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

$\int xe^{cx}\,dx=e^{cx}\left({\frac {cx-1}{c^{2))}\right){\text{ for ))c\neq 0;$ $\int x^{2}e^{cx}\,dx=e^{cx}\left({\frac {x^{2)){c))-{\frac {2x}{c^{2))}+{\frac {2}{c^{3))}\right)$ {\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 was proved by Toyesh Prakash Sharma.[citation needed]

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

## 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)
$\int _{-\infty }^{\infty }xe^{-a(x-b)^{2))\,dx=b{\sqrt {\frac {\pi }{a))}\quad (\operatorname {Re} (a)>0)$ $\int _{-\infty }^{\infty }xe^{-ax^{2}+bx}\,dx={\frac ((\sqrt {\pi ))b}{2a^{3/2))}e^{\frac {b^{2)){4a))\quad (\operatorname {Re} (a)>0)$ $\int _{-\infty }^{\infty }x^{2}e^{-ax^{2))\,dx={\frac {1}{2)){\sqrt {\pi \over a^{3))}\quad (a>0)$ $\int _{-\infty }^{\infty }x^{2}e^{-(ax^{2}+bx)}\,dx={\frac ((\sqrt {\pi ))(2a+b^{2})}{4a^{5/2))}e^{\frac {b^{2)){4a))\quad (\operatorname {Re} (a)>0)$ $\int _{-\infty }^{\infty }x^{3}e^{-(ax^{2}+bx)}\,dx={\frac ((\sqrt {\pi ))(6a+b^{2})b}{8a^{7/2))}e^{\frac {b^{2)){4a))\quad (\operatorname {Re} (a)>0)$ $\int _{0}^{\infty }x^{n}e^{-ax^{2))\,dx={\begin{cases}{\dfrac {\Gamma \left({\frac {n+1}{2))\right)}{2\left(a^{\frac {n+1}{2))\right)))&(n>-1,\ a>0)\\\\{\dfrac {(2k-1)!!}{2^{k+1}a^{k))}{\sqrt {\dfrac {\pi }{a))}&(n=2k,\ k{\text{ integer)),\ a>0)\ {\text{(!! is the double factorial)))\\\\{\dfrac {k!}{2(a^{k+1})))&(n=2k+1,\ k{\text{ integer)),\ a>0)\end{cases))$ (the operator $!!$ is the Double factorial)

$\int _{0}^{\infty }x^{n}e^{-ax}\,dx={\begin{cases}{\dfrac {\Gamma (n+1)}{a^{n+1))}&(n>-1,\ \operatorname {Re} (a)>0)\\\\{\dfrac {n!}{a^{n+1))}&(n=0,1,2,\ldots ,\ \operatorname {Re} (a)>0)\end{cases))$ $\int _{0}^{1}x^{n}e^{-ax}\,dx={\frac {n!}{a^{n+1))}\left[1-e^{-a}\sum _{i=0}^{n}{\frac {a^{i)){i!))\right]$ $\int _{0}^{b}x^{n}e^{-ax}\,dx={\frac {n!}{a^{n+1))}\left[1-e^{-ab}\sum _{i=0}^{n}{\frac {(ab)^{i)){i!))\right]$ $\int _{0}^{\infty }e^{-ax^{b))dx={\frac {1}{b))\ a^{-{\frac {1}{b))}\Gamma \left({\frac {1}{b))\right)$ $\int _{0}^{\infty }x^{n}e^{-ax^{b))dx={\frac {1}{b))\ a^{-{\frac {n+1}{b))}\Gamma \left({\frac {n+1}{b))\right)$ $\int _{0}^{\infty }e^{-ax}\sin bx\,dx={\frac {b}{a^{2}+b^{2))}\quad (a>0)$ $\int _{0}^{\infty }e^{-ax}\cos bx\,dx={\frac {a}{a^{2}+b^{2))}\quad (a>0)$ $\int _{0}^{\infty }xe^{-ax}\sin bx\,dx={\frac {2ab}{(a^{2}+b^{2})^{2))}\quad (a>0)$ $\int _{0}^{\infty }xe^{-ax}\cos bx\,dx={\frac {a^{2}-b^{2)){(a^{2}+b^{2})^{2))}\quad (a>0)$ $\int _{0}^{\infty }{\frac {e^{-ax}\sin bx}{x))\,dx=\arctan {\frac {b}{a))$ $\int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx)){x))\,dx=\ln {\frac {b}{a))$ $\int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx)){x))\sin px\,dx=\arctan {\frac {b}{p))-\arctan {\frac {a}{p))$ $\int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx)){x))\cos px\,dx={\frac {1}{2))\ln {\frac {b^{2}+p^{2)){a^{2}+p^{2)))$ $\int _{0}^{\infty }{\frac {e^{-ax}(1-\cos x)}{x^{2))}\,dx=\operatorname {arccot} a-{\frac {a}{2))\ln {\Big (}{\frac {1}{a^{2))}+1{\Big ))$ $\int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)$ (I0 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,

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

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