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The complete Fermi-Dirac integrals $$ F_s(x) = \frac{1}{\Gamma(s+1)} \int\limits_{0}^{\infty} \frac{t^s}{e^{t-x}+1} \: dt $$ are related to the polylogarithms, see http://dlmf.nist.gov/25.12#iii $$ F_s(x) = -\mathrm{Li}_{s+1}(-e^x) $$ Is there any known closed-form relationship of the incomplete Fermi-Dirac integrals $$ F_j(b,x) = \frac{1}{\Gamma(j+1)} \int\limits_{b}^{\infty} \frac{t^j}{e^{t-x}+1} \: dt, \quad b \ge 0 $$ to polylogarithms (at least for integer orders > 1)? For the $j=1$ case I found a formula with Maple $$ F_1(b,x) = \frac{\pi^2}{6} - \frac{b^2}{2} + \frac{x^2}{2} + b \ln(1+ e^{b-x}) + \mathrm{Li}_{2}(-e^{b-x}). $$ For $j \ge 2$ Maple gives a complicated expression including polylogarithms with limits for $t \rightarrow 0$ but no actual closed form. Wolfram Alpha refuses to give answers (it echos the input).

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  • $\begingroup$ Check this link, please. $\endgroup$ Commented Jan 8, 2015 at 7:10
  • $\begingroup$ Are you still interested in finding a relation? $\endgroup$ Commented Jan 8, 2015 at 7:20
  • $\begingroup$ @Felix Marin: IMO your link does not contain any information about the incomplete integrals. $\endgroup$ Commented Jan 12, 2015 at 9:08
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    $\begingroup$ @Mhenni Benghorbal: About 1.5 years ago I thought about implementing these functions in my special function library. Although this plan has been dropped, I am still interested. $\endgroup$ Commented Jan 12, 2015 at 9:16
  • $\begingroup$ @gammatester Yes. It's true. I was familiar with that link but I didn't realize that 'incomplete' was not covered there. Thanks. $\endgroup$ Commented Jan 12, 2015 at 13:41

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First, several points on polylogarithms:

1) There's a differentiation formula: $$\partial_x Li_s(-e^x)=Li_{s-1}(-e^x).$$

2) the only known exact forms for $Li_s(z)$ are for $s=1,0,-1,\dots$. $$-Li_1(-e^x) = \ln(1+e^x).$$

3) $-Li_s(-1)=\zeta(s)$; $-Li_2(-1) = \frac{\pi^2}{12}$.

In your case, the formula for $F_1$ was obtained via integration by parts (easy to check), the same probably goes for other formulas you've found.

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  • $\begingroup$ I am not fixed to have elementary transcendental functions in the answer, I can accept formulas with polylogarithms (but no unevaluated limits). Does you answer mean, that you expect there are no such formulas? $\endgroup$ Commented Aug 19, 2013 at 11:49
  • $\begingroup$ @gammatester Yes, I wouldn't expect such formulas. Could you post the limits your Maple shows you (I don't have Maple at hand)? $\endgroup$ Commented Aug 19, 2013 at 12:10
  • $\begingroup$ with $P = \mathrm{polylog}$: $$ \lim_{t \rightarrow 0\mathrm{+}} \frac{1}{3} ( - b^3\,t^3 + 3bx^2t^3 + 3\ln(1 + e^{(b - x)})\,t^3b^2 + 6\,P(2, \, - e^{(b - x)})\,t^{3}\,b - 6\,P(3, \, - e^{(b - x)})\,t^{3} - 3\,x ^{2}\,\ln(e^{(b - x)})\,t^{3} + 1 - 3\,\ln(1 + e ^{( - \frac { - 1 + x\,t}{t})})\,t + 6\,\ln(1 + e^{( - \frac { - 1 + x\,t}{t})})\,t ^{2}\,x - 3\,\ln(1 + e^{( - \frac { - 1 + x\,t}{t})})\,t ^{3}\,x^{2} - 6\,P(2, \, - e^{( - \frac { - 1 + x \,t}{t})})\,t^{2} + 6\,P(2, \, - e^{( - \frac { - 1 + x\,t}{ t})})\,t^{3}\,x + 6\,P(3, \, - e^{( - \frac { - 1 + x\,t}{t})})\,t^{3})/t^{3}$$ $\endgroup$ Commented Aug 19, 2013 at 12:35
  • $\begingroup$ Note the above comment is very short of characters, I must use abbreviations, hope I transferred it correctly. It is the output for $F_2(b,x)$ $\endgroup$ Commented Aug 19, 2013 at 12:38
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    $\begingroup$ @gammatester On the first sight, this formula can be simplified and some of the limits evaluated; if I manage to obtain something useful, I'll leave a message. $\endgroup$ Commented Aug 19, 2013 at 14:07

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