Exponential integral

Plot of ${\displaystyle E_{1}}$ function (top) and ${\displaystyle \operatorname {Ei} }$ function (bottom).

For real or complex arguments off the negative real axis, ${\displaystyle E_{1}(z)}$ can be expressed as[5]

${\displaystyle E_{1}(z)=-\gamma -\ln z-\sum _{k=1}^{\infty }{\frac {(-z)^{k}}{k\;k!}}\qquad (\left|\operatorname {Arg} (z)\right|<\pi )}$

where ${\displaystyle \gamma }$ is the Euler–Mascheroni constant. The sum converges for all complex ${\displaystyle z}$, and we take the usual value of the complex logarithm having a branch cut along the negative real axis.

This formula can be used to compute ${\displaystyle E_{1}(x)}$ with floating point operations for real ${\displaystyle x}$ between 0 and 2.5. For ${\displaystyle x>2.5}$, the result is inaccurate due to cancellation.

A faster converging series was found by Ramanujan:

${\displaystyle {\rm {Ei}}(x)=\gamma +\ln x+\exp {(x/2)}\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}x^{n}}{n!\,2^{n-1}}}\sum _{k=0}^{\lfloor (n-1)/2\rfloor }{\frac {1}{2k+1}}}$

These alternating series can also be used to give good asymptotic bounds for small x, e.g. :

${\displaystyle 1-{\frac {3x}{4}}\leq {\rm {Ei}}(x)-\gamma -\ln x\leq 1-{\frac {3x}{4}}+{\frac {11x^{2}}{36}}}$

for ${\displaystyle x\geq 0}$.

Relative error of the asymptotic approximation for different number ${\displaystyle ~N~}$ of terms in the truncated sum

Unfortunately, the convergence of the series above is slow for arguments of larger modulus. For example, more than 40 terms are required to get an answer correct to three significant figures for ${\displaystyle E_{1}(10)}$.[6] However, for positive values of x, there is a divergent series approximation that can be obtained by integrating ${\displaystyle xe^{x}E_{1}(x)}$ by parts:[7]

${\displaystyle E_{1}(x)={\frac {\exp(-x)}{x}}\left(\sum _{n=0}^{N-1}{\frac {n!}{(-x)^{n}}}+O(N!x^{-N})\right)}$

The relative error of the approximation above is plotted on the figure to the right for various values of ${\displaystyle N}$, the number of terms in the truncated sum (${\displaystyle N=1}$ in red, ${\displaystyle N=5}$ in pink).

Using integration by parts, we can obtain an explicit formula[8]

$${\displaystyle \operatorname {Ei} (z)={\frac {e^{z}}{z}}\left(\sum _{k=0}^{n}{\frac {k!}{z^{k}}}+e_{n}(z)\right),\quad e_{n}(z)\equiv (n+1)!\ ze^{-z}\int _{-\infty }^{z}{\frac {e^{t}}{t^{n+2}}}\,dt}$$

For any fixed ${\displaystyle z}$, the absolute value of the error term ${\displaystyle |e_{n}(z)|}$ decreases, then increases. The minimum occurs at ${\displaystyle n\sim |z|}$, at which point ${\displaystyle \vert e_{n}(z)\vert \leq {\sqrt {\frac {2\pi }{\vert z\vert }}}e^{-\vert z\vert }}$. This bound is said to be "asymptotics beyond all orders".

Bracketing of ${\displaystyle E_{1}}$ by elementary functions

From the two series suggested in previous subsections, it follows that ${\displaystyle E_{1}}$ behaves like a negative exponential for large values of the argument and like a logarithm for small values. For positive real values of the argument, ${\displaystyle E_{1}}$ can be bracketed by elementary functions as follows:[9]

${\displaystyle {\frac {1}{2}}e^{-x}\,\ln \!\left(1+{\frac {2}{x}}\right)<E_{1}(x)<e^{-x}\,\ln \!\left(1+{\frac {1}{x}}\right)\qquad x>0}$

The left-hand side of this inequality is shown in the graph to the left in blue; the central part ${\displaystyle E_{1}(x)}$ is shown in black and the right-hand side is shown in red.

Both ${\displaystyle \operatorname {Ei} }$ and ${\displaystyle E_{1}}$ can be written more simply using the entire function ${\displaystyle \operatorname {Ein} }$[10] defined as

${\displaystyle \operatorname {Ein} (z)=\int _{0}^{z}(1-e^{-t}){\frac {dt}{t}}=\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}z^{k}}{k\;k!}}}$

(note that this is just the alternating series in the above definition of ${\displaystyle \mathrm {E} _{1}}$). Then we have

${\displaystyle E_{1}(z)\,=\,-\gamma -\ln z+{\rm {Ein}}(z)\qquad \left|\operatorname {Arg} (z)\right|<\pi }$

${\displaystyle \operatorname {Ei} (x)\,=\,\gamma +\ln {\left|x\right|}-\operatorname {Ein} (-x)\qquad x\neq 0}$

Kummer's equation

${\displaystyle z{\frac {d^{2}w}{dz^{2}}}+(b-z){\frac {dw}{dz}}-aw=0}$

is usually solved by the confluent hypergeometric functions ${\displaystyle M(a,b,z)}$ and ${\displaystyle U(a,b,z).}$ But when ${\displaystyle a=0}$ and ${\displaystyle b=1,}$ that is,

${\displaystyle z{\frac {d^{2}w}{dz^{2}}}+(1-z){\frac {dw}{dz}}=0}$

we have

${\displaystyle M(0,1,z)=U(0,1,z)=1}$

for all z. A second solution is then given by E₁(−z). In fact,

${\displaystyle E_{1}(-z)=-\gamma -i\pi +{\frac {\partial [U(a,1,z)-M(a,1,z)]}{\partial a}},\qquad 0<{\rm {Arg}}(z)<2\pi }$

with the derivative evaluated at ${\displaystyle a=0.}$ Another connexion with the confluent hypergeometric functions is that E₁ is an exponential times the function U(1,1,z):

${\displaystyle E_{1}(z)=e^{-z}U(1,1,z)}$

The exponential integral is closely related to the logarithmic integral function li(x) by the formula

${\displaystyle \operatorname {li} (e^{x})=\operatorname {Ei} (x)}$

for non-zero real values of ${\displaystyle x}$.

The exponential integral may also be generalized to

${\displaystyle E_{n}(x)=\int _{1}^{\infty }{\frac {e^{-xt}}{t^{n}}}\,dt,}$

which can be written as a special case of the upper incomplete gamma function:[11]

${\displaystyle E_{n}(x)=x^{n-1}\Gamma (1-n,x).}$

The generalized form is sometimes called the Misra function[12] ${\displaystyle \varphi _{m}(x)}$, defined as

${\displaystyle \varphi _{m}(x)=E_{-m}(x).}$

Many properties of this generalized form can be found in the NIST Digital Library of Mathematical Functions.

Including a logarithm defines the generalized integro-exponential function[13]

${\displaystyle E_{s}^{j}(z)={\frac {1}{\Gamma (j+1)}}\int _{1}^{\infty }\left(\log t\right)^{j}{\frac {e^{-zt}}{t^{s}}}\,dt.}$

The indefinite integral:

${\displaystyle \operatorname {Ei} (a\cdot b)=\iint e^{ab}\,da\,db}$

is similar in form to the ordinary generating function for ${\displaystyle d(n)}$, the number of divisors of ${\displaystyle n}$:

${\displaystyle \sum \limits _{n=1}^{\infty }d(n)x^{n}=\sum \limits _{a=1}^{\infty }\sum \limits _{b=1}^{\infty }x^{ab}}$

The derivatives of the generalised functions ${\displaystyle E_{n}}$ can be calculated by means of the formula [14]

${\displaystyle E_{n}'(z)=-E_{n-1}(z)\qquad (n=1,2,3,\ldots )}$

Note that the function ${\displaystyle E_{0}}$ is easy to evaluate (making this recursion useful), since it is just ${\displaystyle e^{-z}/z}$.[15]

${\displaystyle E_{1}(ix)}$ against ${\displaystyle x}$; real part black, imaginary part red.

If ${\displaystyle z}$ is imaginary, it has a nonnegative real part, so we can use the formula

${\displaystyle E_{1}(z)=\int _{1}^{\infty }{\frac {e^{-tz}}{t}}\,dt}$

to get a relation with the trigonometric integrals ${\displaystyle \operatorname {Si} }$ and ${\displaystyle \operatorname {Ci} }$:

${\displaystyle E_{1}(ix)=i\left[-{\tfrac {1}{2}}\pi +\operatorname {Si} (x)\right]-\operatorname {Ci} (x)\qquad (x>0)}$

The real and imaginary parts of ${\displaystyle \mathrm {E} _{1}(ix)}$ are plotted in the figure to the right with black and red curves.

There have been a number of approximations for the exponential integral function. These include:

• The Swamee and Ohija approximation[16]
  $${\displaystyle E_{1}(x)=\left(A^{-7.7}+B\right)^{-0.13},}$$

  
  where
  $${\displaystyle {\begin{aligned}A&=\ln \left[\left({\frac {0.56146}{x}}+0.65\right)(1+x)\right]\\B&=x^{4}e^{7.7x}(2+x)^{3.7}\end{aligned}}}$$

  
• The Allen and Hastings approximation [16][17]
  $${\displaystyle E_{1}(x)={\begin{cases}-\ln x+{\textbf {a}}^{T}{\textbf {x}}_{5},&x\leq 1\\{\frac {e^{-x}}{x}}{\frac {{\textbf {b}}^{T}{\textbf {x}}_{3}}{{\textbf {c}}^{T}{\textbf {x}}_{3}}},&x\geq 1\end{cases}}}$$

  
  where
  $${\displaystyle {\begin{aligned}{\textbf {a}}&\triangleq [-0.57722,0.99999,-0.24991,0.05519,-0.00976,0.00108]^{T}\\{\textbf {b}}&\triangleq [0.26777,8.63476,18.05902,8.57333]^{T}\\{\textbf {c}}&\triangleq [3.95850,21.09965,25.63296,9.57332]^{T}\\{\textbf {x}}_{k}&\triangleq [x^{0},x^{1},\dots ,x^{k}]^{T}\end{aligned}}}$$

  
• The continued fraction expansion [17]
  $${\displaystyle E_{1}(x)={\cfrac {e^{-x}}{x+{\cfrac {1}{1+{\cfrac {1}{x+{\cfrac {2}{1+{\cfrac {2}{x+{\cfrac {3}{\ddots }}}}}}}}}}}}.}$$

  
• The approximation of Barry et al. [18]
  $${\displaystyle E_{1}(x)={\frac {e^{-x}}{G+(1-G)e^{-{\frac {x}{1-G}}}}}\ln \left[1+{\frac {G}{x}}-{\frac {1-G}{(h+bx)^{2}}}\right],}$$

  
  where:
  $${\displaystyle {\begin{aligned}h&={\frac {1}{1+x{\sqrt {x}}}}+{\frac {h_{\infty }q}{1+q}}\\q&={\frac {20}{47}}x^{\sqrt {\frac {31}{26}}}\\h_{\infty }&={\frac {(1-G)(G^{2}-6G+12)}{3G(2-G)^{2}b}}\\b&={\sqrt {\frac {2(1-G)}{G(2-G)}}}\\G&=e^{-\gamma }\end{aligned}}}$$

  
  with ${\displaystyle \gamma }$ being the Euler–Mascheroni constant.