# Approximation of Fermi-Dirac integral $\int \text{d}x \frac{f(x,\beta)}{1+e^{\beta x}}$

In physics is quite common to find integrals of the type \begin{align} I(\beta) = \int_{-\infty}^{\infty}\text{d}x \frac{f(x)}{1+e^{\beta x}} \tag{1} \end{align} where $f(x)$ is some quantity we want to average over the Fermi-Dirac distribution $n(x) = \left(1+e^{\beta x}\right)^{-1}$, and $\beta >0$ is a positive real parameter representing the the 'inverse of the temperature'. Since many times physicists are only interested in the 'low temperature' regime $\beta\gg 1$, it is common to consider the following Sommerfeld approximation: \begin{align} I(\beta)\underset{\beta \gg 1}{=} \int_{-\infty}^{0}\text{d} x~f(x)+\frac{\pi^2}{6\beta^2} f'(0)+O(\beta^{-4}). \tag{2} \end{align} Which sometimes appears in a mnemonic fashion as an expansion for $n(x,\beta)$ itself, \begin{align} \frac{1}{1+e^{\beta x}}\underset{\beta\gg1}{=} \theta(-x)-\frac{\pi^2} {6\beta^2}\delta'(x)+O(\beta^{-4}) \tag{3} \end{align}

My question is

Can we understand Eq. (3) in a rigorous fashion? E.g. as the expansion of a distribution. If yes, can we understand the convergence of Eq.(2) as the condition for a series expansion under integral sign? E.g. dominated convergence.

My motivation

I need to study an integral similar to Eq. (1) but with the crucial difference that f(x) is also a function of the parameter $\beta$, and wanted to make a similar expansion as in Eq. (2). My idea was to understand Eq. (2) as a series expansion Eq.(3) under integrals sign, and also expand my $f$ in $\beta$. To check if this is safe, I wanted to use dominated convergence. But I am not confident this makes sense.

References for rigorous discussions of Eq. (2) or Eq. (3) are welcome.

• It would be probably easier to answer by knowing which kind of $f(x)$ or $f_\beta(x)$ you need to deal with. – Jack D'Aurizio Jul 8 '17 at 16:54
• @JackD'Aurizio, in my particular case I am dealing with functions which are the complex argument of a complex function defined by its power series. Thus something that boils down to $\tan^{-1}\left(\frac{A x + B x^2 +\dots}{C x + D x^2 + \dots}\right)$ where $A,B,C,D$ are all functions of $\beta$, and the whole thing bounded by $\pm\pi$. But I would be interested also in a generic discussion for, say, bounded functions or compactly supported bounded functions. – kurtachovo Jul 8 '17 at 17:13

To justify the Sommerfeld expansion let assume that $f$ is a suitable testfunction (something like a Lorentzian should be good for now: integrable at $0$ and a sufficent decay at infinity). Furthermore let us denote the Fermi distribution by $n_{\beta}(x)$.

We want to explore the integral

$$I(\beta)=\int_{\mathbb{R}} n_{\beta}(x)f(x)dx$$

in the limit of $\beta\gg1$.

Now, let us rescale $x\rightarrow \beta x$ and split the range of integration at the origin

$$\beta I(\beta)=I_{+}(\beta)+I_{-}(\beta)=\int_0^{\infty}\frac{f(x/\beta)}{e^x+1}dx+\int_{-\infty}^0\frac{f(x/\beta)}{e^x+1}dx$$

Since $e^{x}<1$ on $\mathbb{R}_-$ we find

$$I_-(\beta)=\int_{-\infty}^0f(x/\beta)dx+\int_{-\infty}^0dx(f(0)+\frac{x}{\beta}f'(0)+O{(\beta^{-2})})\sum_{n\geq 1}(-1)^ne^{n x}=\\\int_{-\infty}^0f(x/\beta)dx-\log(2)f(0)+\frac{\pi^2}{12\beta}f'(0)+O(\beta^{-2})$$

Since on $\mathbb{R}_+$ the integral converges pretty fast we can expand $f$ directly and find

$$I_+(\beta)=\int_0^{\infty}\frac{f(0)+\frac{x}{\beta}f'(0)+O{(\beta^{-2})}}{e^x+1}dx=\log(2)f(0)+\frac{\pi^2}{12\beta}f'(0)+O(\beta^{-2})$$

which yields (after undoing the rescaling from the beginning)

$$I(\beta)=\int_{-\infty}^{0}f(x)dx+\frac{\pi^2}{6\beta^2}f'(0)+O(\beta^{-3})$$

which is in the sense of distributions can indeed be interpreted as

$$n_{\beta}(x)\sim \theta(-x)-\frac{\pi^2}{6\beta^2}\delta'(x)+O(\beta^{-3})\quad \text{as}\,\, \beta >>1$$

that $O(\beta^{-3})$ can be replaced by an $O(\beta^{-4})$ estimate is straightforward to include

Remark: If we allow a temperature dependence for $f$, so $f_{\beta}(x)$ Sommerfeld should be valid as long as we can assure that a Taylor expansion around the origin (in $x$) is valid, so $f_\beta(x/\beta)\sim f_{\beta}(0)+\frac{x}{\beta}f_{\beta}'(0)...$ where $f_{\beta}^{(n)}(0)$ are at least as such that $f_{\beta}^{(n)}(0)/\beta^n \rightarrow 0$ as $\beta\rightarrow \infty$

• Thus the conclusion is that, as long as the coefficients of the Taylor expansion of $f_\beta(x)$ around $x=0$ decay fast enough as $\beta\to\infty$ - and this will depend on the specific functional dependence of $f_{\beta}(x)$ in $\beta$ - Sommerfeld should hold? – kurtachovo Jul 8 '17 at 18:40
• @kurtachovo this is what i strongley suspect...it should be not too difficult to make this rigorous – tired Jul 8 '17 at 20:25

I don't know if this will be of use, but ...

If we assume that $f$ belongs to the space of test functions that are smooth and of compact support, then certainly we have

$$f(x)-f(-x)=2f'(0)x+O(x^3)$$

Hence,

\begin{align} \int_0^\infty \frac{f(x)-f(-x)}{1+e^{\beta x}}\,dx&=2f'(0)\int_0^\infty \frac{x}{1+e^{\beta x}}\,dx+O\left(\int_0^\infty \frac{x^3}{1+e^{\beta x}}\,dx\right)\\\\ &=\frac{2f'(0)}{\beta^2}\int_0^\infty \frac{x}{1+e^x}\,dx+O\left(\frac{1}{\beta^4}\int_0^\infty \frac{x^3}{1+e^{\beta x}}\,dx\right)\\\\ &=\frac{\pi^2}{6\beta^2}f'(0)+O\left(\frac{7\pi^4}{120\beta^4}\right)\\\\ &=\frac{\pi^2}{6\beta^2}f'(0)+O\left(\frac{1}{\beta^4}\right) \end{align}

Therefore, we can write

$$\int_{-\infty}^\infty \frac{f(x)}{1+e^{\beta x}}\,dx=\int_{-\infty}^0 f(x)\,dx+\frac{\pi^2}{6\beta^2}f'(0)+O\left(\frac{1}{\beta^4}\right)$$

Now note that $\frac{1}{1+e^{\beta x}}\sim \Theta(-x)-\frac{\pi^2}{6\beta^2}\delta'(x) +O(\beta^{-4})$ in the sense that

$$\langle f, \frac{1}{1+e^{\beta x}}\rangle =\langle f,\Theta(-x)-\frac{\pi^2}{6\beta^2}\delta'(x) )\rangle +O(\beta^{-4})$$

• better then mine $+1$ – tired Jul 8 '17 at 17:35
• I like your answer better (+1) back – Mark Viola Jul 8 '17 at 18:57
• @MarkViola, thank you for your answer! I like the way you show the Sommerfeld expansion. It is particularly clear. I was just inclined to tired answer because he explicitly address the issue of temperature dependence of $f$. I think your answer accommodates also the generic case, but I am not confident enough. – kurtachovo Jul 8 '17 at 22:53


Note that $\ds{{1 \over \expo{\beta x} + 1} = \left\{\begin{array}{lcl} \ds{\Theta\pars{-x} + {\mrm{sgn}\pars{x} \over \expo{\beta\,\verts{x}} + 1}} & \mbox{if} & \ds{x \not= 0} \\[2mm] \ds{1 \over 2} & \mbox{if} & \ds{x = 0} \end{array}\right.}$

$\ds{\Theta}$ is the Heaviside Step Function.

Then, \begin{align} \left.\vphantom{\Large A}\mrm{I}\pars{\beta}\right\vert_{\ \beta\ >\ 0} & = \int_{-\infty}^{\infty}{\mrm{f}\pars{x} \over \expo{\beta x} + 1}\,\dd x = \int_{-\infty}^{0}\mrm{f}\pars{x}\,\dd x + \int_{-\infty}^{\infty}{\mrm{sgn}\pars{x} \over \expo{\beta\,\verts{x}} + 1}\,\mrm{f}\pars{x}\,\dd x \\[5mm] & = \int_{-\infty}^{0}\mrm{f}\pars{x}\,\dd x + \int_{0}^{\infty} {\mrm{f}\pars{x} - \mrm{f}\pars{-x} \over \expo{\beta x} + 1}\,\dd x \\[5mm] & = \int_{-\infty}^{0}\mrm{f}\pars{x}\,\dd x + 2\sum_{n = 0}^{\infty} {\mrm{f}^{\pars{2n + 1}}\pars{0} \over \pars{2n + 1}!}\int_{0}^{\infty} {x^{2n + 1} \over \expo{\beta x} + 1}\,\dd x \\[5mm] & = \int_{-\infty}^{0}\mrm{f}\pars{x}\,\dd x + 2\sum_{n = 0}^{\infty} {\mrm{f}^{\pars{2n + 1}}\pars{0} \over \pars{2n + 1}!} \,\beta^{-\pars{2n + 2}}\int_{0}^{\infty}{x^{2n + 1} \over \expo{x} + 1}\,\dd x \\[5mm] & = \bbx{\int_{-\infty}^{0}\mrm{f}\pars{x}\,\dd x + \sum_{n = 0}^{\infty} \bracks{2\,\mrm{f}^{\pars{2n + 1}}\pars{0} \pars{1 - 2^{-2n - 1}}\zeta\pars{2n + 2}}\,\beta^{-\pars{2n + 2}}} \end{align}