# Writing three-fold summation over n_x, n_y, n_z as single summation over n?

1. Nov 17, 2011

### PerUlven

1. The problem statement, all variables and given/known data
On this page they're going from a three-fold summation to a single summation (cubed) (equation 58-59), something like this:

$\sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-a\left(n_x^2 + n_y^2 + n_z^2\right)} = \left( \sum_n e^{-an^2} \right)^3$, where $a = \frac{h^2}{8kTmL^2}$

Now, I've run into a similar problem, only I have the square root of the n's, like this:

$Z = \sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-c\sqrt{n_x^2 + n_y^2 + n_z^2}}$, where $c = \frac{hc}{2kTL}$

Could anyone show me how they do the first transition, or some hints to how I should start when trying to do a similar transition on my equation? I'm supposed to show that the sum equals

$Z(\text{high T}) = 8\pi L^3\left(\frac{kT}{hc}\right)^3$

in the high-temperature limit. I know that I can approximate the sum with an integral, but I need to convert it to a single-sum first.

2. Relevant equations

$\sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-a\left(n_x^2 + n_y^2 + n_z^2\right)} = \left( \sum_n e^{-an^2} \right)^3 = \left(\int_0^\infty e^{-an^2} dn\right)^3$

$\int_0^\infty e^{-ax}dx = \frac{1}{a}$

$\int_0^\infty e^{-ax^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a}}$

3. The attempt at a solution
I've tried letting $\sqrt{n_x^2+n_y^2+n_z^2} = n$, but then I end up with just $n$ and not $n^2$ in the exponent, and then I don't get the factor $\pi$ when solving the integral (see the two last equations for solutions to the integrals).

I might be going at this problem all wrong, so I'm open to all suggestions!

EDIT: Just found the solution myself. I'll show it in detail if anyone wants it, just ask and I'll write it out here.

Last edited: Nov 17, 2011