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

${\displaystyle \int xe^{cx}\,dx=e^{cx}\left({\frac {cx-1}{c^{2))}\right){\text{ for ))c\neq 0;}$

${\displaystyle \int x^{2}e^{cx}\,dx=e^{cx}\left({\frac {x^{2)){c))-{\frac {2x}{c^{2))}+{\frac {2}{c^{3))}\right)}$

${\displaystyle {\begin{aligned}\int x^{n}e^{cx}\,dx&={\frac {1}{c))x^{n}e^{cx}-{\frac {n}{c))\int x^{n-1}e^{cx}\,dx\\&=\left({\frac {\partial }{\partial c))\right)^{n}{\frac {e^{cx)){c))\\&=e^{cx}\sum _{i=0}^{n}(-1)^{i}{\frac {n!}{(n-i)!c^{i+1))}x^{n-i}\\&=e^{cx}\sum _{i=0}^{n}(-1)^{n-i}{\frac {n!}{i!c^{n-i+1))}x^{i}\end{aligned))}$

${\displaystyle \int {\frac {e^{cx)){x))\,dx=\ln |x|+\sum _{n=1}^{\infty }{\frac {(cx)^{n)){n\cdot n!))}$

${\displaystyle \int {\frac {e^{cx)){x^{n))}\,dx={\frac {1}{n-1))\left(-{\frac {e^{cx)){x^{n-1))}+c\int {\frac {e^{cx)){x^{n-1))}\,dx\right)\qquad {\text{(for ))n\neq 1{\text{)))}$

Integrals involving only exponential functions

${\displaystyle \int f'(x)e^{f(x)}\,dx=e^{f(x)))$

${\displaystyle \int e^{cx}\,dx={\frac {1}{c))e^{cx))$

${\displaystyle \int a^{cx}\,dx={\frac {1}{c\cdot \ln a))a^{cx}\qquad {\text{ for ))a>0,\ a\neq 1}$

Integrals involving the error function

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

${\displaystyle \int e^{cx}\ln x\,dx={\frac {1}{c))\left(e^{cx}\ln |x|-\operatorname {Ei} (cx)\right)}$

${\displaystyle \int xe^{cx^{2))\,dx={\frac {1}{2c))e^{cx^{2))}$

${\displaystyle \int e^{-cx^{2))\,dx={\sqrt {\frac {\pi }{4c))}\operatorname {erf} ({\sqrt {c))x)}$

${\displaystyle \int xe^{-cx^{2))\,dx=-{\frac {1}{2c))e^{-cx^{2))}$

${\displaystyle \int {\frac {e^{-x^{2))}{x^{2))}\,dx=-{\frac {e^{-x^{2))}{x))-{\sqrt {\pi ))\operatorname {erf} (x)}$

${\displaystyle \int ((\frac {1}{\sigma {\sqrt {2\pi ))))e^{-{\frac {1}{2))\left({\frac {x-\mu }{\sigma ))\right)^{2))}\,dx={\frac {1}{2))\operatorname {erf} \left({\frac {x-\mu }{\sigma {\sqrt {2))))\right)}$

Other integrals

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

where ${\displaystyle c_{2j}={\frac {1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1))}={\frac {(2j)!}{j!2^{2j+1))}\ .}$

(Note that the value of the expression is independent of the value of n, which is why it does not appear in the integral.)

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

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

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

${\displaystyle \int {\frac {1}{ae^{\lambda x}+b))\,dx={\frac {x}{b))-{\frac {1}{b\lambda ))\ln \left(ae^{\lambda x}+b\right)}$ when ${\displaystyle b\neq 0}$, ${\displaystyle \lambda \neq 0}$, and ${\displaystyle ae^{\lambda x}+b>0.}$

${\displaystyle \int {\frac {e^{2\lambda x)){ae^{\lambda x}+b))\,dx={\frac {1}{a^{2}\lambda ))\left[ae^{\lambda x}+b-b\ln \left(ae^{\lambda x}+b\right)\right]}$ when ${\displaystyle a\neq 0}$, ${\displaystyle \lambda \neq 0}$, and ${\displaystyle ae^{\lambda x}+b>0.}$

${\displaystyle \int {\frac {ae^{cx}-1}{be^{cx}-1))\,dx={\frac {(a-b)\log(1-be^{cx})}{bc))+x.}$

${\displaystyle \int {e^{x}\left(f\left(x\right)+f'\left(x\right)\right){\text{dx))}=e^{x}f\left(x\right)+C}$

the below formulae was proved by Toyesh Prakash Sharma.^{[citation needed]}

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

(if ${\displaystyle n}$ is a positive integer)

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

(if ${\displaystyle n}$ is a positive integer)

Definite integrals

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

The last expression is the logarithmic mean.

${\displaystyle \int _{0}^{\infty }e^{-ax}\,dx={\frac {1}{a))\quad (\operatorname {Re} (a)>0)}$

${\displaystyle \int _{0}^{\infty }e^{-ax^{2))\,dx={\frac {1}{2)){\sqrt {\pi \over a))\quad (a>0)}$ (the Gaussian integral)

${\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2))\,dx={\sqrt {\pi \over a))\quad (a>0)}$

${\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2))e^{-{\frac {b}{x^{2))))\,dx={\sqrt {\frac {\pi }{a))}e^{-2{\sqrt {ab))}\quad (a,b>0)}$

${\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx)}\,dx={\sqrt {\pi \over a))e^{\tfrac {b^{2)){4a))\quad (a>0)}$

${\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\pi \over a))e^((\tfrac {b^{2)){4a))-c}\quad (a>0)}$

${\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2))e^{-2bx}\,dx={\sqrt {\frac {\pi }{a))}e^{\frac {b^{2)){a))\quad (a>0)}$ (see Integral of a Gaussian function)

${\displaystyle \int _{-\infty }^{\infty }xe^{-a(x-b)^{2))\,dx=b{\sqrt {\frac {\pi }{a))}\quad (\operatorname {Re} (a)>0)}$

${\displaystyle \int _{-\infty }^{\infty }xe^{-ax^{2}+bx}\,dx={\frac ((\sqrt {\pi ))b}{2a^{3/2))}e^{\frac {b^{2)){4a))\quad (\operatorname {Re} (a)>0)}$

${\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-ax^{2))\,dx={\frac {1}{2)){\sqrt {\pi \over a^{3))}\quad (a>0)}$

${\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-(ax^{2}+bx)}\,dx={\frac ((\sqrt {\pi ))(2a+b^{2})}{4a^{5/2))}e^{\frac {b^{2)){4a))\quad (\operatorname {Re} (a)>0)}$

${\displaystyle \int _{-\infty }^{\infty }x^{3}e^{-(ax^{2}+bx)}\,dx={\frac ((\sqrt {\pi ))(6a+b^{2})b}{8a^{7/2))}e^{\frac {b^{2)){4a))\quad (\operatorname {Re} (a)>0)}$

${\displaystyle \int _{0}^{\infty }x^{n}e^{-ax^{2))\,dx={\begin{cases}{\dfrac {\Gamma \left({\frac {n+1}{2))\right)}{2\left(a^{\frac {n+1}{2))\right)))&(n>-1,\ a>0)\\\\{\dfrac {(2k-1)!!}{2^{k+1}a^{k))}{\sqrt {\dfrac {\pi }{a))}&(n=2k,\ k{\text{ integer)),\ a>0)\ {\text{(!! is the double factorial)))\\\\{\dfrac {k!}{2(a^{k+1})))&(n=2k+1,\ k{\text{ integer)),\ a>0)\end{cases))}$

(the operator ${\displaystyle !!}$ is the Double factorial)

${\displaystyle \int _{0}^{\infty }x^{n}e^{-ax}\,dx={\begin{cases}{\dfrac {\Gamma (n+1)}{a^{n+1))}&(n>-1,\ \operatorname {Re} (a)>0)\\\\{\dfrac {n!}{a^{n+1))}&(n=0,1,2,\ldots ,\ \operatorname {Re} (a)>0)\end{cases))}$

${\displaystyle \int _{0}^{1}x^{n}e^{-ax}\,dx={\frac {n!}{a^{n+1))}\left[1-e^{-a}\sum _{i=0}^{n}{\frac {a^{i)){i!))\right]}$

${\displaystyle \int _{0}^{b}x^{n}e^{-ax}\,dx={\frac {n!}{a^{n+1))}\left[1-e^{-ab}\sum _{i=0}^{n}{\frac {(ab)^{i)){i!))\right]}$

${\displaystyle \int _{0}^{\infty }e^{-ax^{b))dx={\frac {1}{b))\ a^{-{\frac {1}{b))}\Gamma \left({\frac {1}{b))\right)}$

${\displaystyle \int _{0}^{\infty }x^{n}e^{-ax^{b))dx={\frac {1}{b))\ a^{-{\frac {n+1}{b))}\Gamma \left({\frac {n+1}{b))\right)}$

${\displaystyle \int _{0}^{\infty }e^{-ax}\sin bx\,dx={\frac {b}{a^{2}+b^{2))}\quad (a>0)}$

${\displaystyle \int _{0}^{\infty }e^{-ax}\cos bx\,dx={\frac {a}{a^{2}+b^{2))}\quad (a>0)}$

${\displaystyle \int _{0}^{\infty }xe^{-ax}\sin bx\,dx={\frac {2ab}{(a^{2}+b^{2})^{2))}\quad (a>0)}$

${\displaystyle \int _{0}^{\infty }xe^{-ax}\cos bx\,dx={\frac {a^{2}-b^{2)){(a^{2}+b^{2})^{2))}\quad (a>0)}$

${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}\sin bx}{x))\,dx=\arctan {\frac {b}{a))}$

${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx)){x))\,dx=\ln {\frac {b}{a))}$

${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx)){x))\sin px\,dx=\arctan {\frac {b}{p))-\arctan {\frac {a}{p))}$

${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx)){x))\cos px\,dx={\frac {1}{2))\ln {\frac {b^{2}+p^{2)){a^{2}+p^{2))))$

${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}(1-\cos x)}{x^{2))}\,dx=\operatorname {arccot} a-{\frac {a}{2))\ln {\Big (}{\frac {1}{a^{2))}+1{\Big )))$

${\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)}$ (I₀ is the modified Bessel function of the first kind)

${\displaystyle \int _{0}^{2\pi }e^{x\cos \theta +y\sin \theta }d\theta =2\pi I_{0}\left({\sqrt {x^{2}+y^{2))}\right)}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{s-1)){e^{x}/z-1))\,dx=\operatorname {Li} _{s}(z)\Gamma (s),}$

where ${\displaystyle \operatorname {Li} _{s}(z)}$ is the Polylogarithm.

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

${\displaystyle \int _{0}^{\infty }e^{-x}\ln x\,dx=-\gamma ,}$

where ${\displaystyle \gamma }$ is the Euler–Mascheroni constant which equals the value of a number of definite integrals.

Finally, a well known result,

${\displaystyle \int _{0}^{2\pi }e^{i(m-n)\phi }d\phi =2\pi \delta _{m,n))$ (For integer m, n)

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