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In mathematics, the incomplete Fermi-Dirac integral, named after Enrico Fermi and Paul Dirac, for an index
and parameter
is given by

Its derivative is

and this derivative relationship may be used to find the value of the incomplete Fermi-Dirac integral for non-positive indices
.[1]
This is an alternate definition of the incomplete polylogarithm, since:

Which can be used to prove the identity:

where
is the gamma function and
is the upper incomplete gamma function. Since
, it follows that:

where
is the complete Fermi-Dirac integral.
The closed form of the function exists for
: [1]
