Fermi-Dirac integral of half order

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Please try fermi.m from matlab site matlab file exchange. This will give very accurate values(14 digits minimum) for any half-order Fd integral.

Further, your suggestion that MM "knows" how to perform this integral analytically and does so is an egregious error. The Fermi-Dirac integral of negative half order (used in your example) must also be done numerically (quick google search). Even the answer given by MM, a number and not an expression, suggests that it is not employing a symbolic solution, but is indeed switching to a numerical method

A) unrelated to sage: as I already said in my previous post, I'd be careful making all-out statements. (Even more so based on 'quick' google seraches). So once again, to falsify your claims:

[1] The Analytical Evaluation of the Half-Order Fermi-Dirac Integrals Jerry A. Selvaggi, and Jerry P. Selvaggi The Open Mathematics Journal, 2012, 5, 1-7

[2] Solutions to the Fermi-Dirac Integrals in Semiconductor Physics Using Polylogarithms M. Ulrich, W. Seng, P. Barnes Journal of Computational Electronics 1: 431–434, 2002

and several other papers. You should also realize, that there is a difference between scientific papers, like the preceding, which have undergone peer review, and some unpublished and unrefereed 'notes', which pop up on the internet - like the one you point to @ nanohub.org

From the preceeding, I guess it is quite clear, where or to whom to apply vocabulary like 'egregious'. Moreover, it is Ref. [2] which tells you that things work down even to y>-1. Just pls. read the papers before posting.

B) related to sage: MM is not '..programmed to recognize Fermi-Dirac integrals and respond accordingly..'. Basically it uses eqn. (5) of Ref. [2], which is a paper free to be read with results free to be used by anyone - also outside of academia. Obviously MM has simply cared to include this result into its code - while sage has not. As simple as that.

with my tail between my legs

C) related to this forum I like this forum because it tends to be free of such statements. Pls. keep it that way.

First, when asking questions, it might speed up a potential answer if you would care to give enough detail about your problem ... so, I only guess that this is what you want $$F(y,\mu) = \int_0^\infty \frac{x^{y}}{e^{x+\mu}+1} dx$$

Second, I'd be very careful about www-ing statements like

"...which can only be done numerically..."

since, to my recollection $F(y,\mu)$ is just another way of writing the polylog for all $y>-1$, i.e down to those values of $y$ where the singularity of the integrand remains harmless. Therefore, at least I do not understand your claim, since in 1D the DOS in a semiconductor is $\propto x^{-1/2}$ (which you also did not specify that in your question). So it can be done analytically ...

Third, if still you really want to do this numerically in sage ... just look here http://www.sagemath.org/doc/reference/calculus/sage/gsl/integration.html and e.g. do this

def F(y,mu):
    ul = 100+mu # increase the 100 if you really need
    return numerical_integral(x^y/(e^(x+mu)+1),0,ul)

F(-1/2,0)
(1.072154920510084, 8.286087609477639e-07)

Finally, and only to my very humble understanding: sage is a chain of existing tools, i.e. it is kind of as strong as it's weakest link. Since in plain sage things like Maxima are the link for many of those calculations which 90%-99% of non-mathematician might call 'symbolic', plain sage frequently is of little help if compared to MM or Maple once you come to such 'symbolic' problems.

Applied to your case this means:

sage: var('y,mu')
(y, mu)
sage: assume(y+1>0)
sage: assume(y,'integer')
sage: integral(x^y/(e^(x+mu)+1),x,0,oo)
integrate(x^y/(e^(mu + x) + 1), x, 0, +Infinity)

so, nothing ... or if you try special values

sage: integral(x^(-1/2)/(e^(x+mu)+1),x,0,oo)
integrate(1/((e^(mu + x) + 1)*sqrt(x)), x, 0, +Infinity)

sage: integral(x^1/(e^(x+mu)+1),x,0,oo)
polylog(2, -e^mu) + limit(1/2*x^2 - x*log(e^(mu + x) + 1) - polylog(2, -e^(mu + x)), x, +Infinity, minus)

again, not too helpful. Maybe there are 'tricks' to get further but I don't know.

If however you use MM then, out of the box:

sage: mathematica('Integrate[x^a/ (Exp[x + b] + 1), {x, 0, Infinity}, Assumptions -> {Re[a] > -1}]')
-(a*Gamma[a]*PolyLog[1 + a, -E^(-b)])

so MM 'knows' that your integral can be done analytically and provides the answer, and

sage: mathematica('-(a*Gamma[a]*PolyLog[1 + a, -E^(-b)])/. {a -> -1/2, b -> 0} // N')
1.072154929940191

which is the numerical value we got previously.

In conclusion: To answer if that implies you should better be using MM, there are more powerful experts around here.