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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).

gammatester
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2 Answers2

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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.

TZakrevskiy
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  • 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? – gammatester Aug 19 '13 at 11:49
  • @gammatester Yes, I wouldn't expect such formulas. Could you post the limits your Maple shows you (I don't have Maple at hand)? – TZakrevskiy Aug 19 '13 at 12:10
  • 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}$$
     – gammatester Aug 19 '13 at 12:35
  • Note the above comment is very short of characters, I must use abbreviations, hope I transferred it correctly. It is the output for $F_2(b,x)$ – gammatester Aug 19 '13 at 12:38
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    @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. – TZakrevskiy Aug 19 '13 at 14:07
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Similar to the incomplete Fermi-Dirac integral, you can define the incomplete Polylogarithm by replacing the lower integration limit $0$ by $b$. In these incomplete Polylogarithms you can then use a substitution and the binomial formula to replace one incomplete integral by a finite number of complete ones: \begin{align} \mathrm{Li}_n(b, z) &= \frac{1}{\Gamma(n)}\int\limits_b^\infty\frac{x^{n-1}\mathrm{d}x}{\mathrm{e}^x / z - 1} \\&= \begin{vmatrix} t &=& x - b \\ \mathrm{d}t &=& \mathrm{d}x \end{vmatrix} = \frac{1}{(n - 1)!}\int\limits_0^\infty\frac{(t + b)^{n-1}\mathrm{d}t}{\mathrm{e}^t\mathrm{e}^{b} / z - 1} \\&= \frac{1}{(n - 1)!}\int\limits_0^\infty\frac{\sum\limits_{k=0}^{n-1}\binom{n-1}{k}t^{(k+1) - 1}b^{n-1-k}\mathrm{d}t}{\mathrm{e}^t / (z\cdot \mathrm{e}^{-b}) - 1} \\&= \frac{1}{(n - 1)!}\sum\limits_{k=0}^{n-1}\binom{n-1}{k}b^{n-1-k}\,k!\,\mathrm{Li}_{k+1}(z\cdot\mathrm{e}^{-b}) \end{align} and therefore \begin{align} \mathrm{Li}_n(b, z) &= \sum\limits_{k=0}^{n-1}\frac{b^{n-k-1}}{\Gamma(n-k)}\mathrm{Li}_{k+1}(z\cdot\mathrm{e}^{-b}) =\sum\limits_{k=0}^{n-1}\frac{b^{k}}{k!}\mathrm{Li}_{n - k}(z\cdot\mathrm{e}^{-b}). \end{align}

So for example \begin{align} \mathrm{Li}_2(b, z) &= b\cdot\mathrm{Li}_{1}(z\cdot\mathrm{e}^{-b}) + \mathrm{Li}_{2}(z\cdot\mathrm{e}^{-b}) \end{align} or \begin{align} \mathrm{Li}_3(b, z) &= \frac{1}{2}b^2\cdot\mathrm{Li}_{1}(z\cdot\mathrm{e}^{-b}) + b\mathrm{Li}_{2}(z\cdot\mathrm{e}^{-b}) + \mathrm{Li}_{3}(z\cdot\mathrm{e}^{-b}). \end{align}

You can then express your incomplete Fermi-Dirac integral by the same formula: \begin{align} \mathrm{F}_j(b, x) &= -\mathrm{Li}_{j + 1}(b, -\mathrm{e}^{x}) \\&= \sum\limits_{k=0}^{j}\frac{b^{k}}{k!}(-1)\mathrm{Li}_{j + 1 - k}(-\mathrm{e}^{x-b}) \\&= \sum\limits_{k=0}^{j}\frac{b^{k}}{k!}\mathrm{F}_{j - k}(x - b) \end{align} Is this what you had in mind with 'closed relationship'?

ricma
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