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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?

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.

Thank you.

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?

NoVa
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    What is $\varepsilon_n$? – saulspatz Apr 16 '18 at 18:28
  • In this context, $\varepsilon$ is the energy of my system. $\varepsilon = \frac{\hbar^2 \pi^2}{2m} \frac{(n_x^2 +x_y^2)}{L^2}$. Is the exact definition of that term necessary in the transformation? – NoVa Apr 16 '18 at 18:31
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    Yes, I should think so. The formula isn't true for arbitrary functions. – saulspatz Apr 16 '18 at 18:35
  • @saulspatz, does my edit clarify anything? – NoVa Apr 16 '18 at 18:53
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    So basically, the sum is $$\sum_{n=0}^{\infty}{e^{Cn^2}}$$ for some constant $C$? – saulspatz Apr 16 '18 at 19:01
  • Yeah, I suppose. – NoVa Apr 16 '18 at 19:02
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    Now that we know the form of $\varepsilon_n$ the integral is elementary. It's of the form $\int{xe^{-kx^2}},\mathrm dx$ I don't see how to handle the sum, though. – saulspatz Apr 16 '18 at 19:09
  • Yes, evaluating the problem using the integral is significantly easier. That transformation from the sum to the integral is what I don't understand. They just give me that expression in the book with no background. – NoVa Apr 16 '18 at 19:11

3 Answers3

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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.

saulspatz
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  • $\tau$ is the temperature, which can be considered constant here. $\lambda$ is another constant here. I'll try it out and see if I can make anything from it, thank you. – NoVa Apr 16 '18 at 19:25
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    But I don't see where the $\pi/2$ is going to come from. You're in a much better position to try this than I am. – saulspatz Apr 16 '18 at 19:27
  • I did find this page: https://math.stackexchange.com/questions/1356472/turning-infinite-sum-into-integral , which might be my answer. – NoVa Apr 16 '18 at 19:28
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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 $$\int_0^\infty \exp(-cx^2)\, x^2 \, \mathrm dx=\frac1{2c\sqrt c}\int_0^\infty e^{-x}\, x^{1/2} \, \mathrm dx=\frac{\sqrt\pi}{4c\sqrt c}$$

Did
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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, $$\mathrm d V_2(n) = \mathrm d\left( \frac {\pi n^2} 4 \right) = \frac \pi 2 n \mathrm dn, \\ \mathrm d V_3(n) = \mathrm d\left( \frac {4 \pi n^3 / 3} 8 \right) = \frac \pi 2 n^2 \mathrm dn.$$

Maxim
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    "valid for small α, assuming some conditions on f hold." This seems frighteningly vague, wouldn't you say? – Did Apr 17 '18 at 18:12
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    @Did Yes, without knowing more about $f$ and about what kind of asymptotic we're talking about in the actual problem, things remain vague. This certainly holds: $$\sum_{n = 0}^\infty e^{-\alpha n^2} \sim \int_0^\infty e^{-\alpha n^2} dn = \frac 1 2 \sqrt \frac \pi \alpha, \quad \alpha \to +0.$$ But for the multidimensional case $f$ is probably something more complicated, otherwise we would just have the one-dimensional sum/integral to the power of $d$. – Maxim Apr 17 '18 at 19:36