Integral of holomorphic function in 2 variables is holomorphic

In summary, it seems that the claim in your notes that I_\rho is holomorphic is straightforward from the hypotheses.
  • #1
Giraffro
4
0
Define [tex]\forall \rho \in (0,\pi), C_\rho[/tex] to be contour traveling 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
I may be mistaking, but doesn't this follow from 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
Landau said:
I may be mistaking, but doesn't this follow from 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
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
Landau said:
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
The OP still seems to ask something else than the above:
Giraffro said:
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
Landau said:
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
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.
 

1. What is a holomorphic function in two variables?

A holomorphic function in two variables is a complex-valued function that is differentiable at every point in its domain. In other words, it is a function that can be approximated by a linear transformation at every point in its domain.

2. What does it mean for an integral to be holomorphic?

An integral is said to be holomorphic if it is differentiable at every point in its domain. This means that the function being integrated can be approximated by a linear transformation at every point in its domain.

3. Why is the integral of a holomorphic function in two variables also holomorphic?

This is because the integral of a holomorphic function is itself a holomorphic function. This can be proven using the Cauchy-Riemann equations, which state that if a function is holomorphic, its partial derivatives must satisfy a set of conditions. These conditions are also satisfied by the partial derivatives of the integral, making it a holomorphic function.

4. Can a non-holomorphic function have a holomorphic integral?

Yes, it is possible for a non-holomorphic function to have a holomorphic integral. This can happen if the function has a removable singularity at a point, meaning that it can be extended to be holomorphic at that point. In this case, the integral will also be holomorphic.

5. How is the integral of a holomorphic function in two variables related to complex analysis?

The study of holomorphic functions, including their integrals, is an important part of complex analysis. This field of mathematics deals with functions of complex variables and their properties. The integral of a holomorphic function in two variables is a fundamental concept in complex analysis, and it is used to solve many problems in this field.

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