$\sum_{n=0}^\infty\exp(-\varepsilon_n / \tau) =\frac{\pi}{2}\int_0^\infty\exp(-\varepsilon_n/\tau) n dn$

Question

I'm working on a problem for my thermal physics course and got stuck when trying to evaluate a specific sum. Looking at the solutions, they transformed the sum into an integral. For some reason, I have never come across this in my physics career. The specific transformation I'm referring to is this:

$$\sum_{n=0}^\infty \lambda \exp(-\varepsilon_n / \tau) = \frac{\pi}{2}\int_0^\infty \lambda \exp(-\varepsilon_n / \tau) n \ \mathrm dn $$

What are the conditions to be able to evaluate a sum in this manner and what is the method to be able to find this integral?

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EDIT: It seems I should be a little more specific regarding the problem. The exact question I'm working on is:

Find the chemical potential of an ideal monatomic gas in two dimensions, with $N$ atoms confined to a square area $A=L^2$. The spin is zero.

For this system, $$\varepsilon_n = \frac{\hbar^2 \pi^2}{2m} \frac{(n_x^2 + n_y^2)}{L^2} = \alpha \frac{n^2}{A} $$

For $\alpha = \frac{\hbar^2 \pi^2}{2m}$

I did find the formula for the transformation in the book, but there is no derivation. They give it as fact, which I don't really like. Any help with why this is possible would be appreciated.

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Thank you.

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EDIT 2:

I believe I found a mistake in my book. In one area they use the expression I listed above, but in another there is an $n^2$ instead of an $n$. i.e.,

$$\sum_{n=0}^\infty \lambda \exp(-\varepsilon_n / \tau) = \frac{\pi}{2}\int_0^\infty \lambda \exp(-\varepsilon_n / \tau) n^2 \ \mathrm dn $$

Does this fix any issues with my question?

Answer 1

I believe the general idea goes as follows. Let $n^2 = n_1^2 + \ldots + n_d^2$. First make the approximation $$\sum_{n_1 \gt 0, ..., n_d \gt 0} f(n) e^{-\alpha n^2} \approx \int_0^\infty \dots \int_0^\infty f(n) e^{-\alpha n^2} \mathrm d n_1 \mathrm \dots \mathrm d n_d,$$ valid for small $\alpha$, assuming some conditions on $f$ hold.

Then convert the integral to spherical coordinates. Integrating over all spherical angles leaves $\mathrm d V_d(n)$, where $V_d(n)$ is the volume of the part of the $d$-dimensional ball of radius $n$ where all coordinates are positive. The result is $$\int_0^\infty f(n) e^{-\alpha n^2} \mathrm d V_d(n) = \frac {d \pi^{d/2}} {2^d \Gamma\left( \frac d 2 + 1 \right)} \int_0^\infty f(n) n^{d - 1} e^{-\alpha n^2} \mathrm d n.$$ In particular, $$\mathrm d V_2(n) = \mathrm d\left( \frac {\pi n^2} 4 \right) = \frac \pi 2 n \mathrm dn, \\ \mathrm d V_3(n) = \mathrm d\left( \frac {4 \pi n^3 / 3} 8 \right) = \frac \pi 2 n^2 \mathrm dn.$$

Answer 2

I think what you are supposed to do is write $$ \int_0^{\infty}{n\exp(-\varepsilon_n/\tau)}\, \mathrm dn=\sum_{k=0}^\infty{\int_k^{k+1}{n\exp(-\varepsilon_n/\tau)}\, \mathrm dn} $$ and show that the integral is equal to the individual term in the original sum.

I didn't try this in detail, because there are still a couple of symbols whose value I don't know, like $\tau$ and $\lambda$. Still, the general form looks about right.

Give it a shot and see what happens.

Answer 3

If the question is to know whether, for $c$ positive, $$\sum_{n=0}^\infty \exp(-cn^2) = \frac{\pi}{2}\int_0^\infty \exp(-cx^2)\, x \, \mathrm dx$$ then the answer is: No, since the RHS is $$\frac\pi2\,\frac1{2c}$$ while the LHS involves a Theta function, and has no simple form.

Edit: If one replaces $x$ by $x^2$ in the integral on the RHS, things do not become better since $$\int_0^\infty \exp(-cx^2)\, x^2 \, \mathrm dx=\frac1{2c\sqrt c}\int_0^\infty e^{-x}\, x^{1/2} \, \mathrm dx=\frac{\sqrt\pi}{4c\sqrt c}$$