Integral of holomorphic function in 2 variables is holomorphic

  • Thread starter Giraffro
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Define [tex]\forall \rho \in (0,\pi), C_\rho[/tex] to be contour travelling from [tex]+\infty + \pi i/2[/tex] to [tex]\rho i[/tex], then a semicircle to [tex]-\rho i[/tex] then a straight line to [tex]+\infty -\rho i[/tex]. Also define:
[tex]I_\rho : \mathbb{C} \to \mathbb{C}, s \mapsto \int_{C_\rho} \frac{z^{s-1}}{e^z - 1} dz[/tex]
I've shown that this function is well-defined, independent of the value of [tex]\rho[/tex] and [tex]\forall \rho \in (0, \pi), \forall s \in \mathbb{C}[/tex] with [tex]\Re(s) > 1, I_\rho(s) = (e^{2 \pi i s} - 1) \Gamma(s) \zeta(s)[/tex] - This is part of a proof of the functional equation for the Riemann zeta function in my lecture notes. However, my notes claim I can show that [tex]I_\rho[/tex] is holomorphic on [tex]\mathbb{C}[/tex] by showing [tex]\forall R > 0, \exists M > 0[/tex] such that [tex]\forall s \in \mathbb{C}[/tex] with [tex]|s| \leq R[/tex]:
[tex]\int_{C_\rho} \left| \frac{z^{s-1}}{e^z-1} \right| dz \leq M [/tex]
I can't find a reference that shows this gives you a holomorphic function.

Can anyone help?
 
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  • #2
Landau
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I may be mistaking, but doesn't this follow from [URL [Broken] Theorem[/url]? If f is bounded on every disc, then I guess its integral over every closed curve must vanish.
 
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  • #3
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I may be mistaking, but doesn't this follow from [URL [Broken] Theorem[/url]? If f is bounded on every disc, then I guess its integral over every closed curve must vanish.

Edit: Misread you're article and thought it was referring to the reverse implication AKA Cauchy's residue formula.
 
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  • #4
Landau
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I was suggesting that the claim of your notes implies that the hypotheses of Morera's theorem are satisfied. Forget about your I_\rho for a moment; we are trying to prove the following:

Suppose [itex]f:\mathbb{C}\to \mathbb{C}[/itex] has the property that [tex]\forall R > 0, \exists M > 0[/tex] such that [tex]\forall s \in \mathbb{C}[/tex] with [tex]|s| \leq R, |f(s)| \leq M[/tex]. Then f is holomorphic.
 
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  • #5
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I was suggesting that the claim of your notes implies that the hypotheses of Morera's theorem are satisfied. Forget about your I_\rho for a moment; we are trying to prove the following:

Suppose [itex]f:\mathbb{C}\to \mathbb{C}[/itex] has the property that [tex]\forall R > 0, \exists M > 0[/tex] such that [tex]\forall s \in \mathbb{C}[/tex] with [tex]|s| \leq R, |I_\rho(s)| \leq M[/tex]. Then f is holomorphic.

I don't know whether you caught the edit, but I misread you're article and I've updated my OP to a stronger condition. I'm not particularly worried about the bound since I have one, so the claim I suppose we're trying to prove is:

If [tex]g : \mathbb{C}^2 \to \mathbb{C}[/tex] is holomorphic in both variables and define:
[tex]f : \mathbb{C} \to \mathbb{C}, s \mapsto \int_C g(s, z) dz[/tex]

If [tex]\forall R > 0, \exists M > 0 : \forall s \in \mathbb{C}[/tex] with [tex]|s| \leq R[/tex]:
[tex]\int_C |g(s, z)| dz \leq M[/tex]

then [tex]g[/tex] is holomorphic.
 
  • #6
Landau
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The OP still seems to ask something else than the above:
However, my notes claim I can show that [tex]I_\rho[/tex] is holomorphic on [tex]\mathbb{C}[/tex] by showing (...)
So you agree that [itex]I_\rho:\mathbb{C}\to \mathbb{C}[/itex] is holomorphic for every \rho? But you want to show that in fact

[tex]\mathbb{C}^2\to \mathbb{C}[/tex]
[tex](\rho,s)\mapsto I_{\rho}(s)[/tex]

is holomorphic?
 
  • #7
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The OP still seems to ask something else than the above:

So you agree that [itex]I_\rho:\mathbb{C}\to \mathbb{C}[/itex] is holomorphic for every \rho? But you want to show that in fact

[tex]\mathbb{C}^2\to \mathbb{C}[/tex]
[tex](\rho,s)\mapsto I_{\rho}(s)[/tex]

is holomorphic?

No, the z and s variables coincide with the integrand in the definition of [tex]I[/tex], which is independent of [tex]\rho[/tex]. So here, [tex]g : \mathbb{C}^2 \to \mathbb{C}, (s, z) \mapsto z^{s-1} / (e^z - 1)[/tex]. Actually just noticed this is undefined at the origin, but our contour doesn't pass through it, so we should be okay. In any case, [tex]g[/tex] is holomorphic in both [tex]s[/tex] and [tex]z[/tex], where it's defined. My notes claim that [tex]I_\rho[/tex] is holomorphic because the uniform bound property I stated holds, but I don't see why it follows.
 
  • #8
Landau
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There still seems to be miscommunication. You probably made a typo in the last sentence of your previous post, where you say you want to show that g is holomorphic. So my understanding is that you DO want to prove that, for every rho, I_rho is holomorphic. Yes? Please be clear about this.

If so, I stand by my very first reaction, because the change you made in the OP is only a weaker condition:

if

[tex]\int_{C_\rho} \left| \frac{z^{s-1}}{e^z-1} \right| dz \leq M [/tex]

then certainly

[tex]|I_\rho(s)| \leq M [/tex]

by the triangle inequality for integrals. So then we're back at post 4 where I suggest that I_\rho satisfies the hypothesis of Morera's Theorem.
 

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