Exponenciális függvények integráljainak listája

Az alábbi lista az exponenciális függvények integráljait tartalmazza. A c egy tetszőleges konstans (c ≠ 0).

Határozatlan integrálok

e x d x = e x {\displaystyle \int e^{x}\;\mathrm {d} x=e^{x}}
e c x d x = 1 c e c x {\displaystyle \int e^{cx}\;\mathrm {d} x={\frac {1}{c}}e^{cx}}
a c x d x = 1 c ln a a c x ( a > 0 ,   a 1 ) {\displaystyle \int a^{cx}\;\mathrm {d} x={\frac {1}{c\cdot \ln a}}a^{cx}\qquad (a>0,\ a\neq 1)}
x e c x d x = e c x c 2 ( c x 1 ) {\displaystyle \int xe^{cx}\;\mathrm {d} x={\frac {e^{cx}}{c^{2}}}(cx-1)}
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}\;\mathrm {d} x=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 {\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}}}
e c x x d x = ln | x | + n = 1 ( c x ) n n n ! {\displaystyle \int {\frac {e^{cx}}{x}}\;\mathrm {d} x=\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 ) ( n 1 ) {\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 (n\neq 1)}
e c x ln x d x = 1 c e c x ln | x | Ei ( c x ) {\displaystyle \int e^{cx}\ln x\;\mathrm {d} x={\frac {1}{c}}e^{cx}\ln |x|-\operatorname {Ei} \,(cx)}
e c x sin b x d x = e c x c 2 + b 2 ( c sin b x b cos b x ) {\displaystyle \int e^{cx}\sin bx\;\mathrm {d} x={\frac {e^{cx}}{c^{2}+b^{2}}}(c\sin bx-b\cos bx)}
e c x cos b x d x = e c x c 2 + b 2 ( c cos b x + b sin b x ) {\displaystyle \int e^{cx}\cos bx\;\mathrm {d} x={\frac {e^{cx}}{c^{2}+b^{2}}}(c\cos bx+b\sin bx)}
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\;\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}
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\;\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}
x e c x 2 d x = 1 2 c e c x 2 {\displaystyle \int xe^{cx^{2}}\;\mathrm {d} x={\frac {1}{2c}}\;e^{cx^{2}}}
e c x 2 d x = π 4 c erf ( c x ) {\displaystyle \int e^{-cx^{2}}\;\mathrm {d} x={\sqrt {\frac {\pi }{4c}}}{\mbox{erf}}({\sqrt {c}}x)} {\displaystyle \qquad } (erf a Gauss-féle hibafüggvény)
x e c x 2 d x = 1 2 c e c x 2 {\displaystyle \int xe^{-cx^{2}}\;\mathrm {d} x=-{\frac {1}{2c}}e^{-cx^{2}}}
1 σ 2 π e ( x μ ) 2 / 2 σ 2 d x = 1 2 ( erf x + μ σ 2 ) {\displaystyle \int {1 \over \sigma {\sqrt {2\pi }}}\,e^{-{(x-\mu )^{2}/2\sigma ^{2}}}\;\mathrm {d} x=-{\frac {1}{2}}\left({\mbox{erf}}\,{\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 ( n > 0 ) {\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 (n>0)}
ahol 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}}}}
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 ) ( 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 (x>0)}
ahol a m n = { 1 ha  n = 0 , 1 n ! ha  m = 1 , 1 n j = 1 n j a m , n j a m 1 , j 1 ha  n 0 ,   m 1 {\displaystyle a_{mn}={\begin{cases}1&{\text{ha }}n=0,\\{\frac {1}{n!}}&{\text{ha }}m=1,\\{\frac {1}{n}}\sum _{j=1}^{n}ja_{m,n-j}a_{m-1,j-1}&{\text{ha }}n\neq 0,\ m\neq 1\\\end{cases}}}
1 a e λ x + b d x = x b 1 b λ ln ( a e λ x + b ) ( b , λ 0 ,   a e λ x + b > 0 ) {\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)\qquad (b,\lambda \neq 0,\ ae^{\lambda x}+b>0)}
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 ) ] ( a , λ 0 ,   a e λ 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]\qquad (a,\lambda \neq 0,\ ae^{\lambda x}+b>0)}

Néhány speciális határozott integrál

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 \int \limits _{0}^{1}e^{x\cdot \ln a+(1-x)\cdot \ln b}\;\mathrm {d} x=\int \limits _{0}^{1}\left({\frac {a}{b}}\right)^{x}\cdot b\;\mathrm {d} x=\int \limits _{0}^{1}a^{x}\cdot b^{1-x}\;\mathrm {d} x={\frac {a-b}{\ln a-\ln b}}\qquad (a>0,\ b>0,\ a\neq b)} ami a logaritmikus átlag
0 e a x d x = 1 a ( a > 0 ) {\displaystyle \int \limits _{0}^{\infty }e^{-ax}\,\mathrm {d} x={\frac {1}{a}}\qquad (a>0)}
0 e a x 2 d x = 1 2 π a ( a > 0 ) {\displaystyle \int \limits _{0}^{\infty }e^{-ax^{2}}\,\mathrm {d} x={\frac {1}{2}}{\sqrt {\pi \over a}}\qquad (a>0)} (lásd Gauss-integrál)
e a x 2 d x = π a ( a > 0 ) {\displaystyle \int \limits _{-\infty }^{\infty }e^{-ax^{2}}\,\mathrm {d} x={\sqrt {\pi \over a}}\qquad (a>0)}
e a x 2 e 2 b x d x = π a e b 2 a ( a > 0 ) {\displaystyle \int \limits _{-\infty }^{\infty }e^{-ax^{2}}e^{-2bx}\,\mathrm {d} x={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\qquad (a>0)} (lásd Gauss-függvény integrálja)
x e a ( x b ) 2 d x = b π a {\displaystyle \int \limits _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,\mathrm {d} x=b{\sqrt {\frac {\pi }{a}}}}
x 2 e a x 2 d x = 1 2 π a 3 ( a > 0 ) {\displaystyle \int \limits _{-\infty }^{\infty }x^{2}e^{-ax^{2}}\,\mathrm {d} x={\frac {1}{2}}{\sqrt {\pi \over a^{3}}}\qquad (a>0)}
0 x n e a x 2 d x = { 1 2 Γ ( n + 1 2 ) / a n + 1 2 ( n > 1 ,   a > 0 ) ( 2 k 1 ) ! ! 2 k + 1 a k π a ( k Z ,   n = 2 k ,   a > 0 ) k ! 2 a k + 1 ( k Z ,   n = 2 k + 1 ,   a > 0 ) {\displaystyle \int \limits _{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}}}&(k\in \mathbb {Z} ,\ n=2k,\ a>0)\\{\frac {k!}{2a^{k+1}}}&(k\in \mathbb {Z} ,\ n=2k+1,\ a>0)\end{cases}}} (!! a dupla faktoriális)
0 x n e a x d x = { Γ ( n + 1 ) a n + 1 ( n > 1 , a > 0 ) n ! a n + 1     ( n = 0 , 1 , 2 , , a > 0 ) {\displaystyle \int \limits _{0}^{\infty }x^{n}e^{-ax}\,\mathrm {d} x={\begin{cases}{\frac {\Gamma (n+1)}{a^{n+1}}}\qquad (n>-1,a>0)\\{\frac {n!}{a^{n+1}}}\qquad \ \ (n=0,1,2,\ldots ,a>0)\\\end{cases}}}
0 e a x sin b x d x = b a 2 + b 2 ( a > 0 ) {\displaystyle \int \limits _{0}^{\infty }e^{-ax}\sin bx\,\mathrm {d} x={\frac {b}{a^{2}+b^{2}}}\qquad (a>0)}
0 e a x cos b x d x = a a 2 + b 2 ( a > 0 ) {\displaystyle \int \limits _{0}^{\infty }e^{-ax}\cos bx\,\mathrm {d} x={\frac {a}{a^{2}+b^{2}}}\qquad (a>0)}
0 x e a x sin b x d x = 2 a b ( a 2 + b 2 ) 2 ( a > 0 ) {\displaystyle \int \limits _{0}^{\infty }xe^{-ax}\sin bx\,\mathrm {d} x={\frac {2ab}{(a^{2}+b^{2})^{2}}}\qquad (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 \limits _{0}^{\infty }xe^{-ax}\cos bx\,\mathrm {d} x={\frac {a^{2}-b^{2}}{(a^{2}+b^{2})^{2}}}\qquad (a>0)}
0 2 π e x cos θ d θ = 2 π I 0 ( x ) {\displaystyle \int \limits _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)} ( I 0 {\displaystyle I_{0}} az elsőfajú nulladrendű módosított Bessel-függvény)
0 2 π e x cos θ + y sin θ d θ = 2 π I 0 ( x 2 + y 2 ) {\displaystyle \int \limits _{0}^{2\pi }e^{x\cos \theta +y\sin \theta }d\theta =2\pi I_{0}\left({\sqrt {x^{2}+y^{2}}}\right)}

Fordítás

Ez a szócikk részben vagy egészben a List of integrals of exponential functions című angol Wikipédia-szócikk ezen változatának fordításán alapul. Az eredeti cikk szerkesztőit annak laptörténete sorolja fel. Ez a jelzés csupán a megfogalmazás eredetét és a szerzői jogokat jelzi, nem szolgál a cikkben szereplő információk forrásmegjelöléseként.

További információk

  • V. H. Moll, The Integrals in Gradshteyn and Ryzhik