Матеріал з Вікіпедії — вільної енциклопедії.
Визначені інтеграли без явних первісних
Деякі функції, чиї первісні не можуть бути представлені явно, тим не менш їхні деякі визначені інтеграли можуть бути обчислені.
∫
0
∞
x
n
−
1
e
−
x
d
x
=
∫
0
1
(
ln
1
x
)
n
−
1
d
x
=
Γ
(
n
)
=
(
n
−
1
)
!
{\displaystyle \int _{0}^{\infty }{x^{n-1}\,e^{-x}\,dx}=\int _{0}^{1}{\left(\ln {\frac {1}{x}}\right)^{n-1}\,dx}=\Gamma (n)=(n-1)!}
Γ
(
x
)
{\displaystyle \Gamma (x)\!}
Гамма-функція )
∫
0
∞
x
e
−
x
d
x
=
Γ
(
3
2
)
=
π
2
{\displaystyle \int _{0}^{\infty }{{\sqrt {x}}\,e^{-x}\,dx}=\Gamma \left({\frac {3}{2}}\right)={\frac {\sqrt {\pi }}{2}}}
∫
0
∞
x
n
e
−
a
2
x
2
d
x
=
Γ
(
n
+
1
2
)
2
a
n
+
1
{\displaystyle \int _{0}^{\infty }{x^{n}e^{-a^{2}x^{2}}\,dx}={\frac {\Gamma ({\frac {n+1}{2}})}{2a^{n+1}}}}
a
>
0
{\displaystyle a>0\!}
Гамма-функція )
∫
−
∞
∞
(
1
+
x
2
/
ν
)
−
(
ν
+
1
)
/
2
d
x
=
ν
π
Γ
(
ν
/
2
)
Γ
(
(
ν
+
1
)
/
2
)
{\displaystyle \int _{-\infty }^{\infty }{(1+x^{2}/\nu )^{-(\nu +1)/2}dx}={\frac {{\sqrt {\nu \pi }}\ \Gamma (\nu /2)}{\Gamma ((\nu +1)/2)}}\,}
ν
>
0
{\displaystyle \nu >0\,}
густини ймовірності для T-розподілу Стьюдента
∫
0
1
x
m
−
1
(
1
−
x
)
n
−
1
d
x
=
Γ
(
m
)
Γ
(
n
)
Γ
(
m
+
n
)
=
B
(
m
,
n
)
{\displaystyle \int _{0}^{1}{x^{m-1}\,(1-x)^{n-1}\,dx}={\frac {\Gamma (m)\,\Gamma (n)}{\Gamma (m+n)}}=B(m,n)}
m
,
n
>
0
{\displaystyle m,n\,>\,0\!}
B
(
m
,
n
)
{\displaystyle B(m,n)\!}
Бета-функція
∫
0
a
x
m
−
1
(
a
−
x
)
n
−
1
d
x
=
a
m
+
n
−
1
Γ
(
m
)
Γ
(
n
)
Γ
(
m
+
n
)
=
a
m
+
n
−
1
B
(
m
,
n
)
{\displaystyle \int _{0}^{a}{x^{m-1}\,(a-x)^{n-1}\,dx}=a^{m+n-1}\,{\frac {\Gamma (m)\Gamma (n)}{\Gamma (m+n)}}=a^{m+n-1}\,B(m,n)}
a
,
m
,
n
>
0
{\displaystyle a,m,n\,>\,0\!}
B
(
m
,
n
)
{\displaystyle B(m,n)\!}
∫
0
1
x
m
−
1
+
x
n
−
1
(
1
+
x
)
m
+
n
d
x
=
Γ
(
m
)
Γ
(
n
)
Γ
(
m
+
n
)
=
B
(
m
,
n
)
{\displaystyle \int _{0}^{1}{{\frac {x^{m-1}+x^{n-1}}{(1+x)^{m+n}}}\,dx}={\frac {\Gamma (m)\,\Gamma (n)}{\Gamma (m+n)}}=B(m,n)}
m
,
n
>
0
{\displaystyle m,n\,>\,0\!}
∫
0
1
x
m
1
−
x
2
d
x
{\displaystyle \int _{0}^{1}{{\frac {x^{m}}{\sqrt {1-x^{2}}}}\,dx}}
=
π
2
Γ
(
m
+
1
2
)
Γ
(
m
+
2
2
)
{\displaystyle ={\frac {\sqrt {\pi }}{2}}\,{\frac {\Gamma \left({\frac {m+1}{2}}\right)}{\Gamma \left({\frac {m+2}{2}}\right)}}}
m
>
−
1
{\displaystyle m\,>\,-1\!}
m
{\displaystyle m\!}
=
2
⋅
4
⋅
6
⋅
.
.
.
⋅
(
m
−
1
)
3
⋅
5
⋅
7
⋅
.
.
.
⋅
m
{\displaystyle ={\frac {2\cdot 4\cdot 6\cdot ...\cdot (m-1)}{3\cdot 5\cdot 7\cdot ...\cdot m}}}
m
>
1
{\displaystyle m\,>\,1\!}
m
{\displaystyle m\!}
непарне число ;
=
1
⋅
3
⋅
5
⋅
.
.
.
⋅
(
m
−
1
)
2
⋅
4
⋅
6
⋅
.
.
.
⋅
m
π
2
{\displaystyle ={\frac {1\cdot 3\cdot 5\cdot ...\cdot (m-1)}{2\cdot 4\cdot 6\cdot ...\cdot m}}\,{\frac {\pi }{2}}}
m
>
0
{\displaystyle m\,>\,0\!}
m
{\displaystyle m\!}
парне число ;
∫
0
1
x
m
+
1
−
x
2
d
x
=
1
m
+
2
∫
0
1
x
m
1
−
x
2
d
x
{\displaystyle \int _{0}^{1}{x^{m}+{\sqrt {1-x^{2}}}\,dx}={\frac {1}{m+2}}\,\int _{0}^{1}{{\frac {x^{m}}{\sqrt {1-x^{2}}}}\,dx}}
m
{\displaystyle m\!}
m
>
−
1
{\displaystyle m\,>\,-1\!}
∫
0
1
d
x
1
−
x
n
=
π
n
Γ
(
1
n
)
Γ
(
1
n
+
1
2
)
{\displaystyle \int _{0}^{1}{\frac {dx}{\sqrt {1-x^{n}}}}={\frac {\sqrt {\pi }}{n}}\,{\frac {\Gamma \left({\frac {1}{n}}\right)}{\Gamma \left({\frac {1}{n}}+{\frac {1}{2}}\right)}}}
n
>
0
{\displaystyle n\,>\,0\!}
∫
0
1
x
m
d
x
1
−
x
n
=
π
n
Γ
(
m
+
1
n
)
Γ
(
m
+
1
n
+
1
2
)
{\displaystyle \int _{0}^{1}{\frac {x^{m}\,dx}{\sqrt {1-x^{n}}}}={\frac {\sqrt {\pi }}{n}}\,{\frac {\Gamma \left({\frac {m+1}{n}}\right)}{\Gamma \left({\frac {m+1}{n}}+{\frac {1}{2}}\right)}}}
m
+
1
,
n
>
0
{\displaystyle m+1,n\,>\,0\!}
∫
0
1
x
m
(
1
−
x
2
)
p
d
x
=
Γ
(
p
+
1
)
Γ
(
m
+
1
2
)
2
Γ
(
p
+
m
+
3
2
)
{\displaystyle \int _{0}^{1}{x^{m}\,(1-x^{2})^{p}\,dx}={\frac {\Gamma (p+1)\Gamma \left({\frac {m+1}{2}}\right)}{2\Gamma \left(p+{\frac {m+3}{2}}\right)}}}
p
+
1
,
m
+
1
>
0
{\displaystyle p+1,m+1\,>\,0\!}
∫
0
1
x
m
(
1
−
x
n
)
p
d
x
=
Γ
(
p
+
1
)
Γ
(
m
+
1
n
)
n
Γ
(
p
+
1
+
m
+
1
n
)
{\displaystyle \int _{0}^{1}{x^{m}\,(1-x^{n})^{p}\,dx}={\frac {\Gamma (p+1)\Gamma \left({\frac {m+1}{n}}\right)}{n\Gamma \left(p+1+{\frac {m+1}{n}}\right)}}}
p
+
1
,
m
+
1
,
n
>
0
{\displaystyle p+1,m+1,n\,>\,0\!}
∫
0
1
x
n
d
x
1
+
x
=
(
−
1
)
n
[
ln
2
−
1
+
1
2
−
1
3
+
.
.
.
+
(
−
1
)
n
n
]
{\displaystyle \int _{0}^{1}{\frac {x^{n}\,dx}{1+x}}=(-1)^{n}\left[\ln {2}-1+{\frac {1}{2}}-{\frac {1}{3}}+...+{\frac {(-1)^{n}}{n}}\right]}
n
=
1
,
2
,
3
,
.
.
.
{\displaystyle n\,=\,1,2,3,...\!}
∫
0
1
x
n
−
1
d
x
1
+
x
m
=
1
n
−
1
n
+
m
+
1
n
+
2
m
−
1
n
+
3
m
+
.
.
.
{\displaystyle \int _{0}^{1}{\frac {x^{n-1}\,dx}{1+x^{m}}}={\frac {1}{n}}-{\frac {1}{n+m}}+{\frac {1}{n+2m}}-{\frac {1}{n+3m}}+...}
m
,
n
>
0
{\displaystyle m,n\,>\,0\!}
∫
0
1
d
x
1
+
x
+
x
2
=
π
3
3
{\displaystyle \int _{0}^{1}{\frac {dx}{1+x+x^{2}}}={\frac {\pi }{3{\sqrt {3}}}}}
∫
0
1
d
x
1
−
x
+
x
2
=
2
π
3
3
{\displaystyle \int _{0}^{1}{\frac {dx}{1-x+x^{2}}}={\frac {2\pi }{3{\sqrt {3}}}}}
∫
0
1
x
p
−
1
d
x
(
1
−
x
)
p
=
π
sin
p
π
{\displaystyle \int _{0}^{1}{\frac {x^{p-1}\,dx}{(1-x)^{p}}}={\frac {\pi }{\sin {p\pi }}}}
0
<
p
<
1
{\displaystyle 0\,<\,p\,<\,1\!}
∫
0
1
x
p
+
x
−
p
1
+
x
2
d
x
=
π
2
cos
p
π
2
{\displaystyle \int _{0}^{1}{{\frac {x^{p}+x^{-p}}{1+x^{2}}}\,dx}={\frac {\pi }{2\cos {\frac {p\pi }{2}}}}}
−
1
<
p
<
1
{\displaystyle -1\,<\,p\,<\,1\!}
∫
0
∞
e
−
x
2
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
Гаусовий інтеграл )
∫
0
∞
x
e
−
a
2
x
2
d
x
=
1
2
a
2
{\displaystyle \int _{0}^{\infty }{xe^{-a^{2}x^{2}}\,dx}={\frac {1}{2a^{2}}}}
∫
0
∞
x
2
e
−
a
2
x
2
d
x
=
π
4
a
3
{\displaystyle \int _{0}^{\infty }{x^{2}e^{-a^{2}x^{2}}\,dx}={\frac {\sqrt {\pi }}{4a^{3}}}}
a
>
0
{\displaystyle a>0\!}
∫
0
∞
x
2
i
+
1
e
−
a
2
x
2
d
x
=
a
!
2
a
2
i
+
2
{\displaystyle \int _{0}^{\infty }{x^{2i+1}e^{-a^{2}x^{2}}\,dx}={\frac {a!}{2a^{2i+2}}}}
a
>
0
;
i
=
1
,
2
,
3...
{\displaystyle a>0;\;\;\!i=1,2,3...}
∫
0
∞
x
2
i
+
1
e
−
a
2
x
2
d
x
=
1
⋅
3
⋅
5
⋅
.
.
.
⋅
(
2
i
−
1
)
2
i
+
1
a
2
i
+
1
π
{\displaystyle \int _{0}^{\infty }{x^{2i+1}e^{-a^{2}x^{2}}\,dx}={\frac {1\cdot 3\cdot 5\cdot ...\cdot (2i-1)}{2^{i+1}a^{2i+1}}}{\sqrt {\pi }}}
a
>
0
;
i
=
1
,
2
,
3...
{\displaystyle a>0;\;\;\!i=1,2,3...}
∫
0
∞
x
e
x
−
1
d
x
=
π
2
6
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{x}-1}}\,dx}={\frac {\pi ^{2}}{6}}}
числа Бернуллі )
∫
0
∞
x
3
e
x
−
1
d
x
=
π
4
15
{\displaystyle \int _{0}^{\infty }{{\frac {x^{3}}{e^{x}-1}}\,dx}={\frac {\pi ^{4}}{15}}}
∫
0
∞
1
e
a
x
+
1
d
x
=
ln
2
a
{\displaystyle \int _{0}^{\infty }{{\frac {1}{e^{ax}+1}}\,dx}={\frac {\ln {2}}{a}}}
a
>
0
{\displaystyle a>0\!}
∫
0
∞
x
e
a
x
+
1
d
x
=
π
2
12
a
2
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{ax}+1}}\,dx}={\frac {\pi ^{2}}{12a^{2}}}}
a
>
0
{\displaystyle a>0\!}
∫
0
∞
x
3
e
a
x
+
1
d
x
=
7
120
π
4
a
4
{\displaystyle \int _{0}^{\infty }{{\frac {x^{3}}{e^{ax}+1}}\,dx}={\frac {7}{120}}{\frac {\pi ^{4}}{a^{4}}}}
a
>
0
{\displaystyle a>0\!}
∫
0
∞
sin
(
x
)
x
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}}
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
1
⋅
3
⋅
5
⋅
⋯
⋅
(
n
−
1
)
2
⋅
4
⋅
6
⋅
⋯
⋅
n
π
2
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}{x}\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}{x}\,dx={\frac {1\cdot 3\cdot 5\cdot \cdots \cdot (n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot n}}{\frac {\pi }{2}}}
n парне число і
n
≥
2
{\displaystyle \scriptstyle {n\geq 2}}
∫
0
π
2
sin
n
x
d
x
=
∫
0
π
2
cos
n
x
d
x
=
2
⋅
4
⋅
6
⋅
⋯
⋅
(
n
−
1
)
3
⋅
5
⋅
7
⋅
⋯
⋅
n
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}{x}\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}{x}\,dx={\frac {2\cdot 4\cdot 6\cdot \cdots \cdot (n-1)}{3\cdot 5\cdot 7\cdot \cdots \cdot n}}}
n
{\displaystyle \scriptstyle {n}}
n
≥
3
{\displaystyle \scriptstyle {n\geq 3}}
∫
−
π
π
cos
(
α
x
)
cos
n
(
β
x
)
d
x
=
{
2
π
2
n
(
n
m
)
|
α
|
=
|
β
(
2
m
−
n
)
|
0
otherwise
{\displaystyle \int _{-\pi }^{\pi }\cos(\alpha x)\cos ^{n}(\beta x)dx=\left\{{\begin{array}{cc}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&|\alpha |=|\beta (2m-n)|\\0&{\mbox{otherwise}}\\\end{array}}\right.}
α
,
β
,
m
,
n
{\displaystyle \scriptstyle \alpha ,\beta ,m,n}
β
≠
0
{\displaystyle \scriptstyle \beta \neq 0}
m
,
n
≥
0
{\displaystyle \scriptstyle m,n\geq 0}
Біноміальний коефіцієнт )
∫
−
π
π
sin
(
α
x
)
cos
n
(
β
x
)
d
x
=
0
{\displaystyle \int _{-\pi }^{\pi }\sin(\alpha x)\cos ^{n}(\beta x)dx=0}
α
,
β
{\displaystyle \scriptstyle \alpha ,\beta }
n
{\displaystyle \scriptstyle n}
Симетрія )
∫
−
π
π
sin
(
α
x
)
sin
n
(
β
x
)
d
x
=
{
(
−
1
)
(
n
+
1
)
/
2
(
−
1
)
m
2
π
2
n
(
n
m
)
n
odd
,
α
=
β
(
2
m
−
n
)
0
otherwise
{\displaystyle \int _{-\pi }^{\pi }\sin(\alpha x)\sin ^{n}(\beta x)dx=\left\{{\begin{array}{cc}(-1)^{(n+1)/2}(-1)^{m}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&n{\mbox{ odd}},\ \alpha =\beta (2m-n)\\0&{\mbox{otherwise}}\\\end{array}}\right.}
α
,
β
,
m
,
n
{\displaystyle \scriptstyle \alpha ,\beta ,m,n}
β
≠
0
{\displaystyle \scriptstyle \beta \neq 0}
m
,
n
≥
0
{\displaystyle \scriptstyle m,n\geq 0}
Біноміальний коефіцієнт )
∫
−
π
π
cos
(
α
x
)
sin
n
(
β
x
)
d
x
=
{
(
−
1
)
n
/
2
(
−
1
)
m
2
π
2
n
(
n
m
)
n
even
,
|
α
|
=
|
β
(
2
m
−
n
)
|
0
otherwise
{\displaystyle \int _{-\pi }^{\pi }\cos(\alpha x)\sin ^{n}(\beta x)dx=\left\{{\begin{array}{cc}(-1)^{n/2}(-1)^{m}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&n{\mbox{ even}},\ |\alpha |=|\beta (2m-n)|\\0&{\mbox{otherwise}}\\\end{array}}\right.}
α
,
β
,
m
,
n
{\displaystyle \scriptstyle \alpha ,\beta ,m,n}
β
≠
0
{\displaystyle \scriptstyle \beta \neq 0\!}
m
,
n
≥
0
{\displaystyle \scriptstyle m,n\geq 0\!}
Біноміальний коефіцієнт )
∫
0
∞
sin
2
x
x
2
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}{x}}{x^{2}}}\,dx={\frac {\pi }{2}}}
∫
−
∞
∞
e
−
(
a
x
2
+
b
x
+
c
)
d
x
=
π
a
exp
[
b
2
−
4
a
c
4
a
]
{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\frac {\pi }{a}}}\exp \left[{\frac {b^{2}-4ac}{4a}}\right]}
exp
[
u
]
{\displaystyle \exp[u]\!}
експонента
e
u
{\displaystyle e^{u}}
a
>
0
{\displaystyle a>0\!}
∫
0
2
π
e
x
cos
θ
d
θ
=
2
π
I
0
(
x
)
{\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)}
I
0
(
x
)
{\displaystyle I_{0}(x)\!}
Функція Бесселя першого роду)
∫
0
2
π
e
x
cos
θ
+
y
sin
θ
d
θ
=
2
π
I
0
(
x
2
+
y
2
)
{\displaystyle \int _{0}^{2\pi }e^{x\cos \theta +y\sin \theta }d\theta =2\pi I_{0}\left({\sqrt {x^{2}+y^{2}}}\right)}
∫
0
π
2
ln
sin
x
d
x
=
−
π
2
ln
2
{\displaystyle \int _{0}^{\frac {\pi }{2}}\ln \sin x\,dx=-\,{\frac {\pi }{2}}\ln 2}
Для загального випадку, якщо первісної не існує, застосовується метод вичерпання :
∫
a
b
f
(
x
)
d
x
=
(
b
−
a
)
∑
n
=
1
∞
∑
m
=
1
2
n
−
1
(
−
1
)
m
+
1
2
−
n
f
(
a
+
m
(
b
−
a
)
2
−
n
)
.
{\displaystyle \int _{a}^{b}{f(x)\,dx}=(b-a)\sum \limits _{n=1}^{\infty }{\sum \limits _{m=1}^{2^{n}-1}{\left({-1}\right)^{m+1}}}2^{-n}f(a+m\left({b-a}\right)2^{-n}).}
∫
0
1
[
ln
(
1
/
x
)
]
p
d
x
=
p
!
{\displaystyle \int _{0}^{1}[\ln(1/x)]^{p}\,dx=p!}
![{\displaystyle \int _{0}^{1}{\frac {x^{n}\,dx}{1+x}}=(-1)^{n}\left[\ln {2}-1+{\frac {1}{2}}-{\frac {1}{3}}+...+{\frac {(-1)^{n}}{n}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/842fdbcf7e5cf55a92275ab578df0b4712065752)
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![{\displaystyle \int _{0}^{1}[\ln(1/x)]^{p}\,dx=p!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e5028566f81ea22ae449458f859edfc08a532cd)