# Ascertaining a mathematical equality to derive a partition function

we have an equation like this:

$$\mathcal N(x)=\sum_{q=1}^\infty (\psi(x,q) \log(q)) \qquad (1)$$

while $\psi(x)$ is the function for some oscillations (may contain complex part), $x\in \Bbb R$ and $q \in \Bbb N$.

It is further ensured that:

$$\mathcal N(x) = \left\{ \begin{array}{l l} x & \quad \text{if x\in \Bbb N}\\ 0 & \quad \text{otherwise} \end{array} \right.$$

I need your help, even if it sounds simple, whether the following equality is mathematically correct:

$$n^{-\psi^*}=\mathcal N(n)^{-\psi(n,q)}$$

with $n \in \mathbb{N}$ and $\psi^*=\psi(n,q)$.

I would really appreciate your help. I need this for constructing a partition function.

If the above equality as such is not correct, how is the correct formalism?

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The second part about $\mathcal{N}(x)$ doesn't make sense. What is $n$, where does it come from? – Raskolnikov Jun 13 '13 at 8:59
the second part looks trivial but is true. See the first two equations as a spectrum of particles $\mathcal N$ points then to the location of the particles which is exactly the location of positive integers on real number line. The second equation is implicit in the first. Does thta help? – al-Hwarizmi Jun 13 '13 at 9:07
@Raskolnikov imagine $\mathcal N$ as a Fourier that points out the natural numbers. – al-Hwarizmi Jun 13 '13 at 9:10
I've made some edits. But it now seems to me your expression is trivially true. – Raskolnikov Jun 13 '13 at 9:10
Dear al-Hwarizmi, the Fourier transform of your function that is zero almost everywhere yet finite is zero. To get a nontrivial function in the L2 or Fourier sense, you would have to have delta-functions located at the integer values of $x$, like $\sum_k k\delta(x-k)$, not just a finite value. – Luboš Motl Jun 13 '13 at 13:27