# Integral of holomorphic function in 2 variables is holomorphic

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

Can anyone help?

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## Answers and Replies

Landau
Science Advisor
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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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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Landau
Science Advisor
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 $f:\mathbb{C}\to \mathbb{C}$ has the property that $$\forall R > 0, \exists M > 0$$ such that $$\forall s \in \mathbb{C}$$ with $$|s| \leq R, |f(s)| \leq M$$. Then f is holomorphic.

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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 $f:\mathbb{C}\to \mathbb{C}$ has the property that $$\forall R > 0, \exists M > 0$$ such that $$\forall s \in \mathbb{C}$$ with $$|s| \leq R, |I_\rho(s)| \leq M$$. 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 $$g : \mathbb{C}^2 \to \mathbb{C}$$ is holomorphic in both variables and define:
$$f : \mathbb{C} \to \mathbb{C}, s \mapsto \int_C g(s, z) dz$$

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

then $$g$$ is holomorphic.

Landau
Science Advisor
The OP still seems to ask something else than the above:
However, my notes claim I can show that $$I_\rho$$ is holomorphic on $$\mathbb{C}$$ by showing (...)
So you agree that $I_\rho:\mathbb{C}\to \mathbb{C}$ is holomorphic for every \rho? But you want to show that in fact

$$\mathbb{C}^2\to \mathbb{C}$$
$$(\rho,s)\mapsto I_{\rho}(s)$$

is holomorphic?

The OP still seems to ask something else than the above:

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

$$\mathbb{C}^2\to \mathbb{C}$$
$$(\rho,s)\mapsto I_{\rho}(s)$$

is holomorphic?

No, the z and s variables coincide with the integrand in the definition of $$I$$, which is independent of $$\rho$$. So here, $$g : \mathbb{C}^2 \to \mathbb{C}, (s, z) \mapsto z^{s-1} / (e^z - 1)$$. 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, $$g$$ is holomorphic in both $$s$$ and $$z$$, where it's defined. My notes claim that $$I_\rho$$ is holomorphic because the uniform bound property I stated holds, but I don't see why it follows.

Landau
Science Advisor
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

$$\int_{C_\rho} \left| \frac{z^{s-1}}{e^z-1} \right| dz \leq M$$

then certainly

$$|I_\rho(s)| \leq M$$

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.