در این عبارت‌ها،
ϕ
(
x
)
=
1
2
π
e
−
1
2
x
2
{\displaystyle \phi (x)={\tfrac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}x^{2}}}
توزیع نرمال تابع چگالی احتمال است و
Φ
(
x
)
=
∫
−
∞
x
ϕ
(
t
)
d
t
=
1
2
(
1
+
erf
(
x
2
)
)
{\displaystyle \textstyle \Phi (x)=\int _{-\infty }^{x}\phi (t)dt={\frac {1}{2}}{\big (}1+\operatorname {erf} {\big (}{\frac {x}{\sqrt {2}}}{\big )}{\big )}}
تابع توزیع تجمعی مربوط به آن است که در آن erf ، تابع خطا و
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}
در پایین چند نمونه انتگرال آمده است:
انتگرال‌های نامعین [ ویرایش ]
∫
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 ۱]
∫
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}
(در این انتگرال‌ها، n !!n های زوج برابر است با حاصل ضرب 2 تا n و برای n های فرد برابر است با حاصل ضرب 1 تا n همچنین فرض می‌شود که 0!! = (−1)!! = 1
∫
ϕ
(
x
)
2
d
x
=
1
2
π
Φ
(
x
2
)
+
C
{\displaystyle \int \phi (x)^{2}\,dx={\tfrac {1}{2{\sqrt {\pi }}}}\Phi (x{\sqrt {2}})+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={\tfrac {1}{t}}\phi (a/t)\Phi (tx+ab/t)+C,\quad t={\sqrt {1+b^{2}}}}
[nb ۲]
∫
x
ϕ
(
a
+
b
x
)
d
x
=
−
1
b
2
ϕ
(
a
+
b
x
)
−
a
b
2
Φ
(
a
+
b
x
)
+
C
{\displaystyle \int x\phi (a+bx)\,dx=-{\tfrac {1}{b^{2}}}\phi (a+bx)-{\tfrac {a}{b^{2}}}\Phi (a+bx)+C}
∫
x
2
ϕ
(
a
+
b
x
)
d
x
=
a
2
+
1
b
3
Φ
(
a
+
b
x
)
+
a
−
b
x
b
3
ϕ
(
a
+
b
x
)
+
C
{\displaystyle \int x^{2}\phi (a+bx)\,dx={\tfrac {a^{2}+1}{b^{3}}}\Phi (a+bx)+{\frac {a-bx}{b^{3}}}\phi (a+bx)+C}
∫
ϕ
(
a
+
b
x
)
n
d
x
=
(
2
π
)
−
(
n
−
1
)
/
2
b
n
Φ
(
n
(
a
+
b
x
)
)
+
C
{\displaystyle \int \phi (a+bx)^{n}\,dx={\frac {(2\pi )^{-(n-1)/2}}{b{\sqrt {n}}}}\Phi {\big (}{\sqrt {n}}(a+bx){\big )}+C}
∫
Φ
(
a
+
b
x
)
d
x
=
1
b
(
a
+
b
x
)
Φ
(
a
+
b
x
)
+
1
b
ϕ
(
a
+
b
x
)
+
C
{\displaystyle \int \Phi (a+bx)\,dx={\tfrac {1}{b}}(a+bx)\Phi (a+bx)+{\tfrac {1}{b}}\phi (a+bx)+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={\tfrac {1}{2b^{2}}}{\big (}(b^{2}x^{2}-a^{2}-1)\Phi (a+bx)+(bx-a)\phi (a+bx){\big )}+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={\tfrac {1}{3b^{3}}}{\big (}(b^{3}x^{3}+a^{3}+3a)\Phi (a+bx)+(b^{2}x^{2}-abx+a^{2}+2)\phi (a+bx){\big )}+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={\tfrac {1}{n+1}}{\Big (}(x^{n+1}-nx^{n-1})\Phi (x)+x^{n}\phi (x)+n(n-1)\int x^{n-2}\Phi (x)\,dx{\Big )}+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={\tfrac {b}{t}}\phi (a/t)\Phi (xt+ab/t)-\phi (x)\Phi (a+bx)+C,\quad 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)-{\tfrac {1}{\sqrt {\pi }}}\Phi (x{\sqrt {2}})+C}
∫
e
c
x
ϕ
(
b
x
)
n
d
x
=
1
b
n
(
2
π
)
n
−
1
e
c
2
/
(
2
n
b
2
)
Φ
(
b
x
n
−
c
b
n
)
+
C
,
b
≠
0
,
n
>
0
{\displaystyle \int e^{cx}\phi (bx)^{n}\,dx={\frac {1}{b{\sqrt {n(2\pi )^{n-1}}}}}e^{c^{2}/(2nb^{2})}\Phi (bx{\sqrt {n}}-{\tfrac {c}{b{\sqrt {n}}}})+C,\quad b\neq 0,n>0}
انتگرال‌های معین [ ویرایش ]
∫
−
∞
∞
x
2
ϕ
(
x
)
n
d
x
=
(
n
3
/
2
(
2
π
)
(
n
−
1
)
/
2
)
−
1
{\displaystyle \int _{-\infty }^{\infty }x^{2}\phi (x)^{n}\,\,dx={\Big (}n^{3/2}(2\pi )^{(n-1)/2}{\Big )}^{-1}}
∫
−
∞
0
ϕ
(
a
x
)
Φ
(
b
x
)
d
x
=
(
2
π
a
)
−
1
arctan
(
a
/
b
)
{\displaystyle \int _{-\infty }^{0}\phi (ax)\Phi (bx)dx=(2\pi a)^{-1}\arctan(a/b)}
∫
0
∞
ϕ
(
a
x
)
Φ
(
b
x
)
d
x
=
(
2
π
a
)
−
1
(
π
2
−
arctan
(
b
/
a
)
)
{\displaystyle \int _{0}^{\infty }\phi (ax)\Phi (bx)\,dx=(2\pi a)^{-1}{\big (}{\tfrac {\pi }{2}}-\arctan(b/a){\big )}}
∫
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 }}}}{\bigg (}1+{\frac {b}{\sqrt {1+b^{2}}}}{\bigg )}}
∫
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 }}{\bigg (}{\frac {b}{1+b^{2}}}+\arctan b{\bigg )}}
∫
x
ϕ
(
x
)
2
Φ
(
x
)
d
x
=
1
4
π
3
{\displaystyle \int x\phi (x)^{2}\Phi (x)\,dx={\frac {1}{4\pi {\sqrt {3}}}}}
∫
0
∞
Φ
(
b
x
)
2
ϕ
(
x
)
d
x
=
(
2
π
)
−
1
(
arctan
b
+
arctan
1
+
2
b
2
)
{\displaystyle \int _{0}^{\infty }\Phi (bx)^{2}\phi (x)\,dx=(2\pi )^{-1}{\big (}\arctan b+\arctan {\sqrt {1+2b^{2}}}{\big )}}
∫
−
∞
∞
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 {\bigg (}{\frac {a}{\sqrt {1+b^{2}}}}{\bigg )}-2T{\bigg (}{\frac {a}{\sqrt {1+b^{2}}}},{\frac {1}{\sqrt {1+2b^{2}}}}{\bigg )}}
∫
−
∞
∞
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 (a/t)\Phi {\bigg (}{\frac {a}{{\sqrt {1+b^{2}}}{\sqrt {1+2b^{2}}}}}{\bigg )}}
[nb ۳]
∫
−
∞
∞
Φ
(
b
x
)
2
ϕ
(
x
)
d
x
=
π
−
1
arctan
1
+
2
b
2
{\displaystyle \int _{-\infty }^{\infty }\Phi (bx)^{2}\phi (x)\,dx=\pi ^{-1}\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 {\big (}a/{\sqrt {1+b^{2}}}{\big )}}
∫
−
∞
∞
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=(b/t)\phi (a/t),\quad t={\sqrt {1+b^{2}}}}
∫
0
∞
x
Φ
(
a
+
b
x
)
ϕ
(
x
)
d
x
=
(
b
/
t
)
ϕ
(
a
/
t
)
Φ
(
−
a
b
/
t
)
+
(
2
π
)
−
1
/
2
Φ
(
a
)
,
t
=
1
+
b
2
{\displaystyle \int _{0}^{\infty }x\Phi (a+bx)\phi (x)\,dx=(b/t)\phi (a/t)\Phi (-ab/t)+(2\pi )^{-1/2}\Phi (a),\quad 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}){\tfrac {1}{\sigma }}\phi {\big (}{\tfrac {x}{\sigma }}{\big )}\,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 (2nd ed.). CRC Press. ISBN 0-8247-9342-0 . Owen, D. (1980). "A table of normal integrals". Communications in Statistics: Simulation and Computation B9 . pp. 389 – 419.
خطای یادکرد: خطای یادکرد: برچسب <ref> برای گروهی به نام «nb» وجود دارد، اما برچسب <references group="nb"/> متناظر پیدا نشد.
