En-Function

The $\mathop{\rm E}\nolimits_n(x)$ function is defined by the integral

$\begin{displaymath} \mathop{\rm E}\nolimits_n(x) \equiv \int_1^\infty {e^{-xt}\,dt\over t^n} \end{displaymath}$

(1)

and is given by the Mathematica ${}^{\scriptstyle\circledRsymbol}$ (Wolfram Research, Champaign, IL) function ExpIntegralE[n,x]. Defining $t\equiv \eta^{-1}$ so that $dt = -\eta^{-2}d\eta$ ,

$\begin{displaymath} \mathop{\rm E}\nolimits_n(x) = \int^1_0 e^{-x/\eta} \eta^{n-2}\,d\eta \end{displaymath}$

(2)

$\begin{displaymath} \mathop{\rm E}\nolimits_n(0) = {1\over n-1}. \end{displaymath}$

(3)

The function satisfies the Recurrence Relations

$\begin{displaymath} \mathop{\rm E}\nolimits_n'(x) = -\mathop{\rm E}\nolimits_{n-1}(x) \end{displaymath}$

(4)

$\begin{displaymath} n\mathop{\rm E}\nolimits_{n+1}(x) = e^{-x}-x\mathop{\rm E}\nolimits_n(x). \end{displaymath}$

(5)

Equation (4) can be derived from

$\displaystyle \mathop{\rm E}\nolimits_n(x)$	$\textstyle =$	$\displaystyle \int_1^\infty {e^{-t x}\over t^n}\,dt$	(6)
$\displaystyle \mathop{\rm E}\nolimits_n'(x)$	$\textstyle =$	$\displaystyle {d\over dx} \int_1^\infty {e^{- t x}\over t^n}\,dt =\int_1^\infty {d\over dx}\left({e^{- t x}\over t^n}\right)\,dt$
	$\textstyle =$	$\displaystyle -\int_1^\infty t{e^{-tx}\over t^n}\,dt$
	$\textstyle =$	$\displaystyle -\int_1^\infty {e^{-tx}\over t^{n-1}}\,dt = -\mathop{\rm E}\nolimits_{n-1}(x),$	(7)

and (5) using integrating by parts, letting

$\begin{displaymath} u={1\over t^n} \qquad dv=e^{- t x}\,d t \end{displaymath}$

(8)

$\begin{displaymath} du=-{n\over t^{n+1}}\,d t \qquad v=-{e^{- t x}\over x} \end{displaymath}$

(9)

gives

$\displaystyle \mathop{\rm E}\nolimits_n(x)$	$\textstyle =$	$\displaystyle \int_1^\infty u\,dv = [uv]_1^\infty-\int_1^\infty v\,du$
	$\textstyle =$	$\displaystyle \left[{-{e^{-tx}\over xt^n}}\right]_{t=1}^\infty-{n\over x}\int_1^\infty {e^{-tx}\over t^{n+1}}\,dt$
	$\textstyle =$	$\displaystyle \left[{0-\left({-{e^{-x}\over x}}\right)}\right]-{n\over x}\int_1^\infty {e^{-tx}\over t^{n+1}}\,dt$
	$\textstyle =$	$\displaystyle {e^{-x}\over x}-{n\over x}\mathop{\rm E}\nolimits_{n+1}(x).$	(10)

Solving (10) for $n\mathop{\rm E}\nolimits_{n+1}(x)$ then gives (5). An asymptotic expansion gives

$\begin{displaymath} (n-1)!\,\mathop{\rm E}\nolimits_n(x) = (-x)^{n-1}\mathop{\rm E}\nolimits_1(x)+e^{-x} \sum_{s=0}^n-2 (n-s-2)!(-x)^s, \end{displaymath}$

(11)

$\begin{displaymath} \mathop{\rm E}\nolimits_n(x) = {e^{-x}\over x} \left[{1 - {n\over x} + {n(n+1)\over x^2} + \ldots}\right]. \end{displaymath}$

(12)

The special case gives

$\begin{displaymath} \mathop{\rm E}\nolimits_1(x) \equiv -\mathop{\rm ei}\nolimit... ...nfty {e^{-tx}\,dt\over t} = \int_x^\infty {e^{-u}\,du\over u}, \end{displaymath}$

(13)

where $\mathop{\rm ei}\nolimits (x)$ is the Exponential Integral, which is also equal to

$\begin{displaymath} \mathop{\rm E}\nolimits_1(x) = -\gamma - \ln x - \sum_{n=1}^\infty {(-1)^n x^n\over n!n}, \end{displaymath}$

(14)

where $\gamma$ is the Euler-Mascheroni Constant.

$\displaystyle \mathop{\rm E}\nolimits_1(0)$	$\textstyle =$	$\displaystyle \infty$	(15)
$\displaystyle \mathop{\rm E}\nolimits_1(ix)$	$\textstyle =$	$\displaystyle -\mathop{\rm ci}\nolimits (x)+i \mathop{\rm si}\nolimits (x),$	(16)

where $\mathop{\rm ci}\nolimits (x)$ and $\mathop{\rm si}\nolimits (x)$ are the Cosine Integral and Sine Integral.

References

Abramowitz, M. and Stegun, C. A. (Eds.). ``Exponential Integral and Related Functions.'' Ch. 5 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 227-233, 1972.

Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and Vetterling, W. T. ``Exponential Integrals.'' §6.3 in Numerical Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. Cambridge, England: Cambridge University Press, pp. 215-219, 1992.

Spanier, J. and Oldham, K. B. ``The Exponential Integral Ei() and Related Functions.'' Ch. 37 in An Atlas of Functions. Washington, DC: Hemisphere, pp. 351-360, 1987.