List of integrals of exponential functions

The following is a list of integrals of exponential functions. For a complete list of integral functions, please see the list of integrals.

Indefinite integral

Indefinite integrals are antiderivative functions. A constant (the constant of integration) may be added to the right hand side of any of these formulas, but has been suppressed here in the interest of brevity.

Integrals of polynomials

$\int xe^{cx}\,dx=e^{cx}\left({\frac {cx-1}{c^{2}}}\right)\qquad {\text{ for }}c\neq 0;$
$\int x^{2}e^{cx}\,dx=e^{cx}\left({\frac {x^{2}}{c}}-{\frac {2x}{c^{2}}}+{\frac {2}{c^{3}}}\right)$
${\begin{aligned}\int x^{n}e^{cx}\,dx&={\frac {1}{c}}x^{n}e^{cx}-{\frac {n}{c}}\int x^{n-1}e^{cx}\,dx\\&=\left({\frac {\partial }{\partial c}}\right)^{n}{\frac {e^{cx}}{c}}\\&=e^{cx}\sum _{i=0}^{n}(-1)^{i}{\frac {n!}{(n-i)!c^{i+1}}}x^{n-i}\\&=e^{cx}\sum _{i=0}^{n}(-1)^{n-i}{\frac {n!}{i!c^{n-i+1}}}x^{i}\end{aligned}}$
$\int {\frac {e^{cx}}{x}}\,dx=\ln |x|+\sum _{n=1}^{\infty }{\frac {(cx)^{n}}{n\cdot n!}}$
$\int {\frac {e^{cx}}{x^{n}}}\,dx={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,dx\right)\qquad {\text{(for }}n\neq 1{\text{)}}$

Integrals involving only exponential functions

$\int f'(x)e^{f(x)}\,dx=e^{f(x)}$
$\int e^{cx}\,dx={\frac {1}{c}}e^{cx}$
$\int a^{cx}\,dx={\frac {1}{c\cdot \ln a}}a^{cx}\qquad {\text{ for }}a>0,\ a\neq 1$

Integrals involving the error function

In the following formulas, $erf$ is the error function and $Ei$ is the exponential integral.

$\int e^{cx}\ln x\,dx={\frac {1}{c}}\left(e^{cx}\ln |x|-\operatorname {Ei} (cx)\right)$
$\int xe^{cx^{2}}\,dx={\frac {1}{2c}}e^{cx^{2}}$
$\int e^{-cx^{2}}\,dx={\sqrt {\frac {\pi }{4c}}}\operatorname {erf} ({\sqrt {c}}x)$
$\int xe^{-cx^{2}}\,dx=-{\frac {1}{2c}}e^{-cx^{2}}$
$\int {\frac {e^{-x^{2}}}{x^{2}}}\,dx=-{\frac {e^{-x^{2}}}{x}}-{\sqrt {\pi }}\operatorname {erf} (x)$
$\int {{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}}\,dx={\frac {1}{2}}\operatorname {erf} \left({\frac {x-\mu }{\sigma {\sqrt {2}}}}\right)$

Other integrals

$\int e^{x^{2}}\,dx=e^{x^{2}}\left(\sum _{j=0}^{n-1}c_{2j}{\frac {1}{x^{2j+1}}}\right)+(2n-1)c_{2n-2}\int {\frac {e^{x^{2}}}{x^{2n}}}\,dx\quad {\text{valid for any }}n>0,$
where $c_{2j}={\frac {1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1}}}={\frac {(2j)!}{j!2^{2j+1}}}\ .$

(Note that the value of the expression is independent of the value of

n

, which is why it does not appear in the integral.)

${\int \underbrace {x^{x^{\cdot ^{\cdot ^{x}}}}} _{m}dx=\sum _{n=0}^{m}{\frac {(-1)^{n}(n+1)^{n-1}}{n!}}\Gamma (n+1,-\ln x)+\sum _{n=m+1}^{\infty }(-1)^{n}a_{mn}\Gamma (n+1,-\ln x)\qquad {\text{(for }}x>0{\text{)}}}$

where $a_{mn}={\begin{cases}1&{\text{if }}n=0,\\\\{\dfrac {1}{n!}}&{\text{if }}m=1,\\\\{\dfrac {1}{n}}\sum _{j=1}^{n}ja_{m,n-j}a_{m-1,j-1}&{\text{otherwise}}\end{cases}}$

and $Γ(x, y)$ is the upper incomplete gamma function.

$\int {\frac {1}{ae^{\lambda x}+b}}\,dx={\frac {x}{b}}-{\frac {1}{b\lambda }}\ln \left(ae^{\lambda x}+b\right)$ when $b\neq 0$ , $\lambda \neq 0$ , and $ae^{\lambda x}+b>0.$
$\int {\frac {e^{2\lambda x}}{ae^{\lambda x}+b}}\,dx={\frac {1}{a^{2}\lambda }}\left[ae^{\lambda x}+b-b\ln \left(ae^{\lambda x}+b\right)\right]$ when $a\neq 0$ , $\lambda \neq 0$ , and $ae^{\lambda x}+b>0.$
$\int {\frac {ae^{cx}-1}{be^{cx}-1}}\,dx={\frac {(a-b)\log(1-be^{cx})}{bc}}+x.$
$\int {e^{x}\left(f\left(x\right)+f'\left(x\right)\right){\text{dx}}}=e^{x}f\left(x\right)+C$

The below formulae were proved by Toyesh Prakash Sharma (for $n\in \mathbb {Z} ^{+}$ ):

$\int {e^{x}\left(f\left(x\right)-\left(-1\right)^{n}{\frac {d^{n}f\left(x\right)}{dx^{n}}}\right)\,dx}=e^{x}\sum _{k=1}^{n}{\left(-1\right)^{k-1}{\frac {d^{k-1}f\left(x\right)}{dx^{k-1}}}}+C$
$\int {e^{-x}\left(f\left(x\right)-{\frac {d^{n}f\left(x\right)}{dx^{n}}}\right)\,dx}=-e^{-x}\sum _{k=1}^{n}{\frac {d^{k-1}f\left(x\right)}{dx^{k-1}}}+C$

In 2023, Toyesh Prakash Sharma and Etisha Sharma found:

$\int {e^{ax}\left(\left(a\right)^{n}f\left(x\right)-\left(-1\right)^{n}{\frac {d^{n}f\left(x\right)}{dx^{n}}}\right)\,dx}=e^{ax}\sum _{k=1}^{n}{\left(a\right)^{n-k}\left(-1\right)^{k-1}{\frac {d^{k-1}f\left(x\right)}{dx^{k-1}}}}+C$

Definite integrals

${\begin{aligned}\int _{0}^{1}e^{x\cdot \ln a+(1-x)\cdot \ln b}\,dx&=\int _{0}^{1}\left({\frac {a}{b}}\right)^{x}\cdot b\,dx\\&=\int _{0}^{1}a^{x}\cdot b^{1-x}\,dx\\&={\frac {a-b}{\ln a-\ln b}}\qquad {\text{for }}a>0,\ b>0,\ a\neq b\end{aligned}}$

The last expression is the logarithmic mean.

$\int _{0}^{\infty }e^{-ax}\,dx={\frac {1}{a}}\quad (\operatorname {Re} (a)>0)$
$\int _{0}^{\infty }e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\pi \over a}}\quad (a>0)$ (the Gaussian integral)
$\int _{-\infty }^{\infty }e^{-ax^{2}}\,dx={\sqrt {\pi \over a}}\quad (a>0)$
$\int _{-\infty }^{\infty }e^{-ax^{2}}e^{-{\frac {b}{x^{2}}}}\,dx={\sqrt {\frac {\pi }{a}}}e^{-2{\sqrt {ab}}}\quad (a,b>0)$
$\int _{-\infty }^{\infty }e^{-(ax^{2}+bx)}\,dx={\sqrt {\pi \over a}}e^{\tfrac {b^{2}}{4a}}\quad (a>0)$
$\int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\pi \over a}}e^{{\tfrac {b^{2}}{4a}}-c}\quad (a>0)$
$\int _{-\infty }^{\infty }e^{-ax^{2}}e^{-2bx}\,dx={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\quad (a>0)$ (see Integral of a Gaussian function)
$\int _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,dx=b{\sqrt {\frac {\pi }{a}}}\quad (\operatorname {Re} (a)>0)$
$\int _{-\infty }^{\infty }xe^{-ax^{2}+bx}\,dx={\frac {{\sqrt {\pi }}b}{2a^{3/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)$
$\int _{-\infty }^{\infty }x^{2}e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\pi \over a^{3}}}\quad (a>0)$
$\int _{-\infty }^{\infty }x^{2}e^{-(ax^{2}+bx)}\,dx={\frac {{\sqrt {\pi }}(2a+b^{2})}{4a^{5/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)$
$\int _{-\infty }^{\infty }x^{3}e^{-(ax^{2}+bx)}\,dx={\frac {{\sqrt {\pi }}(6a+b^{2})b}{8a^{7/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)$
$\int _{0}^{\infty }x^{n}e^{-ax^{2}}\,dx={\begin{cases}{\dfrac {\Gamma \left({\frac {n+1}{2}}\right)}{2\left(a^{\frac {n+1}{2}}\right)}}&(n>-1,\ a>0)\\{\dfrac {(2k-1)!!}{2^{k+1}a^{k}}}{\sqrt {\dfrac {\pi }{a}}}&(n=2k,\ k{\text{ integer}},\ a>0)\\{\dfrac {k!}{2(a^{k+1})}}&(n=2k+1,\ k{\text{ integer}},\ a>0)\end{cases}}$

(the operator

{\displaystyle

is the Double factorial)

$\int _{0}^{\infty }x^{n}e^{-ax}\,dx={\begin{cases}{\dfrac {\Gamma (n+1)}{a^{n+1}}}&(n>-1,\ \operatorname {Re} (a)>0)\\\\{\dfrac {n!}{a^{n+1}}}&(n=0,1,2,\ldots ,\ \operatorname {Re} (a)>0)\end{cases}}$
$\int _{0}^{1}x^{n}e^{-ax}\,dx={\frac {n!}{a^{n+1}}}\left[1-e^{-a}\sum _{i=0}^{n}{\frac {a^{i}}{i!}}\right]$
$\int _{0}^{b}x^{n}e^{-ax}\,dx={\frac {n!}{a^{n+1}}}\left[1-e^{-ab}\sum _{i=0}^{n}{\frac {(ab)^{i}}{i!}}\right]$
$\int _{0}^{\infty }e^{-ax^{b}}dx={\frac {1}{b}}\ a^{-{\frac {1}{b}}}\Gamma \left({\frac {1}{b}}\right)$
$\int _{0}^{\infty }x^{n}e^{-ax^{b}}dx={\frac {1}{b}}\ a^{-{\frac {n+1}{b}}}\Gamma \left({\frac {n+1}{b}}\right)$
$\int _{0}^{\infty }e^{-ax}\sin bx\,dx={\frac {b}{a^{2}+b^{2}}}\quad (a>0)$
$\int _{0}^{\infty }e^{-ax}\cos bx\,dx={\frac {a}{a^{2}+b^{2}}}\quad (a>0)$
$\int _{0}^{\infty }xe^{-ax}\sin bx\,dx={\frac {2ab}{(a^{2}+b^{2})^{2}}}\quad (a>0)$
$\int _{0}^{\infty }xe^{-ax}\cos bx\,dx={\frac {a^{2}-b^{2}}{(a^{2}+b^{2})^{2}}}\quad (a>0)$
$\int _{0}^{\infty }{\frac {e^{-ax}\sin bx}{x}}\,dx=\arctan {\frac {b}{a}}$
$\int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x}}\,dx=\ln {\frac {b}{a}}$
$\int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x}}\sin px\,dx=\arctan {\frac {b}{p}}-\arctan {\frac {a}{p}}$
$\int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x}}\cos px\,dx={\frac {1}{2}}\ln {\frac {b^{2}+p^{2}}{a^{2}+p^{2}}}$
$\int _{0}^{\infty }{\frac {e^{-ax}(1-\cos x)}{x^{2}}}\,dx=\operatorname {arccot} a-{\frac {a}{2}}\ln {\Big (}{\frac {1}{a^{2}}}+1{\Big )}$
$\int _{-\infty }^{\infty }e^{ax^{4}+bx^{3}+cx^{2}+dx+f}\,dx=e^{f}\sum _{n,m,p=0}^{\infty }{\frac {b^{4n}}{(4n)!}}{\frac {c^{2m}}{(2m)!}}{\frac {d^{4p}}{(4p)!}}{\frac {\Gamma (3n+m+p+{\frac {1}{4}})}{a^{3n+m+p+{\frac {1}{4}}}}}$ (appears in several models of extended superstring theory in higher dimensions)
$\int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)$ ( $I 0$ is the modified Bessel function of the first kind)
$\int _{0}^{2\pi }e^{x\cos \theta +y\sin \theta }d\theta =2\pi I_{0}\left({\sqrt {x^{2}+y^{2}}}\right)$
$\int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}/z-1}}\,dx=\operatorname {Li} _{s}(z)\Gamma (s),$

where

\operatorname {Li} _{s}(z)

is the Polylogarithm.

$\int _{0}^{\infty }{\frac {\sin mx}{e^{2\pi x}-1}}\,dx={\frac {1}{4}}\coth {\frac {m}{2}}-{\frac {1}{2m}}$
$\int _{0}^{\infty }e^{-x}\ln x\,dx=-\gamma ,$

where

\gamma

is the Euler–Mascheroni constant which equals the value of a number of definite integrals.

Finally, a well known result,

\int _{0}^{2\pi }e^{i(m-n)\phi }d\phi =2\pi \delta _{m,n}\qquad {\text{for }}m,n\in \mathbb {Z}

where $\delta _{m,n}$ is the Kronecker delta.

References

Toyesh Prakash Sharma, Etisha Sharma, "Putting Forward Another Generalization Of The Class Of Exponential Integrals And Their Applications.," International Journal of Scientific Research in Mathematical and Statistical Sciences, Vol.10, Issue.2, pp.1-8, 2023.

External links

Wolfram Mathematica Online Integrator
Moll, Victor Hugo. "List with the formulas and proofs in GR". Retrieved 2016-02-12.

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