아래 목록은 지수함수의 적분이다.

각 적분식에서 적분상수 ${\displaystyle C}$는 생략하였다.

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

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

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

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

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

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

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

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

${\displaystyle {\begin{aligned}\int e^{cx}\sin bx\,dx&={\frac {e^{cx}}{c^{2}+b^{2}}}(c\sin bx-b\cos bx)\\&={\frac {e^{cx}}{\sqrt {c^{2}+b^{2}}}}\sin(bx-\phi )\end{aligned}}}$ (이때 ${\displaystyle \cos(\phi )={\frac {c}{\sqrt {c^{2}+b^{2}}}}}$)

${\displaystyle {\begin{aligned}\int e^{cx}\cos bx\,dx&={\frac {e^{cx}}{c^{2}+b^{2}}}(c\cos bx+b\sin bx)\\&={\frac {e^{cx}}{\sqrt {c^{2}+b^{2}}}}\cos(bx-\phi )\end{aligned}}}$ (이때 ${\displaystyle \cos(\phi )={\frac {c}{\sqrt {c^{2}+b^{2}}}}}$)

${\displaystyle \int e^{cx}\sin ^{n}x\,dx={\frac {e^{cx}\sin ^{n-1}x}{c^{2}+n^{2}}}(c\sin x-n\cos x)+{\frac {n(n-1)}{c^{2}+n^{2}}}\int e^{cx}\sin ^{n-2}x\,dx}$

${\displaystyle \int e^{cx}\cos ^{n}x\,dx={\frac {e^{cx}\cos ^{n-1}x}{c^{2}+n^{2}}}(c\cos x+n\sin x)+{\frac {n(n-1)}{c^{2}+n^{2}}}\int e^{cx}\cos ^{n-2}x\,dx}$

다음 식들에서 erf는 오차 함수이고, Ei는 지수 적분 함수이다.

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

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

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

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

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

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

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

(이때 ${\displaystyle c_{2j}={\frac {1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1}}}={\frac {(2j)!}{j!2^{2j+1}}}}$이고, 모든 ${\displaystyle n>0}$에 대해 성립한다.)

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

(이때 ${\displaystyle a_{mn}={\begin{cases}1&{\text{if }}n=0,\\\\{\dfrac {1}{n!}}&{\text{if }}m=1,\\\\{\dfrac {1}{n}}\sum _{j=1}^{n}ja_{m,n-j}a_{m-1,j-1}&{\text{otherwise}}\end{cases}}}$이고, Γ(x,y)는 불완전 감마 함수이다.)

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

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

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

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

위 적분식의 마지막 값은 로그 평균을 뜻한다.

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

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

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

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

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

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

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

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

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

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

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

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

(이때 ${\displaystyle k}$는 정수, ${\displaystyle !!}$는 이중계승이다.)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)}$ (I₀는 제1종 변형 베셀 함수이다.)

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

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

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

${\displaystyle \int _{0}^{\infty }e^{-x}\ln x\,dx=-\gamma ,}$ (${\displaystyle \gamma }$는 오일러-마스케로니 상수)

• 지수함수
• 적분표