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Let $\{ \lambda_i | \ i \in \mathbb{N}_0\} \subset \mathbb{R}^+$ the spectrum of a Laplacian on a compact closed Riemannian manifold.

I have to show that the function

$ f(s) := \frac{1}{\Gamma(s)} \int_0^1 t^{s-1} ( \sum_{i=1}^\infty e^{-\lambda_1t}) \mathrm{dt}$

is holomorphic. In a book I found the following statement:

Since the Gamma-function is never zero and f is of the form $ \frac{1}{\Gamma(s)} \int_0^1 t^{s-1} O(e^{-\lambda_1t}) \mathrm{dt}$, where $\lambda_ 1$ refers to the first nonzero Eigenvalue, f is holomorphic"

Can someone explain exactly how this argumentation proves the holomorphy?

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1 Answer 1

up vote 7 down vote accepted

There are four statements here: (1) the function $t^s$ is holomorphic with respect to $s$, for any fixed $t\ge 0$. (2) The integral of $t^s$ multiplied by any bounded measurable function of $t$, and taken over a bounded interval, is holomorphic wrt $s$. (3) The Gamma function is holomorphic. (4) The ratio of two holomorphic functions is again holomorphic, provided that the denominator does not vanish.

The least obvious part is (2). The right tool for its proof is Morera's theorem. It breaks (2) further into (2a) the integral is continuous with respect to $s$, which is true by the uniform continuity of $s\mapsto t^s$. (2b) Fubini's theorem can be applied to evaluate the integral over a closed curve: first, integrate $t^s\, ds$ and get zero, then do the $dt$ integration (and still have zero).

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Thanks a lot for your detailed description. When trying to write down your steps I got a little bit confused when evaluating the integral of over a closed curve $c$: Let $f_i$ be one summand of f, when one takes the sum outside the integral. Then: $\oint_c f_i(s) ds = \int_a^b \int_0^1 t^{c(u)} e^{-t\lambda_i} \mathrm{d c(u)} |c'(u)|\mathrm{du}$. But why is this zero? Or did I miss something? –  Robin Neumann Sep 26 '12 at 11:18
    
@Robin The Cauchy theorem –  user31373 Sep 26 '12 at 11:40
    
Ohhhh of course! Thank you again. –  Robin Neumann Sep 26 '12 at 18:15
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