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

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

\mathop{\rm E}\nolimits_n(0) = {1\over n-1}.
\end{displaymath} (3)

The function satisfies the Recurrence Relations
\mathop{\rm E}\nolimits_n'(x) = -\mathop{\rm E}\nolimits_{n-1}(x)
\end{displaymath} (4)

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
u={1\over t^n} \qquad dv=e^{- t x}\,d t
\end{displaymath} (8)

du=-{n\over t^{n+1}}\,d t \qquad v=-{e^{- t x}\over x}
\end{displaymath} (9)

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

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

(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)

\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 $n=1$ gives

\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
\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.

See also Cosine Integral, Et-Function, Exponential Integral, Gompertz Constant, Sine Integral


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($x$) and Related Functions.'' Ch. 37 in An Atlas of Functions. Washington, DC: Hemisphere, pp. 351-360, 1987.

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© 1996-9 Eric W. Weisstein