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For Fermi-Dirac Integrals

$$\mathcal{F}_{j}(x)=\frac{1}{\Gamma(j+1)}\int_{0}^{\infty}\frac{t^{j}}{e^{t-x}+1}~\mathrm{d}t,$$

we know that for $x\gg1,$ we can approximate

$$\mathcal{F}_{0}(x) \approx x$$

and

$$\mathcal{F}_{1}(x) \approx\frac{1}{2}x^{2}.$$

However, using numerical integration for obtaining the integrals, the function $2\mathcal{F}_{0}(x)-\mathcal{F}_{0}(x)^{2}$ converges to a constant threshold.

How can that be?

Jenny Reininger
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  • Your integral does not involve $x$. Is the denominator $e^{t-x}+1$ as in the Wikipedia article? Also, is your last expression $2\mathcal{F}_1(x)-\mathcal{F}_0(x)^2$? – Sangchul Lee Jun 17 '21 at 08:20
  • There is an asymptotic expansion $$ \mathcal{F}p (x) \sim \cos (\pi p)\mathcal{F}_p ( - x) + 2x^{p + 1} \sum\limits{n = 0}^\infty {\frac{{(1 - 2^{1 - 2n} )\zeta (2n)}}{{\Gamma (p - 2n + 2)}}\frac{1}{{x^{2n} }}} $$ for large positive $x$, which may be used to obtain more accurate information about the behaviour of the difference you are looking at. Note that $$ \mathcal{F}p ( - x) = - \operatorname{Li}{p + 1} ( - e^{ - x} ) \sim e^{ - x} $$ is an exponentially small contribution. – Gary Jun 17 '21 at 08:31
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    It follows from the above asymptotics that $\mathcal{F}_0 (x) = x + \mathcal{O}(e^{ - x} )$ and $\mathcal{F}_1 (x) = \frac{1}{2}x^2 + \frac{{\pi ^2 }}{6} + \mathcal{O}(e^{ - x} )$ for large positive $x$. Hence, $2\mathcal{F}_1 (x) - \mathcal{F}_0^2 (x) = \frac{{\pi ^2 }}{3} + \mathcal{O}(e^{ - x} )$. Thus, your constant treshold is $\frac{{\pi ^2 }}{3}$. – Gary Jun 17 '21 at 08:44
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    A lightweight proof of Gary's conclusion follows by noting that $$\mathcal{F}0(x)=\log(1+e^x)=x+\underbrace{\log(1+e^{-x})}{=\mathcal{O}(e^{-x})\text{ as }x\to\infty}$$ and $$\mathcal{F}1(x)=\int{-\infty}^{x}\mathcal{F}0(t),\mathrm{d}t=2C+\frac{x^2}{2}-\underbrace{\int{x}^{\infty}\log(1+e^{-t}),\mathrm{d}t}{=\mathcal{O}(e^{-x})\text{ as }x\to\infty},$$ where $C=\mathcal{F}_1(0)=\int{0}^{\infty}\log(1+e^{-t}),\mathrm{d}t=\frac{1}{2}\zeta(2)$. – Sangchul Lee Jun 17 '21 at 09:01

0 Answers0