exp(-t)\, dt\\ 3. \qquad& \sum_{n=1}^ \infty n^{-z} \end{align}$$

For number 2 I think it is analytic at all of the complex numbers except the integers, but that is just random guess. No idea for number 1. Any suggestion would help. Thanks.

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What do you mean exactly by $z>1$? –  Jonas Meyer Dec 28 '12 at 5:34
    
I mean for $z=1$ you will have harmonic series which diverges, but when you have $z>1$, then by p-test, #3 converges. –  Deepak Dec 28 '12 at 7:32
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The first one can be related to the error function, the second one is the Gamma function $\Gamma(z)$, and the third one is the Riemann zeta function $\zeta(z)$. –  Mhenni Benghorbal Dec 29 '12 at 18:54
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Use Morera's theorem for the first. Consider the integral of that function over an arbitrary curve, switch the order of integration, and show it vanishes. The others are standard functions whose properties you can look up: the Gamma and Zeta functions respectively. –  Potato Dec 29 '12 at 21:46
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See here for proving the first function is entire. –  Mhenni Benghorbal Dec 29 '12 at 22:12
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1 Answer

The question can be interpreted in two ways:

  • find the maximal region in which the integral defines a holomorphic function
  • find the maximal region to which the aforementioned function can be extended holomorphically (such an extension need not be given by the stated integral)

With the first interpretation the answers are $\mathbb C$, $\{z\in\mathbb C:\operatorname{Re}z>0\}$, and $\{z\in\mathbb C:\operatorname{Re}z>1\}$. The second and third integrals diverge outside of the stated regions, and therefore do not define a holomorphic function there.

With the second interpretation the answers are $\mathbb C$, $\mathbb C\setminus \{0,-1,-2,-3,\dots\}$, and $ \mathbb C\setminus \{1\}$. This is not trivial, and requires the consideration of relevant functional equations for $\Gamma$ function and $\zeta$ function. (Pointed out by Mhenni Benghorbal).

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