지수함수 적분표

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)

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