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

[PerUlven]

Homework Statement

On http://www.chem.arizona.edu/~salzmanr/480b/statt02/statt02.html 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.

Homework Equations

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

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

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

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.

 

[mark726]

Thank you for bringing this interesting problem to our attention. I am always excited to see others exploring and learning about mathematical concepts.

I have looked into the problem you mentioned and I believe I have found a solution that may help you. First, let's recap what we know:

We have the following equation:

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}

And we want to show that in the high-temperature limit, this equation becomes:

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

To do this, we will use the following identities:

1. \sum_{n=1}^\infty e^{-an^2} = \frac{1}{2}\sqrt{\frac{\pi}{a}} \text{ (from integral identity 2)}

2. \sum_{n=0}^\infty e^{-an^2} = \frac{1}{2}\sqrt{\frac{\pi}{a}} \text{ (from integral identity 1)}

Now, let's start by rewriting the original equation:

Z = \sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-c\sqrt{n_x^2 + n_y^2 + n_z^2}}
= \sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-c\sqrt{(n_x^2 + n_y^2 + n_z^2)^2}}
= \sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-c(n_x^2 + n_y^2 + n_z^2)}
= \sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-c(n_x^2 + n_y^2 + n_z^2)} \cdot e^{-c(n_x^2 + n_y^2 + n_z^2)}
= \left(\sum_{n_x} \sum_{n_y} \sum_{n_z} e^{-c(n_x^2 + n_y^2 + n_z^2)}\right)^2

Now,