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; $$ \int x^n e^{cx}\,dx &= \frac{1}{c} x^n e^{cx} - \frac{n}{c}\int x^{n-1} e^{cx} \,dx \\ &= \left( \frac{\partial}{\partial c} \right)^n \frac{e^{cx}}{c} \\ &= e^{cx}\sum_{i=0}^n (-1)^i\frac{n!}{(n-i)!c^{i+1}}x^{n-i} \\ &= e^{cx}\sum_{i=0}^n (-1)^{n-i}\frac{n!}{i!c^{n-i+1}}x^i \end{align}$$
 * $$\int x^2 e^{cx}\,dx = e^{cx}\left(\frac{x^2}{c}-\frac{2x}{c^2}+\frac{2}{c^3}\right)$$
 * $$\begin{align}
 * $$\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

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

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


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

Other integrals
$$ where $$ c_{2j}=\frac{ 1 \cdot 3 \cdot 5 \cdots (2j-1)}{2^{j+1}}=\frac{(2j)!}{j!2^{2j+1}} \. $$  (Note that the value of the expression is independent of the value of $n$, which is why it does not appear in the integral.) where $$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 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,
 * $$ {\int \underbrace{x^{x^{\cdot^{\cdot^{x}}}}}_mdx= \sum_{n=0}^m\frac{(-1)^n(n+1)^{n-1}}{n!}\Gamma(n+1,- \ln x) + \sum_{n=m+1}^\infty(-1)^na_{mn}\Gamma(n+1,-\ln x) \qquad\text{(for }x> 0\text{)}}$$
 * $$\int \frac{1}{ae^{\lambda x} + b} \,dx = \frac{x}{b} - \frac{1}{b \lambda} \ln\left(a e^{\lambda x} + b \right) $$ when $$b \neq 0$$, $$\lambda \neq 0$$, and $$ae^{\lambda x} + b > 0.$$
 * $$\int \frac{e^{2\lambda x}}{ae^{\lambda x} + b} \,dx = \frac{1}{a^2 \lambda} \left[a e^{\lambda x} + b - b \ln\left(a e^{\lambda x} + b \right) \right] $$ when $$a \neq 0$$, $$\lambda \neq 0$$, and $$ae^{\lambda x} + b > 0.$$
 * $$\int \frac{ae^{cx}-1}{be^{cx}-1}\,dx=\frac{(a-b)\log(1-be^{cx})}{bc}+x.$$
 * $$\int{e^{x}\left( f\left( x \right) + f'\left( x \right) \right)\text{dx}} = e^{x}f\left( x \right) + C$$


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


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

Definite integrals
\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{align}$$ The last expression is the logarithmic mean. \begin{cases} \dfrac{\Gamma \left(\frac{n+1}{2}\right)}{2\left(a^\frac{n+1}{2}\right) } & (n>-1,\ a>0) \\ \dfrac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\dfrac{\pi}{a}} & (n=2k,\ k \text{ integer},\ a>0) \\ \dfrac{k!}{2(a^{k+1})} & (n=2k+1,\ k \text{ integer},\ a>0) \end{cases}$$ (the operator $$!!$$ is the Double factorial) \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}$$ \frac{n!}{a^{n+1}}\left[1-e^{-a}\sum_{i=0}^{n} \frac{a^i}{i!}\right]$$ \frac{n!}{a^{n+1}}\left[1-e^{-ab}\sum_{i=0}^{n} \frac{(ab)^i}{i!}\right]$$ = e^f \sum_{n,m,p=0}^\infty \frac{ b^{4n}}{(4n)!} \frac{c^{2m}}{(2m)!} \frac{d^{4p}}{(4p)!} \frac{ \Gamma(3n+m+p+\frac14) }{a^{3n+m+p+\frac14} } $$ (appears in several models of extended superstring theory in higher dimensions) where $$\operatorname{Li}_{s}(z)$$ is the Polylogarithm. where $$\gamma$$ is the Euler–Mascheroni constant which equals the value of a number of definite integrals.
 * $$\begin{align}
 * $$\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} x e^{-a(x-b)^2}\,dx= b \sqrt{\frac{\pi}{a}} \quad (\operatorname{Re}(a)>0)$$
 * $$\int_{-\infty}^{\infty} x e^{-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 =
 * $$\int_0^{\infty} x^n e^{-ax}\,dx =
 * $$\int_0^{1} x^n e^{-ax}\,dx =
 * $$\int_0^{b} x^n e^{-ax}\,dx =
 * $$\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=\arccot a - \frac{a}{2}\ln \Big(\frac{1}{a^2}+1\Big)$$
 * $$ \int_{-\infty}^\infty e^{a x^4+b x^3+c x^2+d x+f} \, dx
 * $$\int_0^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_0(x)$$ ($I_{0}$ is the modified Bessel function of the first kind)
 * $$\int_0^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_0 \left( \sqrt{x^2 + y^2} \right)$$
 * $$\int_0^\infty\frac{x^{s-1}}{e^x/z-1} \,dx = \operatorname{Li}_{s}(z)\Gamma(s), $$
 * $$\int_0^\infty\frac{\sin mx}{e^{2 \pi x}-1} \,dx = \frac{1}{4} \coth \frac{m}{2} - \frac{1}{2m} $$
 * $$\int_0^\infty e^{-x} \ln x\, dx = - \gamma, $$

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.