Határozatlan integrálok
szerkesztés
∫
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
szerkesztés
∫
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)}
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)}
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)}
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}}}
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}}
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)}
Ez a szócikk részben vagy egészben a List of integrals of exponential functions ezen változatának fordításán alapul.
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c3194f4bfa4fc6c3217b6bd5d45d431f6354fb4)
