Exponential integral

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Plot of the exponential integral function E n(z) with n=2 in the complex plane from -2-2i to 2+2i with colors created with Mathematica 13.1 function ComplexPlot3D

In mathematics, the exponential integral Ei is a special function on the complex plane.

It is defined as one particular definite integral of the ratio between an exponential function and its argument.

For real non-zero values of x, the exponential integral Ei(x) is defined as

${\displaystyle \operatorname {Ei} (x)=-\int _{-x}^{\infty }{\frac {e^{-t}}{t}},円dt=\int _{-\infty }^{x}{\frac {e^{t}}{t}},円dt.}${\displaystyle \operatorname {Ei} (x)=-\int _{-x}^{\infty }{\frac {e^{-t}}{t}},円dt=\int _{-\infty }^{x}{\frac {e^{t}}{t}},円dt.}

The Risch algorithm shows that Ei is not an elementary function. The definition above can be used for positive values of x, but the integral has to be understood in terms of the Cauchy principal value due to the singularity of the integrand at zero.

For complex values of the argument, the definition becomes ambiguous due to branch points at 0 and ${\displaystyle \infty }${\displaystyle \infty }.^[1] Instead of Ei, the following notation is used,^[2]

${\displaystyle E_{1}(z)=\int _{z}^{\infty }{\frac {e^{-t}}{t}},円dt,\qquad |{\rm {Arg}}(z)|<\pi }${\displaystyle E_{1}(z)=\int _{z}^{\infty }{\frac {e^{-t}}{t}},円dt,\qquad |{\rm {Arg}}(z)|<\pi }

Plot of the exponential integral function Ei(z) in the complex plane from -2-2i to 2+2i with colors created with Mathematica 13.1 function ComplexPlot3D

For positive values of x, we have ${\displaystyle -E_{1}(x)=\operatorname {Ei} (-x)}${\displaystyle -E_{1}(x)=\operatorname {Ei} (-x)}.

In general, a branch cut is taken on the negative real axis and E₁ can be defined by analytic continuation elsewhere on the complex plane.

For positive values of the real part of ${\displaystyle z}${\displaystyle z}, this can be written^[3]

${\displaystyle E_{1}(z)=\int _{1}^{\infty }{\frac {e^{-tz}}{t}},円dt=\int _{0}^{1}{\frac {e^{-z/u}}{u}},円du,\qquad \Re (z)\geq 0.}${\displaystyle E_{1}(z)=\int _{1}^{\infty }{\frac {e^{-tz}}{t}},円dt=\int _{0}^{1}{\frac {e^{-z/u}}{u}},円du,\qquad \Re (z)\geq 0.}

The behaviour of E₁ near the branch cut can be seen by the following relation:^[4]

${\displaystyle \lim _{\delta \to 0+}E_{1}(-x\pm i\delta )=-\operatorname {Ei} (x)\mp i\pi ,\qquad x>0.}${\displaystyle \lim _{\delta \to 0+}E_{1}(-x\pm i\delta )=-\operatorname {Ei} (x)\mp i\pi ,\qquad x>0.}

Several properties of the exponential integral below, in certain cases, allow one to avoid its explicit evaluation through the definition above.

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

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

where ${\displaystyle \gamma }${\displaystyle \gamma } is the Euler–Mascheroni constant. The sum converges for all complex ${\displaystyle z}${\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 ${\displaystyle E_{1}(x)}${\displaystyle E_{1}(x)} with floating point operations for real ${\displaystyle x}${\displaystyle x} between 0 and 2.5. For ${\displaystyle x>2.5}${\displaystyle x>2.5}, the result is inaccurate due to cancellation.

A faster converging series was found by Ramanujan:^[6]

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

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 ${\displaystyle E_{1}(10)}${\displaystyle E_{1}(10)}.^[7] However, for positive values of x, there is a divergent series approximation that can be obtained by integrating ${\displaystyle xe^{x}E_{1}(x)}${\displaystyle xe^{x}E_{1}(x)} by parts:^[8]

${\displaystyle E_{1}(x)={\frac {\exp(-x)}{x}}\left(\sum _{n=0}^{N-1}{\frac {n!}{(-x)^{n}}}+O(N!x^{-N})\right)}${\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 ${\displaystyle N}${\displaystyle N}, the number of terms in the truncated sum (${\displaystyle N=1}${\displaystyle N=1} in red, ${\displaystyle N=5}${\displaystyle N=5} in pink).

Using integration by parts, we can obtain an explicit formula^[9] $${\displaystyle \operatorname {Ei} (z)={\frac {e^{z}}{z}}\left(\sum _{k=0}^{n}{\frac {k!}{z^{k}}}+e_{n}(z)\right),\quad e_{n}(z)\equiv (n+1)!\ ze^{-z}\int _{-\infty }^{z}{\frac {e^{t}}{t^{n+2}}},円dt}$${\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}}},円dt} For any fixed ${\displaystyle z}${\displaystyle z}, the absolute value of the error term ${\displaystyle |e_{n}(z)|}${\displaystyle |e_{n}(z)|} decreases, then increases. The minimum occurs at ${\displaystyle n\sim |z|}${\displaystyle n\sim |z|}, at which point ${\displaystyle \vert e_{n}(z)\vert \leq {\sqrt {\frac {2\pi }{\vert z\vert }}}e^{-\vert z\vert }}${\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".

From the two series suggested in previous subsections, it follows that ${\displaystyle E_{1}}${\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, ${\displaystyle E_{1}}${\displaystyle E_{1}} can be bracketed by elementary functions as follows:^[10]

${\displaystyle {\frac {1}{2}}e^{-x},円\ln \!\left(1+{\frac {2}{x}}\right)<E_{1}(x)<e^{-x},円\ln \!\left(1+{\frac {1}{x}}\right)\qquad x>0}${\displaystyle {\frac {1}{2}}e^{-x},円\ln \!\left(1+{\frac {2}{x}}\right)<E_{1}(x)<e^{-x},円\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 ${\displaystyle E_{1}(x)}${\displaystyle E_{1}(x)} is shown in black and the right-hand side is shown in red.

Both ${\displaystyle \operatorname {Ei} }${\displaystyle \operatorname {Ei} } and ${\displaystyle E_{1}}${\displaystyle E_{1}} can be written more simply using the entire function ${\displaystyle \operatorname {Ein} }${\displaystyle \operatorname {Ein} }^[11] defined as

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

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

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

The function ${\displaystyle \operatorname {Ein} }${\displaystyle \operatorname {Ein} } is related to the exponential generating function of the harmonic numbers:

${\displaystyle \operatorname {Ein} (z)=e^{-z},円\sum _{n=1}^{\infty }{\frac {z^{n}}{n!}}H_{n}}${\displaystyle \operatorname {Ein} (z)=e^{-z},円\sum _{n=1}^{\infty }{\frac {z^{n}}{n!}}H_{n}}

Kummer's equation

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

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

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

we have

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

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

${\displaystyle E_{1}(-z)=-\gamma -i\pi +{\frac {\partial [U(a,1,z)-M(a,1,z)]}{\partial a}},\qquad 0<{\rm {Arg}}(z)<2\pi }${\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 ${\displaystyle a=0.}${\displaystyle a=0.} Another connexion with the confluent hypergeometric functions is that E₁ is an exponential times the function U(1,1,z):

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

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

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

The series expansion of the exponential integral immediately gives rise to an expression in terms of the generalized hypergeometric function ${\displaystyle {}_{2}F_{2}}${\displaystyle {}_{2}F_{2}}:

${\displaystyle \operatorname {Ei} (x)=x{}_{2}F_{2}(1,1;2,2;x)+\ln x+\gamma .}${\displaystyle \operatorname {Ei} (x)=x{}_{2}F_{2}(1,1;2,2;x)+\ln x+\gamma .}

The exponential integral may also be generalized to

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

which can be written as a special case of the upper incomplete gamma function:^[12]

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

The generalized form is sometimes called the Misra function^[13] ${\displaystyle \varphi _{m}(x)}${\displaystyle \varphi _{m}(x)}, defined as

${\displaystyle \varphi _{m}(x)=E_{-m}(x).}${\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^[14]

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

The derivatives of the generalised functions ${\displaystyle E_{n}}${\displaystyle E_{n}} can be calculated by means of the formula ^[15]

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

Note that the function ${\displaystyle E_{0}}${\displaystyle E_{0}} is easy to evaluate (making this recursion useful), since it is just ${\displaystyle e^{-z}/z}${\displaystyle e^{-z}/z}.^[16]

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

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

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

${\displaystyle E_{1}(ix)=i\left[-{\tfrac {1}{2}}\pi +\operatorname {Si} (x)\right]-\operatorname {Ci} (x)\qquad (x>0)}${\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 ${\displaystyle \mathrm {E} _{1}(ix)}${\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^[17] $${\displaystyle E_{1}(x)=\left(A^{-7.7}+B\right)^{-0.13},}$${\displaystyle E_{1}(x)=\left(A^{-7.7}+B\right)^{-0.13},} where $${\displaystyle {\begin{aligned}A&=\ln \left[\left({\frac {0.56146}{x}}+0.65\right)(1+x)\right]\\B&=x^{4}e^{7.7x}(2+x)^{3.7}\end{aligned}}}$${\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 ^{[17] [18]} $${\displaystyle E_{1}(x)={\begin{cases}-\ln x+{\textbf {a}}^{T}{\textbf {x}}_{5},&x\leq 1\\{\frac {e^{-x}}{x}}{\frac {{\textbf {b}}^{T}{\textbf {x}}_{3}}{{\textbf {c}}^{T}{\textbf {x}}_{3}}},&x\geq 1\end{cases}}}$${\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 $${\displaystyle {\begin{aligned}{\textbf {a}}&\triangleq [-0.57722,0.99999,-0.24991,0.05519,-0.00976,0.00108]^{T}\\{\textbf {b}}&\triangleq [0.26777,8.63476,18.05902,8.57333]^{T}\\{\textbf {c}}&\triangleq [3.95850,21.09965,25.63296,9.57332]^{T}\\{\textbf {x}}_{k}&\triangleq [x^{0},x^{1},\dots ,x^{k}]^{T}\end{aligned}}}$${\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 ^[18] $${\displaystyle E_{1}(x)={\cfrac {e^{-x}}{x+{\cfrac {1}{1+{\cfrac {1}{x+{\cfrac {2}{1+{\cfrac {2}{x+{\cfrac {3}{\ddots }}}}}}}}}}}}.}$${\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. ^[19] $${\displaystyle E_{1}(x)={\frac {e^{-x}}{G+(1-G)e^{-{\frac {x}{1-G}}}}}\ln \left[1+{\frac {G}{x}}-{\frac {1-G}{(h+bx)^{2}}}\right],}$${\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: $${\displaystyle {\begin{aligned}h&={\frac {1}{1+x{\sqrt {x}}}}+{\frac {h_{\infty }q}{1+q}}\\q&={\frac {20}{47}}x^{\sqrt {\frac {31}{26}}}\\h_{\infty }&={\frac {(1-G)(G^{2}-6G+12)}{3G(2-G)^{2}b}}\\b&={\sqrt {\frac {2(1-G)}{G(2-G)}}}\\G&=e^{-\gamma }\end{aligned}}}$${\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 ${\displaystyle \gamma }${\displaystyle \gamma } being the Euler–Mascheroni constant.

We can express the Inverse function of the exponential integral in power series form:^[20]

${\displaystyle \forall |x|<{\frac {\mu }{\ln(\mu )}},\quad \mathrm {Ei} ^{-1}(x)=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}{\frac {P_{n}(\ln(\mu ))}{\mu ^{n}}}}${\displaystyle \forall |x|<{\frac {\mu }{\ln(\mu )}},\quad \mathrm {Ei} ^{-1}(x)=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}{\frac {P_{n}(\ln(\mu ))}{\mu ^{n}}}}

where ${\displaystyle \mu }${\displaystyle \mu } is the Ramanujan–Soldner constant and ${\displaystyle (P_{n})}${\displaystyle (P_{n})} is polynomial sequence defined by the following recurrence relation:

${\displaystyle P_{0}(x)=x,\ P_{n+1}(x)=x(P_{n}'(x)-nP_{n}(x)).}${\displaystyle P_{0}(x)=x,\ P_{n+1}(x)=x(P_{n}'(x)-nP_{n}(x)).}

For ${\displaystyle n>0}${\displaystyle n>0}, ${\displaystyle \deg P_{n}=n}${\displaystyle \deg P_{n}=n} and we have the formula :

${\displaystyle P_{n}(x)=\left.\left({\frac {\mathrm {d} }{\mathrm {d} t}}\right)^{n-1}\left({\frac {te^{x}}{\mathrm {Ei} (t+x)-\mathrm {Ei} (x)}}\right)^{n}\right|_{t=0}.}${\displaystyle P_{n}(x)=\left.\left({\frac {\mathrm {d} }{\mathrm {d} t}}\right)^{n-1}\left({\frac {te^{x}}{\mathrm {Ei} (t+x)-\mathrm {Ei} (x)}}\right)^{n}\right|_{t=0}.}

• Time-dependent heat transfer
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