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Is it valid to approximate the function $$ Z(t)=\sum_{n}e^{-tE_{n}} ,\ t\ge 0$$

by the integral over phase space: $$ \frac{ 1}{2\sqrt \pi}\int_{0}^{\infty}dxe^{-tV(x)}?$$

For example, in order to evaluate the zeta function over eigenvalues

$ \Gamma (s) \zeta _{H} (s,q) = \int_{0}^{\infty}dtZ(t)\exp(-qt)Z(t)$

with 'q' a positive number, and $ \zeta _{H} (s,q)= \sum_{n}(q+E_{n})^{-s} $

Is this approximation faithful and reliable?

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It really rather depends on what $V(x)$ is, doesn't it? Clearly if $V(x) = \sum_n \delta(E(x)-E_n)$ (with $\delta$=Dirac delta function) then the two expressions will be exactly equal, aside from that odd-looking factor of $\frac{1}{2\sqrt{\pi}}$. Generally if $V$ is some kind of density of energy states with respect to $x$ then you can make this type of approximation as long as the density is not too small. If you can give more of the context I might be able to give a proper answer. –  Nathaniel Apr 15 '12 at 21:42
($E(x)$ probably needs to be strictly increasing in order for the above to work. I probably should just have written "$\delta(x-E_n)$" but was trying to be clever by making the units match.) –  Nathaniel Apr 15 '12 at 21:50
i am talking about a potential that tends to $\infty$ as $ x \rightarrow \infty$ –  Jose Javier Garcia Apr 16 '12 at 10:04
Ah, I see, I think - am I right in thinking your $V(x)$ is a quantum mechanical potential for a single particle moving in 1d, $V(x)$ is symmetrical about 0, the $E_n$ are the energy levels of the eigenstates, and $t=1/T$? In that case I would suspect the approximation is good as long as $t$ is small enough. But I'm not really sure because my expertise is in classical stat. mech. rather than quantum stuff. Perhaps someone more familiar with it can give an answer. It may still help to edit the question to spell out the context a bit more clearly. –  Nathaniel Apr 16 '12 at 10:46
@Nathaniel I'm not sure there is anyone more familiar with it ;-P In any case, it probably wouldn't hurt for you to offer an answer, even if it isn't perfect (if you would like to, of course). –  David Z Apr 19 '12 at 5:58
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