지수함수 적분표

아래 목록은 지수함수의 적분이다. 각 적분식에서 적분상수 C {\displaystyle C} 는 생략하였다. ∫ f ′ ( x ) e f ( x ) d x = e f ( x ) {\displaystyle \int f'(x)e^{f(x)}\,dx=e^{f(x)}} ∫ e c x d x = 1 c e c x {\displaystyle \int e^{cx}\,dx={\frac {1}{c}}e^{cx}} ∫ a c x d x = 1 c ⋅ ln ⁡ a a c x {\displaystyle \int a^{cx}\,dx={\frac {1}{c\cdot \ln a}}a^{cx}} ( a > 0 ,   a ≠ 1 ) {\displaystyle (a>0,\ a\neq 1)} ∫ x e c x d x = e c x ( c x − 1 c 2 ) {\displaystyle \int xe^{cx}\,dx=e^{cx}\left({\frac {cx-1}{c^{2}}}\right)} ∫ x 2 e c x d x = e c x ( x 2 c − 2 x c 2 + 2 c 3 ) {\displaystyle \int x^{2}e^{cx}\,dx=e^{cx}\left({\frac {x^{2}}{c}}-{\frac {2x}{c^{2}}}+{\frac {2}{c^{3}}}\right)} ∫ x n e c x d x = 1 c x n e c x − n c ∫ x n − 1 e c x d x = ( ∂ ∂ c ) n e c x c = e c x ∑ i = 0 n ( − 1 ) i n ! ( n − i ) ! c i + 1 x n − i = e c x ∑ i = 0 n ( − 1 ) n − i n ! i ! c n − i + 1 x i {\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}}} ∫ e c x x d x = ln ⁡ | x | + ∑ n = 1 ∞ ( c x ) n n ⋅ n ! {\displaystyle \int {\frac {e^{cx}}{x}}\,dx=\ln |x|+\sum _{n=1}^{\infty }{\frac {(cx)^{n}}{n\cdot n!}}} ∫ e c x x n d x = 1 n − 1 ( − e c x x n − 1 + c ∫ e c x x n − 1 d x ) {\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)} ( n ≠ 1 ) {\displaystyle (n\neq 1)} ∫ e c x sin ⁡ b x d x = e c x c 2 + b 2 ( c sin ⁡ b x − b cos ⁡ b x ) = e c x c 2 + b 2 sin ⁡ ( b x − ϕ ) {\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 cos ⁡ ( ϕ ) = c c 2 + b 2 {\displaystyle \cos(\phi )={\frac {c}{\sqrt {c^{2}+b^{2}}}}} ) ∫ e c x cos ⁡ b x d x = e c x c 2 + b 2 ( c cos ⁡ b x + b sin ⁡ b x ) = e c x c 2 + b 2 cos ⁡ ( b x − ϕ ) {\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 cos ⁡ ( ϕ ) = c c 2 + b 2 {\displaystyle \cos(\phi )={\frac {c}{\sqrt {c^{2}+b^{2}}}}} ) ∫ e c x sin n ⁡ x d x = e c x sin n − 1 ⁡ x c 2 + n 2 ( c sin ⁡ x − n cos ⁡ x ) + n ( n − 1 ) c 2 + n 2 ∫ e c x sin n − 2 ⁡ x d x {\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} ∫ e c x cos n ⁡ x d x = e c x cos n − 1 ⁡ x c 2 + n 2 ( c cos ⁡ x + n sin ⁡ x ) + n ( n − 1 ) c 2 + n 2 ∫ e c x cos n − 2 ⁡ x d x {\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. ∫ e c x ln ⁡ x d x = 1 c ( e c x ln ⁡ | x | − Ei ⁡ ( c x ) ) {\displaystyle \int e^{cx}\ln x\,dx={\frac {1}{c}}\left(e^{cx}\ln |x|-\operatorname {Ei} (cx)\right)} ∫ x e c x 2 d x = 1 2 c e c x 2 {\displaystyle \int xe^{cx^{2}}\,dx={\frac {1}{2c}}e^{cx^{2}}} ∫ e − c x 2 d x = π 4 c erf ⁡ ( c x ) {\displaystyle \int e^{-cx^{2}}\,dx={\sqrt {\frac {\pi }{4c}}}\operatorname {erf} ({\sqrt {c}}x)} ∫ x e − c x 2 d x = − 1 2 c e − c x 2 {\displaystyle \int xe^{-cx^{2}}\,dx=-{\frac {1}{2c}}e^{-cx^{2}}} ∫ e − x 2 x 2 d x = − e − x 2 x − π erf ⁡ ( x ) {\displaystyle \int {\frac {e^{-x^{2}}}{x^{2}}}\,dx=-{\frac {e^{-x^{2}}}{x}}-{\sqrt {\pi }}\operatorname {erf} (x)} ∫ 1 σ 2 π e − 1 2 ( x − μ σ ) 2 d x = 1 2 erf ⁡ ( x − μ σ 2 ) {\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)} ∫ e x 2 d x = e x 2 ( ∑ j = 0 n − 1 c 2 j 1 x 2 j + 1 ) + ( 2 n − 1 ) c 2 n − 2 ∫ e x 2 x 2 n d x {\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 c 2 j = 1 ⋅ 3 ⋅ 5 ⋯ ( 2 j − 1 ) 2 j + 1 = ( 2 j ) ! j ! 2 2 j + 1 {\displaystyle c_{2j}={\frac {1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1}}}={\frac {(2j)!}{j!2^{2j+1}}}} 이고, 모든 n > 0 {\displaystyle n>0} 에 대해 성립한다.) ∫ x x ⋅ ⋅ x ⏟ m d x = ∑ n = 0 m ( − 1 ) n ( n + 1 ) n − 1 n ! Γ ( n + 1 , − ln ⁡ x ) + ∑ n = m + 1 ∞ ( − 1 ) n a m n Γ ( n + 1 , − ln ⁡ x ) (for  x > 0 ) {\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 a m n = { 1 if  n = 0 , 1 n ! if  m = 1 , 1 n ∑ j = 1 n j a m , n − j a m − 1 , j − 1 otherwise {\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.) ∫ 1 a e λ x + b d x = x b − 1 b λ ln ⁡ ( a e λ x + b ) {\displaystyle \int {\frac {1}{ae^{\lambda x}+b}}\,dx={\frac {x}{b}}-{\frac {1}{b\lambda }}\ln \left(ae^{\lambda x}+b\right)} (이때 b ≠ 0 {\displaystyle b\neq 0} , λ ≠ 0 {\displaystyle \lambda \neq 0} 이고 a e λ x + b > 0 {\displaystyle ae^{\lambda x}+b>0} 이다.) ∫ e 2 λ x a e λ x + b d x = 1 a 2 λ [ a e λ x + b − b ln ⁡ ( a e λ x + b ) ] {\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 a ≠ 0 {\displaystyle a\neq 0} , λ ≠ 0 {\displaystyle \lambda \neq 0} 이고 a e λ x + b > 0 {\displaystyle ae^{\lambda x}+b>0} 이다.) ∫ a e c x − 1 b e c x − 1 d x = ( a − b ) log ⁡ ( 1 − b e c x ) b c + x . {\displaystyle \int {\frac {ae^{cx}-1}{be^{cx}-1}}\,dx={\frac {(a-b)\log(1-be^{cx})}{bc}}+x.} ∫ 0 1 e x ⋅ ln ⁡ a + ( 1 − x ) ⋅ ln ⁡ b d x = ∫ 0 1 ( a b ) x ⋅ b d x = ∫ 0 1 a x ⋅ b 1 − x d x = a − b ln ⁡ a − ln ⁡ b ( a > 0 ,   b > 0 ,   a ≠ b ) {\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. ∫ 0 ∞ e − a x d x = 1 a ( Re ⁡ ( a ) > 0 ) {\displaystyle \int _{0}^{\infty }e^{-ax}\,dx={\frac {1}{a}}\quad (\operatorname {Re} (a)>0)} ∫ 0 ∞ e − a x 2 d x = 1 2 π a ( a > 0 ) {\displaystyle \int _{0}^{\infty }e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\pi \over a}}\quad (a>0)} (가우스 적분) ∫ − ∞ ∞ e − a x 2 d x = π a ( a > 0 ) {\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}\,dx={\sqrt {\pi \over a}}\quad (a>0)} ∫ − ∞ ∞ e − a x 2 e − b x 2 d x = π a e − 2 a b ( a , b > 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)} ∫ − ∞ ∞ e − ( a x 2 + b x ) d x = π a e b 2 4 a ( a > 0 ) {\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx)}\,dx={\sqrt {\pi \over a}}e^{\tfrac {b^{2}}{4a}}\quad (a>0)} ∫ − ∞ ∞ e − a x 2 e − 2 b x d x = π a e b 2 a ( a > 0 ) {\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}e^{-2bx}\,dx={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\quad (a>0)} ∫ − ∞ ∞ x e − a ( x − b ) 2 d x = b π a ( Re ⁡ ( a ) > 0 ) {\displaystyle \int _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,dx=b{\sqrt {\frac {\pi }{a}}}\quad (\operatorname {Re} (a)>0)} ∫ − ∞ ∞ x e − a x 2 + b x d x = π b 2 a 3 / 2 e b 2 4 a ( 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)} ∫ − ∞ ∞ x 2 e − a x 2 d x = 1 2 π a 3 ( a > 0 ) {\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\pi \over a^{3}}}\quad (a>0)} ∫ − ∞ ∞ x 2 e − ( a x 2 + b x ) d x = π ( 2 a + b 2 ) 4 a 5 / 2 e b 2 4 a ( Re ⁡ ( 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)} ∫ − ∞ ∞ x 3 e − ( a x 2 + b x ) d x = π ( 6 a + b 2 ) b 8 a 7 / 2 e b 2 4 a ( 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)} ∫ 0 ∞ x n e − a x 2 d x = { Γ ( n + 1 2 ) 2 ( a n + 1 2 ) ( n > − 1 ,   a > 0 ) ( 2 k − 1 ) ! ! 2 k + 1 a k π a ( n = 2 k ,   a > 0 ) k ! 2 ( a k + 1 ) ( n = 2 k + 1 ,   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 k {\displaystyle k} 는 정수, ! ! {\displaystyle !!} 는 이중계승이다.) ∫ 0 ∞ x n e − a x d x = { Γ ( n + 1 ) a n + 1 ( n > − 1 ,   Re ⁡ ( a ) > 0 ) n ! a n + 1 ( n = 0 , 1 , 2 , … ,   Re ⁡ ( a ) > 0 ) {\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}}} ∫ 0 1 x n e − a x d x = n ! a n + 1 [ 1 − e − a ∑ i = 0 n a i i ! ] {\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]} ∫ 0 b x n e − a x d x = n ! a n + 1 [ 1 − e − a b ∑ i = 0 n ( a b ) i i ! ] {\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]} ∫ 0 ∞ e − a x b d x = 1 b   a − 1 b Γ ( 1 b ) {\displaystyle \int _{0}^{\infty }e^{-ax^{b}}dx={\frac {1}{b}}\ a^{-{\frac {1}{b}}}\Gamma \left({\frac {1}{b}}\right)} ∫ 0 ∞ x n e − a x b d x = 1 b   a − n + 1 b Γ ( n + 1 b ) {\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)} ∫ 0 ∞ e − a x sin ⁡ b x d x = b a 2 + b 2 ( a > 0 ) {\displaystyle \int _{0}^{\infty }e^{-ax}\sin bx\,dx={\frac {b}{a^{2}+b^{2}}}\quad (a>0)} ∫ 0 ∞ e − a x cos ⁡ b x d x = a a 2 + b 2 ( a > 0 ) {\displaystyle \int _{0}^{\infty }e^{-ax}\cos bx\,dx={\frac {a}{a^{2}+b^{2}}}\quad (a>0)} ∫ 0 ∞ x e − a x sin ⁡ b x d x = 2 a b ( a 2 + b 2 ) 2 ( a > 0 ) {\displaystyle \int _{0}^{\infty }xe^{-ax}\sin bx\,dx={\frac {2ab}{(a^{2}+b^{2})^{2}}}\quad (a>0)} ∫ 0 ∞ x e − a x cos ⁡ b x d x = a 2 − b 2 ( a 2 + b 2 ) 2 ( a > 0 ) {\displaystyle \int _{0}^{\infty }xe^{-ax}\cos bx\,dx={\frac {a^{2}-b^{2}}{(a^{2}+b^{2})^{2}}}\quad (a>0)} ∫ 0 ∞ e − a x sin ⁡ b x x d x = arctan ⁡ b a {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}\sin bx}{x}}\,dx=\arctan {\frac {b}{a}}} ∫ 0 ∞ e − a x − e − b x x d x = ln ⁡ b a {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x}}\,dx=\ln {\frac {b}{a}}} ∫ 0 ∞ e − a x − e − b x x sin ⁡ p x d x = arctan ⁡ b p − arctan ⁡ a p {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x}}\sin px\,dx=\arctan {\frac {b}{p}}-\arctan {\frac {a}{p}}} ∫ 0 ∞ e − a x − e − b x x cos ⁡ p x d x = 1 2 ln ⁡ b 2 + p 2 a 2 + p 2 {\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}}}} ∫ 0 ∞ e − a x ( 1 − cos ⁡ x ) x 2 d x = arccot ⁡ a − a 2 ln ⁡ ( a 2 + 1 ) {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}(1-\cos x)}{x^{2}}}\,dx=\operatorname {arccot} a-{\frac {a}{2}}\ln(a^{2}+1)} ∫ 0 2 π e x cos ⁡ θ d θ = 2 π I 0 ( x ) {\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)} (I0는 제1종 변형 베셀 함수이다.) ∫ 0 2 π e x cos ⁡ θ + y sin ⁡ θ d θ = 2 π I 0 ( x 2 + y 2 ) {\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)} ∫ 0 ∞ x s − 1 e x / z − 1 d x = Li s ⁡ ( z ) Γ ( s ) , {\displaystyle \int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}/z-1}}\,dx=\operatorname {Li} _{s}(z)\Gamma (s),} ( Li s ⁡ ( z ) {\displaystyle \operatorname {Li} _{s}(z)} 는 다중로그이다.) ∫ 0 ∞ sin ⁡ m x e 2 π x − 1 d x = 1 4 coth ⁡ m 2 − 1 2 m {\displaystyle \int _{0}^{\infty }{\frac {\sin mx}{e^{2\pi x}-1}}\,dx={\frac {1}{4}}\coth {\frac {m}{2}}-{\frac {1}{2m}}} ∫ 0 ∞ e − x ln ⁡ x d x = − γ , {\displaystyle \int _{0}^{\infty }e^{-x}\ln x\,dx=-\gamma ,} ( γ {\displaystyle \gamma } 는 오일러-마스케로니 상수) 지수함수 적분표