Во следниве изрази,
ϕ
(
x
)
=
1
2
π
e
−
1
2
x
2
{\displaystyle \phi (x)={\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}x^{2}}}
е нормално распределена функција на густината на веројатноста,
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+\operatorname {erf} \left({\frac {x}{\sqrt {2}}}\right)\right)}
е соодветната функција на кумулативната распределба на веројатноста (каде erf е функцијата на грешка ) и
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}
е Овенова Т функција .
Овен[nb 1]
Неопределени интеграли
уреди
∫
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}
[nb 2]
∫
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)\sum _{j=0}^{k}{\frac {(2k+1)!!}{(2j+1)!!}}x^{2j+1}+(2k+1)!!\,\Phi (x)+C}
Во овие интеграли, n !! е двоен факториел : за парни n еднаков е на производот од сите парни броеви од 2 до n , а за непарни n еднаков е на производот на сите непарни броеви од 1 to n ; дополнително претпоставено е дека 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}}}}
[nb 3]
∫
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)\right)+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)\right)+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
+
b
x
)
)
+
C
{\displaystyle \int \Phi (a+bx)\,dx={\frac {1}{b}}\left((a+bx)\Phi (a+bx)+\phi (a+bx)\right)+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)\right)+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
(
b
2
x
n
−
c
b
n
)
+
C
,
b
≠
0
,
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}
Определени интеграли
уреди
∫
−
∞
∞
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
|
a
|
(
π
2
+
arctan
(
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 }}}}\left(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)}
∫
0
∞
x
ϕ
(
x
)
2
Φ
(
x
)
d
x
=
1
4
π
3
{\displaystyle \int _{0}^{\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)}
[nb 4]
∫
−
∞
∞
Φ
(
b
x
)
2
ϕ
(
x
)
d
x
=
1
π
arctan
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}}}}\right)}
∫
−
∞
∞
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). Handbook of the normal distribution (2. изд.). CRC Press. ISBN 0-8247-9342-0 "A table of normal integrals", стр. 389–419. 
