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# 指数関数の原始関数の一覧

## 不定積分

${\displaystyle \int e^{x}\;\mathrm {d} x=e^{x}}$
${\displaystyle \int f'(x)e^{f(x)}\;\mathrm {d} x=e^{f(x)}}$
${\displaystyle \int e^{cx}\;\mathrm {d} x={\frac {1}{c}}e^{cx}}$
${\displaystyle \int a^{cx}\;\mathrm {d} x={\frac {1}{c\cdot \ln a}}a^{cx}}$${\displaystyle a>0,\ a\neq 1}$
${\displaystyle \int xe^{cx}\;\mathrm {d} x={\frac {e^{cx}}{c^{2}}}(cx-1)}$
${\displaystyle \int x^{2}e^{cx}\;\mathrm {d} x=e^{cx}\left({\frac {x^{2}}{c}}-{\frac {2x}{c^{2}}}+{\frac {2}{c^{3}}}\right)}$
${\displaystyle \int x^{n}e^{cx}\;\mathrm {d} x={\frac {1}{c}}x^{n}e^{cx}-{\frac {n}{c}}\int x^{n-1}e^{cx}\mathrm {d} x=\left({\frac {\partial }{\partial c}}\right)^{n}{\frac {e^{cx}}{c}}}$
${\displaystyle \int {\frac {e^{cx}}{x}}\;\mathrm {d} x=\ln |x|+\sum _{n=1}^{\infty }{\frac {(cx)^{n}}{n\cdot n!}}}$
${\displaystyle \int {\frac {e^{cx}}{x^{n}}}\;\mathrm {d} x={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,\mathrm {d} x\right)\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$
${\displaystyle \int e^{cx}\ln x\;\mathrm {d} x={\frac {1}{c}}\left(e^{cx}\ln |x|-\operatorname {Ei} \,(cx)\right)}$
${\displaystyle \int e^{cx}\sin bx\;\mathrm {d} x={\frac {e^{cx}}{c^{2}+b^{2}}}(c\sin bx-b\cos bx)}$
${\displaystyle \int e^{cx}\cos bx\;\mathrm {d} x={\frac {e^{cx}}{c^{2}+b^{2}}}(c\cos bx+b\sin bx)}$
${\displaystyle \int e^{cx}\sin ^{n}x\;\mathrm {d} x={\frac {e^{cx}\sin ^{n-1}x}{c^{2}+n^{2}}}(c\sin x-n\cos x)+{\frac {n(n-1)}{c^{2}+n^{2}}}\int e^{cx}\sin ^{n-2}x\;\mathrm {d} x}$
${\displaystyle \int e^{cx}\cos ^{n}x\;\mathrm {d} x={\frac {e^{cx}\cos ^{n-1}x}{c^{2}+n^{2}}}(c\cos x+n\sin x)+{\frac {n(n-1)}{c^{2}+n^{2}}}\int e^{cx}\cos ^{n-2}x\;\mathrm {d} x}$
${\displaystyle \int xe^{cx^{2}}\;\mathrm {d} x={\frac {1}{2c}}\;e^{cx^{2}}}$
${\displaystyle \int e^{-cx^{2}}\;\mathrm {d} x={\sqrt {\frac {\pi }{4c}}}\operatorname {erf} ({\sqrt {c}}x)}$${\displaystyle \operatorname {erf} }$誤差関数
${\displaystyle \int xe^{-cx^{2}}\;\mathrm {d} x=-{\frac {1}{2c}}e^{-cx^{2}}}$
${\displaystyle \int {\frac {e^{-x^{2}}}{x^{2}}}\;\mathrm {d} x=-{\frac {e^{-x^{2}}}{x}}-{\sqrt {\pi }}\mathrm {erf} (x)}$
${\displaystyle \int {{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}}\;\mathrm {d} x={\frac {1}{2}}\left(\operatorname {erf} \,{\frac {x-\mu }{\sigma {\sqrt {2}}}}\right)}$
${\displaystyle \int e^{x^{2}}\,\mathrm {d} x=e^{x^{2}}\left(\sum _{j=0}^{n-1}c_{2j}\,{\frac {1}{x^{2j+1}}}\right)+(2n-1)c_{2n-2}\int {\frac {e^{x^{2}}}{x^{2n}}}\;\mathrm {d} x\quad {\mbox{(}}n>0{\mbox{)}}}$
ここで ${\displaystyle c_{2j}={\frac {1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1}}}={\frac {(2j)\,!}{j!\,2^{2j+1}}}}$ とする。
${\displaystyle {\int \underbrace {x^{x^{\cdot ^{\cdot ^{x}}}}} _{m}\,dx=\int x\uparrow \uparrow m\,dx=\int x\to m\to 2\,dx=\sum _{n=0}^{m}{\frac {(-1)^{n}(n+1)^{n-1}}{n!}}\Gamma (n+1,-\ln x)+\sum _{n=m+1}^{\infty }(-1)^{n}a_{mn}\Gamma (n+1,-\ln x)\qquad {\mbox{(for }}x>0{\mbox{)}}}}$
ここで ${\displaystyle a_{mn}={\begin{cases}1&n=0,\\{\frac {1}{n!}}m=1,\\{\frac {1}{n}}\sum _{j=1}^{n}ja_{m,n-j}a_{m-1,j-1}&{\text{otherwise}}\end{cases}}}$
${\displaystyle \Gamma (x,y)}$ガンマ関数${\displaystyle \uparrow }$クヌースの矢印表記${\displaystyle \to }$コンウェイのチェーン表記
${\displaystyle \int {\frac {1}{ae^{\lambda x}+b}}\;\mathrm {d} x={\frac {x}{b}}-{\frac {1}{b\lambda }}\ln \left(ae^{\lambda x}+b\right)\,}$${\displaystyle b\neq 0}$, ${\displaystyle \lambda \neq 0}$, かつ ${\displaystyle ae^{\lambda x}+b>0\,}$
${\displaystyle \int {\frac {e^{2\lambda x}}{ae^{\lambda x}+b}}\;\mathrm {d} x={\frac {1}{a^{2}\lambda }}\left[ae^{\lambda x}+b-b\ln \left(ae^{\lambda x}+b\right)\right]\,}$${\displaystyle a\neq 0}$, ${\displaystyle \lambda \neq 0}$, かつ ${\displaystyle ae^{\lambda x}+b>0\,.}$

## 定積分

${\displaystyle \int _{0}^{1}e^{x\cdot \ln a+(1-x)\cdot \ln b}\;\mathrm {d} x=\int _{0}^{1}\left({\frac {a}{b}}\right)^{x}\cdot b\;\mathrm {d} x=\int _{0}^{1}a^{x}\cdot b^{1-x}\;\mathrm {d} x={\frac {a-b}{\ln a-\ln b}}}$${\displaystyle a>0,\ b>0,\ a\neq b}$
${\displaystyle \int _{0}^{\infty }e^{ax}\,\mathrm {d} x={\frac {1}{-a}}\quad (\operatorname {Re} (a)<0)}$
${\displaystyle \int _{0}^{\infty }e^{-ax^{2}}\,\mathrm {d} x={\frac {1}{2}}{\sqrt {\pi \over a}}\quad (a>0)}$ガウス積分
${\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}\,\mathrm {d} x={\sqrt {\pi \over a}}\quad (a>0)}$
${\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}e^{-2bx}\,\mathrm {d} x={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\quad (a>0)}$ガウス関数の積分）
${\displaystyle \int _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,\mathrm {d} x=b{\sqrt {\frac {\pi }{a}}}\quad (\operatorname {Re} (a)>0)}$
${\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-ax^{2}}\,\mathrm {d} x={\frac {1}{2}}{\sqrt {\pi \over a^{3}}}\quad (a>0)}$
${\displaystyle \int _{0}^{\infty }x^{n}e^{-ax^{2}}\,\mathrm {d} x={\begin{cases}{\frac {1}{2}}\Gamma \left({\frac {n+1}{2}}\right)/a^{\frac {n+1}{2}}&(n>-1,a>0)\\{\frac {(2k-1)!!}{2^{k+1}a^{k}}}{\sqrt {\frac {\pi }{a}}}&(n=2k,k\;{\text{integer}},a>0)\\{\frac {k!}{2a^{k+1}}}&(n=2k+1,k\;{\text{integer}},a>0)\end{cases}}}$ （!! は二重階乗
${\displaystyle \int _{0}^{\infty }x^{n}e^{-ax}\,\mathrm {d} x={\begin{cases}{\frac {\Gamma (n+1)}{a^{n+1}}}&(n>-1,a>0)\\{\frac {n!}{a^{n+1}}}&(n=0,1,2,\ldots ,a>0)\\\end{cases}}}$
${\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\,\mathrm {d} x={\frac {b}{a^{2}+b^{2}}}\quad (a>0)}$
${\displaystyle \int _{0}^{\infty }e^{-ax}\cos bx\,\mathrm {d} x={\frac {a}{a^{2}+b^{2}}}\quad (a>0)}$
${\displaystyle \int _{0}^{\infty }xe^{-ax}\sin bx\,\mathrm {d} x={\frac {2ab}{(a^{2}+b^{2})^{2}}}\quad (a>0)}$
${\displaystyle \int _{0}^{\infty }xe^{-ax}\cos bx\,\mathrm {d} x={\frac {a^{2}-b^{2}}{(a^{2}+b^{2})^{2}}}\quad (a>0)}$
${\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)}$${\displaystyle I_{0}}$変形ベッセル関数
${\displaystyle \int _{0}^{2\pi }e^{x\cos \theta +y\sin \theta }d\theta =2\pi I_{0}\left({\sqrt {x^{2}+y^{2}}}\right)}$