I uttrycken i denna artikel,
ϕ ( x ) =
1
2 π
e
−
1 2
x
2
{\displaystyle \phi (x)={\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{ 2}}x^{2}}}
är standardfunktionen för normal sannolikhetstäthet,
Φ ( x ) =
∫
− ∞
x
ϕ ( t ) d t =
1 2
(
1 + erf
(
x
2
)
)
{\displaystyle \Phi (x)=\int _{-\infty }^{x}\ phi (t)\,dt={\frac {1}{2}}\left(1+\operatörsnamn {erf} \left({\frac {x}{\sqrt {2}}}\right)\right )}
är motsvarande kumulativa fördelningsfunktion (där erf är felfunktionen ) och
T ( h , a ) = ϕ ( h )
0
∫
a
ϕ ( h x )
1 +
x
2
d x
{\displaystyle T(h,a)=\phi (h)\int _{0}^{a}{ \frac {\phi (hx)}{1+x^{2}}}\,dx}
är Owens T-funktion .
Owen har en omfattande lista över integraler av Gauss-typ; endast en delmängd anges nedan.
Obestämda integraler
∫ ϕ ( x ) d x = Φ ( x ) + C
{\displaystyle \int \phi (x)\,dx=\Phi (x)+C}
∫ x ϕ ( x ) d x = − ϕ ( x ) + C
{\displaystyle \int x\phi (x)\,dx=-\phi (x)+C}
∫
x
2
ϕ ( x ) d x = Φ ( x ) −
x ϕ ( x ) + C
{\displaystyle \int x^{2}\phi (x)\,dx=\Phi (x)-x\phi (x)+C}
∫
x
2 k + 1
ϕ ( x ) d x = − ϕ ( x )
∑
j =
0
k
( 2 k ) ! !
( 2 j ) ! !
x
2 j
+ C
{\displaystyle \int x^{2k+1}\phi (x)\,dx=-\phi (x)\sum _{j=0}^{k}{\frac {(2k)!!}{ (2j)!!}}x^{2j}+C}
∫
x
2 k + 2
ϕ ( x ) d x = − ϕ ( x )
∑
j =
0
k
( 2 k + 1 ) ! !
( 2 j + 1 ) ! !
x
2 j + 1
+ (
2 k + 1 ) ! ! Φ ( x ) + C
{\displaystyle \int x^{2k+2}\phi (x)\,dx=-\phi (x)\summa _{j=0}^{k}{\frac {( 2k+1)!!}{(2j+1)!!}}x^{2j+1}+(2k+1)!!\,\Phi (x)+C}
I dessa integraler, n !! är dubbelfaktorial : för jämnt n är det lika med produkten av alla jämna tal från 2 till n , och för udda n är det produkten av alla udda tal från 1 till n ; dessutom antas det att 0!! = (−1)!! = 1 .
∫ ϕ ( x
)
2
d x =
1
2
π
Φ
(
x
2
)
+ C
{\displaystyle \int \phi (x)^{2}\,dx={\frac {1}{2{\sqrt {\ pi }}}}\Phi \left(x{\sqrt {2}}\right)+C}
∫ ϕ ( x ) ϕ ( a + b x ) d x =
1 t
ϕ
(
a t
)
Φ
(
t x )
+
a b
t
)
+ C , t =
1 +
b
2
{\displaystyle \int \phi (x)\phi (a+bx)\,dx={\frac {1}{t}}\phi \left( {\frac {a}{t}}\right)\Phi \left(tx+{\frac {ab}{t}}\right)+C,\qquad t={\sqrt {1+b^{2} }}}
∫ x ϕ ( a + b x ) d x = −
1
b
2
(
ϕ ( a + b x ) + a
Φ ( a + b x )
)
+ C
{\displaystyle \int x\phi (a+bx)\,dx=-{\frac {1}{b^{2}}}\left(\phi (a+ bx)+a\Phi (a+bx)\höger)+C}
∫
x
2
ϕ ( a + b x ) d x =
1
b
3
(
(
a
2
+ 1 ) Φ ( a + b x ) + ( a − b
x ) ϕ ( a + b x )
)
+ C
{\displaystyle \int x^{2}\phi (a+bx)\,dx={\frac {1}{b^{3}}}\left( (a^{2}+1)\Phi (a+bx)+(a-bx)\phi (a+bx)\höger)+C}
∫ ϕ ( a + b x
)
n
d x =
1
b
n ( 2 π
)
n − 1
Φ
(
n
( a + b x )
)
+ C
{\displaystyle \int \phi (a+bx)^{n}\,dx={\frac {1}{b{\sqrt {n(2\pi )^{n-1}}}}} \Phi \left({\sqrt {n}}(a+bx)\right)+C}
∫ Φ ( a + b x ) d x =
1 b
(
_
( a + b x ) Φ ( a + b x ) + ϕ ( a + bx )
)
+ C
{\displaystyle \int \Phi (a+bx)\,dx={\frac {1}{b}}\left((a+bx)\Phi (a+bx)+\phi (a+bx)\ höger)+C}
∫ x Φ ( a + b x ) d x =
1
2
b
2
(
(
b
2
x
2
−
a
2
− 1 ) Φ ( a + b x ) + ( b x − a ) ϕ ( a
+ b x )
)
+ C
{\displaystyle \int x\Phi (a+bx)\,dx={\frac {1}{2b^{2}}}\left((b^{2}x^{ 2}-a^{2}-1)\Phi (a+bx)+(bx-a)\phi (a+bx)\höger)+C}
∫
x
2
Φ ( a + b x ) d x =
1
3
b
3
(
(
b
3
x
3
+
a
3
+ 3 a ) Φ ( a + b x
) + (
b
2
x
2
− a b x +
a
2
+ 2 ) ϕ ( a + b x )
)
+ C
{\displaystyle \int x^{2}\Phi (a+bx)\,dx={\ frac {1}{3b^{3}}}\left((b^{3}x^{3}+a^{3}+3a)\Phi (a+bx)+(b^{2}x ^{2}-abx+a^{2}+2)\phi (a+bx)\right)+C}
∫
x
n
Φ ( x ) d x =
1
n + 1
(
(
x
n + 1
− n
x
n − 1
)
Φ ( x ) +
x
n
ϕ ( x ) + n ( n − 1 ) ∫
x
n − 2
Φ ( x ) d x
)
+ C
{\displaystyle \int x^{n}\Phi (x)\,dx={\frac {1}{n+1}}\left(\left(x^{n+1}-nx^{n- 1}\right)\Phi (x)+x^{n}\phi (x)+n(n-1)\int x^{n-2}\Phi (x)\,dx\right)+C }
∫ x ϕ ( x ) Φ ( a + b x ) d x =
b t
ϕ
(
a t
)
Φ
(
x t +
a b
t
)
− ϕ ( x ) Φ ( a +
b x ) + C , t =
1 +
b
2
{\displaystyle \int x\phi (x)\Phi (a+bx)\,dx={\frac {b}{t}}\phi \left({ \frac {a}{t}}\right)\Phi \left(xt+{\frac {ab}{t}}\right)-\phi (x)\Phi (a+bx)+C,\qquad t ={\sqrt {1+b^{2}}}}
∫ Φ ( x
)
2
d x = x Φ ( x
)
2
+ 2 Φ ( x ) ϕ ( x ) −
1
π
Φ
(
x
2
)
+ C
{\displaystyle \int \Phi (x)^{2}\,dx=x\Phi (x)^{2}+2\Phi (x)\phi (x)-{ \frac {1}{\sqrt {\pi }}}\Phi \left(x{\sqrt {2}}\right)+C}
∫
e
c x
ϕ ( b x
)
n
d x =
e
c
2
2 n
b
2
b
n ( 2 π
)
n − 1
Φ
0
(
b
2
x n
− c
b
n
)
+ C , b ≠ , n >
0
{\displaystyle \int e^{cx}\phi (bx)^{n}\,dx={\frac {e^{\frac {c^{2 }}{2nb^{2}}}}{b{\sqrt {n(2\pi )^{n-1}}}}}\Phi \left({\frac {b^{2}xn-c }{b{\sqrt {n}}}}\right)+C,\qquad b\neq 0,n>0}
Bestämda integraler
∫
− ∞
∞
x
2
ϕ ( x
)
n
d x =
1
n
3
( 2 π
)
n − 1
{\displaystyle \int _{-\infty }^{\infty }x^{2}\phi (x) ^{n}\,dx={\frac {1}{\sqrt {n^{3}(2\pi )^{n-1}}}}}
∫
− ∞
0
ϕ ( a x ) Φ ( b x ) d x =
1
2 π
|
a
|
(
π 2
− arctan
(
b
|
a
|
)
)
{\displaystyle \int _{-\infty }^{0}\phi (ax)\Phi (bx)dx={\frac {1}{2\pi |a|}}\left({\frac {\pi }{2}}-\arctan \left({\frac {b}{|a|}}\right)\right)}
0
∫
∞
ϕ ( a x ) Φ ( b x ) d x =
1
2 π
|
en
|
(
π 2
+ arktan
(
b
|
a
|
)
)
{\displaystyle \int _{0}^{\infty }\phi (ax)\Phi (bx)\,dx={\frac {1}{2\pi |a|} }\left({\frac {\pi }{2}}+\arctan \left({\frac {b}{|a|}}\right)\right)}
0
∫
∞
x ϕ ( x ) Φ ( b x ) d x =
1
2
2 π
(
1 +
b
1 +
b
2
)
{\displaystyle \int _{0}^{\infty }x\phi (x)\Phi (bx)\,dx={\frac {1}{2{\sqrt {2\pi }}}}\vänster (1+{\frac {b}{\sqrt {1+b^{2}}}}\right)}
0
∫
∞
x
2
ϕ ( x ) Φ ( b x ) d x =
1 4
+
1
2 π
(
b
1 +
b
2
+ arctan ( b )
)
{\displaystyle \int _{0}^{\infty }x^{2}\phi (x)\Phi (bx)\,dx={\frac {1}{4}}+{\frac {1} {2\pi }}\left({\frac {b}{1+b^{2}}}+\arctan(b)\right)}
∫
− ∞
∞
x ϕ ( x
)
2
Φ ( x ) d x =
1
4 π
3
{\displaystyle \int _{-\infty }^{\infty }x\phi (x)^{2}\Phi (x)\,dx={\frac {1}{4\ pi {\sqrt {3}}}}}
0
∫
∞
Φ ( b x
)
2
ϕ ( x )
d x =
1
2 π
(
arctan ( b ) + arctan
1 + 2
b
2
)
{\displaystyle \int _{0}^{\infty }\Phi (bx)^{2}\phi (x)\ ,dx={\frac {1}{2\pi }}\left(\arctan(b)+\arctan {\sqrt {1+2b^{2}}}\right)}
∫
− ∞
∞
Φ ( a + b x
)
2
ϕ ( x ) d x = Φ
(
a
1 +
b
2
)
− 2 T
(
a
1 +
b
2
,
1
1 + 2
b
2
)
{\displaystyle \int _{-\infty }^{\infty }\Phi (a+bx)^{2} \phi (x)\,dx=\Phi \left({\frac {a}{\sqrt {1+b^{2}}}}\right)-2T\left({\frac {a}{\ sqrt {1+b^{2}}}},{\frac {1}{\sqrt {1+2b^{2}}}}\right)}
∫
− ∞
∞
x Φ ( a + b x
)
2
ϕ (
x ) d x =
2 b
1 +
b
2
ϕ
(
a t
)
Φ
(
a
1 +
b
2
1 + 2
b
2
)
{\displaystyle \int _{-\infty }^{\infty }x\Phi (a +bx)^{2}\phi (x)\,dx={\frac {2b}{\sqrt {1+b^{2}}}}\phi \left({\frac {a}{t} }\right)\Phi \left({\frac {a}{{\sqrt {1+b^{2}}}{\sqrt {1+2b^{2}}}}}\right)}
∫
− ∞
∞
Φ ( b x
)
2
ϕ ( x ) d x =
1 π
arktan
1 + 2
b
2
{\displaystyle \int _{-\infty }^{\infty }\Phi (bx)^{2}\phi (x)\, dx={\frac {1}{\pi }}\arctan {\sqrt {1+2b^{2}}}}
∫
− ∞
∞
x ϕ ( x ) Φ ( b x ) d x =
∫
− ∞
∞
x ϕ ( x )
Φ ( b x
)
2
d x =
b
2 π ( 1 +
b
2
)
{\displaystyle \int _{-\infty }^{\infty }x\phi (x)\Phi (bx)\,dx=\ int _{-\infty }^{\infty }x\phi (x)\Phi (bx)^{2}\,dx={\frac {b}{\sqrt {2\pi (1+b^{ 2})}}}}
∫
− ∞
∞
Φ ( a + b x ) ϕ ( x ) d x = Φ
(
a
1 +
b
2
)
{\displaystyle \int _{-\infty }^{\infty }\Phi (a+bx)\phi (x)\,dx=\Phi \left({\frac {a}{\sqrt { 1+b^{2}}}}\höger)}
∫
− ∞
∞
x Φ ( a + b x ) ϕ ( x ) d x =
b t
ϕ
(
a t
)
, t =
1 +
b
2
{\displaystyle \int _{-\infty }^{\infty }x\Phi (a+bx)\phi (x)\,dx={\frac {b}{t}}\phi \left({\ frac {a}{t}}\right),\qquad t={\sqrt {1+b^{2}}}}
0
∫
∞
x Φ ( a + b x ) ϕ ( x ) d x =
b t
ϕ
(
a t
)
Φ
(
−
a b
t
)
+
1
2 π
Φ ( a ) , t =
1 +
b
2
{\displaystyle \int _{0}^{\infty }x\Phi (a+bx)\phi (x)\,dx={\frac {b}{t}}\phi \left( {\frac {a}{t}}\right)\Phi \left(-{\frac {ab}{t}}\right)+{\frac {1}{\sqrt {2\pi }}}\ Phi (a),\qquad t={\sqrt {1+b^{2}}}}
∫
− ∞
∞
ln (
x
2
)
1 σ
ϕ
(
x σ
)
d x = ln (
σ
2
) − γ − ln 2 ≈
ln (
σ
2
) − 1.27036
{\displaystyle \int _{-\infty }^{\infty }\ln(x^{2}){\frac {1}{\sigma }}\phi \left({ \frac {x}{\sigma }}\right)\,dx=\ln(\sigma ^{2})-\gamma -\ln 2\approx \ln(\sigma ^{2})-1.27036}
Patel , Jagdish K.; Read, Campbell B. (1996). Handbok för normalfördelningen (2:a uppl.). CRC Tryck. ISBN 0-8247-9342-0 .
Owen, D. (1980). "En tabell över normala integraler". Kommunikation i statistik: Simulering och beräkning . B9 (4): 389–419. doi : 10.1080/03610918008812164 .