Во математиката, определениот интеграл :
∫
a
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)\,dx}
е површината од областа на xy -рамнината ограничена од графиконот на f , x -оската и линиите x = a и x = b , така што површината над x -оската се додава на вкупниот збир, а под x -оската се одзема од вкупната сума.
Основната теорема на анализата воспоставува однос меѓу неопределените и определените интеграли и воведува техника за пресметување на определените интеграли.
Ако интервалот е бесконечен, определениот интеграл се нарекува несвојствен интеграл
∫
a
∞
f
(
x
)
d
x
=
lim
b
→
∞
[
∫
a
b
f
(
x
)
d
x
]
{\displaystyle \int _{a}^{\infty }f(x)\,dx=\lim _{b\to \infty }\left[\int _{a}^{b}f(x)\,dx\right]}
Константа, како пи, која може да биде дефинирана со интеграла на алгебарска функција во алгебарски домен е позната како периода.
Следи списокот на најчестите определени интеграли. Списокот на неопределени интеграли е дадени во списокот на неопределени интеграли .
уреди
∫
0
∞
d
x
x
2
+
a
2
=
π
2
a
{\displaystyle \int _{0}^{\infty }{\frac {dx}{x^{2}+a^{2}}}={\frac {\pi }{2a}}}
∫
0
∞
x
m
d
x
x
n
+
a
n
=
π
a
m
−
n
+
1
n
sin
(
m
+
1
n
π
)
for
0
<
m
+
1
<
n
{\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{x^{n}+a^{n}}}={\frac {\pi a^{m-n+1}}{n\sin \left({\dfrac {m+1}{n}}\pi \right)}}\quad {\mbox{for }}0<m+1<n}
∫
0
∞
x
p
−
1
d
x
1
+
x
=
π
sin
(
p
π
)
for
0
<
p
<
1
{\displaystyle \int _{0}^{\infty }{\frac {x^{p-1}dx}{1+x}}={\frac {\pi }{\sin(p\pi )}}\quad {\mbox{for }}0<p<1}
∫
0
∞
x
m
d
x
1
+
2
x
cos
β
+
x
2
=
π
sin
(
m
π
)
⋅
sin
(
m
β
)
sin
(
β
)
{\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{1+2x\cos \beta +x^{2}}}={\frac {\pi }{\sin(m\pi )}}\cdot {\frac {\sin(m\beta )}{\sin(\beta )}}}
∫
0
a
d
x
a
2
−
x
2
=
π
2
{\displaystyle \int _{0}^{a}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}={\frac {\pi }{2}}}
∫
0
a
a
2
−
x
2
d
x
=
π
a
2
4
{\displaystyle \int _{0}^{a}{\sqrt {a^{2}-x^{2}}}dx={\frac {\pi a^{2}}{4}}}
∫
0
a
x
m
(
a
n
−
x
n
)
p
d
x
=
a
m
+
1
+
n
p
Γ
(
m
+
1
n
)
Γ
(
p
+
1
)
n
Γ
(
m
+
1
n
+
p
+
1
)
{\displaystyle \int _{0}^{a}x^{m}(a^{n}-x^{n})^{p}\,dx={\frac {a^{m+1+np}\Gamma \left({\dfrac {m+1}{n}}\right)\Gamma (p+1)}{n\Gamma \left({\dfrac {m+1}{n}}+p+1\right)}}}
∫
0
∞
x
m
d
x
(
x
n
+
a
n
)
r
=
(
−
1
)
r
−
1
π
a
m
+
1
−
n
r
Γ
(
m
+
1
n
)
n
sin
(
m
+
1
n
π
)
(
r
−
1
)
!
Γ
(
m
+
1
n
−
r
+
1
)
for
n
(
r
−
2
)
<
m
+
1
<
n
r
{\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{({x^{n}+a^{n})}^{r}}}={\frac {(-1)^{r-1}\pi a^{m+1-nr}\Gamma \left({\dfrac {m+1}{n}}\right)}{n\sin \left({\dfrac {m+1}{n}}\pi \right)(r-1)!\,\Gamma \left({\dfrac {m+1}{n}}-r+1\right)}}\quad {\mbox{for }}n(r-2)<m+1<nr}
∫
0
π
sin
(
m
x
)
sin
(
n
x
)
d
x
=
{
0
ако
m
≠
n
π
2
ако
m
=
n
за
m
,
n
позитивни цели броеви
{\displaystyle \int _{0}^{\pi }\sin(mx)\sin(nx)dx={\begin{cases}0&{\text{ако }}m\neq n\\\\{\dfrac {\pi }{2}}&{\text{ако }}m=n\end{cases}}\quad {\text{за }}m,n{\text{ позитивни цели броеви}}}
∫
0
π
cos
(
m
x
)
cos
(
n
x
)
d
x
=
{
0
if
m
≠
n
π
2
ако
m
=
n
за
m
,
n
позитивни цели броеви
{\displaystyle \int _{0}^{\pi }\cos(mx)\cos(nx)dx={\begin{cases}0&{\text{if }}m\neq n\\\\{\dfrac {\pi }{2}}&{\text{ако }}m=n\end{cases}}\quad {\text{за }}m,n{\text{ позитивни цели броеви}}}
∫
0
π
sin
(
m
x
)
cos
(
n
x
)
d
x
=
{
0
if
m
+
n
even
2
m
m
2
−
n
2
ако
m
+
n
непарен
за
m
,
n
цели броеви
.
{\displaystyle \int _{0}^{\pi }\sin(mx)\cos(nx)dx={\begin{cases}0&{\text{if }}m+n{\text{ even}}\\\\{\dfrac {2m}{m^{2}-n^{2}}}&{\text{ако }}m+n{\text{ непарен}}\end{cases}}\quad {\text{за }}m,n{\text{ цели броеви}}.}
∫
0
π
2
sin
2
(
x
)
d
x
=
∫
0
π
2
cos
2
(
x
)
d
x
=
π
4
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2}(x)dx=\int _{0}^{\frac {\pi }{2}}\cos ^{2}(x)dx={\frac {\pi }{4}}}
∫
0
π
2
sin
2
m
(
x
)
d
x
=
∫
0
π
2
cos
2
m
(
x
)
d
x
=
1
×
3
×
5
×
⋯
×
(
2
m
−
1
)
2
×
4
×
6
×
⋯
×
2
m
⋅
π
2
for
m
=
1
,
2
,
3
…
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m}(x)dx=\int _{0}^{\frac {\pi }{2}}\cos ^{2m}(x)dx={\frac {1\times 3\times 5\times \cdots \times (2m-1)}{2\times 4\times 6\times \cdots \times 2m}}\cdot {\frac {\pi }{2}}\quad {\mbox{for }}m=1,2,3\ldots }
∫
0
π
2
sin
2
m
+
1
(
x
)
d
x
=
∫
0
π
2
cos
2
m
+
1
(
x
)
d
x
=
2
×
4
×
6
×
⋯
×
2
m
1
×
3
×
5
×
⋯
×
(
2
m
+
1
)
for
m
=
1
,
2
,
3
…
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}(x)dx=\int _{0}^{\frac {\pi }{2}}\cos ^{2m+1}(x)dx={\frac {2\times 4\times 6\times \cdots \times 2m}{1\times 3\times 5\times \cdots \times (2m+1)}}\quad {\mbox{for }}m=1,2,3\ldots }
∫
0
π
2
sin
2
p
−
1
(
x
)
cos
2
q
−
1
(
x
)
d
x
=
Γ
(
p
)
Γ
(
q
)
2
Γ
(
p
+
q
)
=
1
2
B
(
p
,
q
)
{\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2p-1}(x)\cos ^{2q-1}(x)dx={\frac {\Gamma (p)\Gamma (q)}{2\Gamma (p+q)}}={\frac {1}{2}}{\text{B}}(p,q)}
∫
0
∞
sin
(
p
x
)
x
d
x
=
{
π
2
ако
p
>
0
0
ако
p
=
0
−
π
2
ако
p
<
0
{\displaystyle \int _{0}^{\infty }{\frac {\sin(px)}{x}}dx={\begin{cases}{\dfrac {\pi }{2}}&{\text{ако }}p>0\\\\0&{\text{ако }}p=0\\\\-{\dfrac {\pi }{2}}&{\text{ако }}p<0\end{cases}}}
∫
0
∞
sin
p
x
cos
q
x
x
d
x
=
{
0
ако
q
>
p
>
0
π
2
ако
0
<
q
<
p
π
4
ако
p
=
q
>
0
{\displaystyle \int _{0}^{\infty }{\frac {\sin px\cos qx}{x}}\ dx={\begin{cases}0&{\text{ ако }}q>p>0\\\\{\dfrac {\pi }{2}}&{\text{ ако }}0<q<p\\\\{\dfrac {\pi }{4}}&{\text{ ако }}p=q>0\end{cases}}}
∫
0
∞
sin
p
x
sin
q
x
x
2
d
x
=
{
π
p
2
ако
0
<
p
≤
q
π
q
2
ако
0
<
q
≤
p
{\displaystyle \int _{0}^{\infty }{\frac {\sin px\sin qx}{x^{2}}}\ dx={\begin{cases}{\dfrac {\pi p}{2}}&{\text{ ако }}0<p\leq q\\\\{\dfrac {\pi q}{2}}&{\text{ ако }}0<q\leq p\end{cases}}}
∫
0
∞
sin
2
p
x
x
2
d
x
=
π
p
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}px}{x^{2}}}\ dx={\frac {\pi p}{2}}}
∫
0
∞
1
−
cos
p
x
x
2
d
x
=
π
p
2
{\displaystyle \int _{0}^{\infty }{\frac {1-\cos px}{x^{2}}}\ dx={\frac {\pi p}{2}}}
∫
0
∞
cos
p
x
−
cos
q
x
x
d
x
=
ln
q
p
{\displaystyle \int _{0}^{\infty }{\frac {\cos px-\cos qx}{x}}\ dx=\ln {\frac {q}{p}}}
∫
0
∞
cos
p
x
−
cos
q
x
x
2
d
x
=
π
(
q
−
p
)
2
{\displaystyle \int _{0}^{\infty }{\frac {\cos px-\cos qx}{x^{2}}}\ dx={\frac {\pi (q-p)}{2}}}
∫
0
∞
cos
m
x
x
2
+
a
2
d
x
=
π
2
a
e
−
m
a
{\displaystyle \int _{0}^{\infty }{\frac {\cos mx}{x^{2}+a^{2}}}\ dx={\frac {\pi }{2a}}e^{-ma}}
∫
0
∞
x
sin
m
x
x
2
+
a
2
d
x
=
π
2
e
−
m
a
{\displaystyle \int _{0}^{\infty }{\frac {x\sin mx}{x^{2}+a^{2}}}\ dx={\frac {\pi }{2}}e^{-ma}}
∫
0
∞
sin
m
x
x
(
x
2
+
a
2
)
d
x
=
π
2
a
2
(
1
−
e
−
m
a
)
{\displaystyle \int _{0}^{\infty }{\frac {\sin mx}{x(x^{2}+a^{2})}}\ dx={\frac {\pi }{2a^{2}}}\left(1-e^{-ma}\right)}
∫
0
2
π
d
x
a
+
b
sin
x
=
2
π
a
2
−
b
2
{\displaystyle \int _{0}^{2\pi }{\frac {dx}{a+b\sin x}}={\frac {2\pi }{\sqrt {a^{2}-b^{2}}}}}
∫
0
2
π
d
x
a
+
b
cos
x
=
2
π
a
2
−
b
2
{\displaystyle \int _{0}^{2\pi }{\frac {dx}{a+b\cos x}}={\frac {2\pi }{\sqrt {a^{2}-b^{2}}}}}
∫
0
π
2
d
x
a
+
b
cos
x
=
cos
−
1
(
b
a
)
a
2
−
b
2
{\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {dx}{a+b\cos x}}={\frac {\cos ^{-1}\left({\dfrac {b}{a}}\right)}{\sqrt {a^{2}-b^{2}}}}}
∫
0
2
π
d
x
(
a
+
b
sin
x
)
2
=
∫
0
2
π
d
x
(
a
+
b
cos
x
)
2
=
2
π
a
(
a
2
−
b
2
)
3
/
2
{\displaystyle \int _{0}^{2\pi }{\frac {dx}{(a+b\sin x)^{2}}}=\int _{0}^{2\pi }{\frac {dx}{(a+b\cos x)^{2}}}={\frac {2\pi a}{(a^{2}-b^{2})^{3/2}}}}
∫
0
2
π
d
x
1
−
2
a
cos
x
+
a
2
=
2
π
1
−
a
2
for
0
<
a
<
1
{\displaystyle \int _{0}^{2\pi }{\frac {dx}{1-2a\cos x+a^{2}}}={\frac {2\pi }{1-a^{2}}}\quad {\mbox{for }}0<a<1}
∫
0
π
x
sin
x
d
x
1
−
2
a
cos
x
+
a
2
=
{
π
a
ln
|
1
+
a
|
ако
|
a
|
<
1
π
a
ln
|
1
+
1
a
|
ако
|
a
|
>
1
{\displaystyle \int _{0}^{\pi }{\frac {x\sin x\ dx}{1-2a\cos x+a^{2}}}={\begin{cases}{\dfrac {\pi }{a}}\ln \left|1+a\right|&{\text{ако }}|a|<1\\\\{\dfrac {\pi }{a}}\ln \left|1+{\dfrac {1}{a}}\right|&{\text{ако }}|a|>1\end{cases}}}
∫
0
π
cos
m
x
d
x
1
−
2
a
cos
x
+
a
2
=
π
a
m
1
−
a
2
for
a
2
<
1
,
m
=
0
,
1
,
2
,
…
{\displaystyle \int _{0}^{\pi }{\frac {\cos mx\ dx}{1-2a\cos x+a^{2}}}={\frac {\pi a^{m}}{1-a^{2}}}\quad {\mbox{for }}a^{2}<1\ ,\ m=0,1,2,\dots }
∫
0
∞
sin
a
x
2
d
x
=
∫
0
∞
cos
a
x
2
=
1
2
π
2
a
{\displaystyle \int _{0}^{\infty }\sin ax^{2}\ dx=\int _{0}^{\infty }\cos ax^{2}={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}}
∫
0
∞
sin
a
x
n
=
1
n
a
1
/
n
Γ
(
1
n
)
sin
π
2
n
for
n
>
1
{\displaystyle \int _{0}^{\infty }\sin ax^{n}={\frac {1}{na^{1/n}}}\Gamma \left({\frac {1}{n}}\right)\sin {\frac {\pi }{2n}}\quad {\mbox{for }}n>1}
∫
0
∞
cos
a
x
n
=
1
n
a
1
/
n
Γ
(
1
n
)
cos
π
2
n
for
n
>
1
{\displaystyle \int _{0}^{\infty }\cos ax^{n}={\frac {1}{na^{1/n}}}\Gamma \left({\frac {1}{n}}\right)\cos {\frac {\pi }{2n}}\quad {\mbox{for }}n>1}
∫
0
∞
sin
x
x
d
x
=
∫
0
∞
cos
x
x
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin x}{\sqrt {x}}}\ dx=\int _{0}^{\infty }{\frac {\cos x}{\sqrt {x}}}\ dx={\sqrt {\frac {\pi }{2}}}}
∫
0
∞
sin
x
x
p
d
x
=
π
2
Γ
(
p
)
sin
(
p
π
2
)
for
0
<
p
<
1
{\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x^{p}}}\ dx={\frac {\pi }{2\Gamma (p)\sin \left({\dfrac {p\pi }{2}}\right)}}\quad {\mbox{for }}0<p<1}
∫
0
∞
cos
x
x
p
d
x
=
π
2
Γ
(
p
)
cos
(
p
π
2
)
for
0
<
p
<
1
{\displaystyle \int _{0}^{\infty }{\frac {\cos x}{x^{p}}}\ dx={\frac {\pi }{2\Gamma (p)\cos \left({\dfrac {p\pi }{2}}\right)}}\quad {\mbox{for }}0<p<1}
∫
0
∞
sin
a
x
2
cos
2
b
x
d
x
=
1
2
π
2
a
(
cos
b
2
a
−
sin
b
2
a
)
{\displaystyle \int _{0}^{\infty }\sin ax^{2}\cos 2bx\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}\left(\cos {\frac {b^{2}}{a}}-\sin {\frac {b^{2}}{a}}\right)}
∫
0
∞
cos
a
x
2
cos
2
b
x
d
x
=
1
2
π
2
a
(
cos
b
2
a
+
sin
b
2
a
)
{\displaystyle \int _{0}^{\infty }\cos ax^{2}\cos 2bx\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}\left(\cos {\frac {b^{2}}{a}}+\sin {\frac {b^{2}}{a}}\right)}
уреди
∫
0
∞
x
e
−
x
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{\sqrt {x}}\,e^{-x}\,dx={\frac {1}{2}}{\sqrt {\pi }}}
Гама функција )
∫
0
∞
e
−
a
x
cos
b
x
d
x
=
a
a
2
+
b
2
{\displaystyle \int _{0}^{\infty }e^{-ax}\cos bx\,dx={\frac {a}{a^{2}+b^{2}}}}
∫
0
∞
e
−
a
x
sin
b
x
d
x
=
b
a
2
+
b
2
{\displaystyle \int _{0}^{\infty }e^{-ax}\sin bx\,dx={\frac {b}{a^{2}+b^{2}}}}
∫
0
∞
e
−
a
x
sin
b
x
x
d
x
=
tan
−
1
b
a
{\displaystyle \int _{0}^{\infty }{\frac {{}e^{-ax}\sin bx}{x}}\,dx=\tan ^{-1}{\frac {b}{a}}}
∫
0
∞
e
−
a
x
−
e
−
b
x
x
d
x
=
ln
b
a
{\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x}}\,dx=\ln {\frac {b}{a}}}
∫
0
∞
e
−
a
x
2
d
x
=
1
2
π
a
for
a
>
0
{\displaystyle \int _{0}^{\infty }e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}\quad {\mbox{for }}a>0}
Гаусов интеграл )
∫
0
∞
e
−
a
x
2
cos
b
x
d
x
=
1
2
π
a
e
(
−
b
2
4
a
)
{\displaystyle \int _{0}^{\infty }{e^{-ax^{2}}}\cos bx\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}e^{\left({\frac {-b^{2}}{4a}}\right)}}
∫
0
∞
e
−
(
a
x
2
+
b
x
+
c
)
d
x
=
1
2
π
a
e
(
b
2
−
4
a
c
4
a
)
⋅
erfc
b
2
a
,
каде
erfc
(
p
)
=
2
π
∫
p
∞
e
−
x
2
d
x
{\displaystyle \int _{0}^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}e^{\left({\frac {b^{2}-4ac}{4a}}\right)}\cdot \operatorname {erfc} {\frac {b}{2{\sqrt {a}}}},{\text{ каде }}\operatorname {erfc} (p)={\frac {2}{\sqrt {\pi }}}\int _{p}^{\infty }e^{-x^{2}}\,dx}
∫
−
∞
∞
e
−
(
a
x
2
+
b
x
+
c
)
d
x
=
π
a
e
(
b
2
−
4
a
c
4
a
)
{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\ dx={\sqrt {\frac {\pi }{a}}}e^{\left({\frac {b^{2}-4ac}{4a}}\right)}}
∫
0
∞
x
n
e
−
a
x
d
x
=
Γ
(
n
+
1
)
a
n
+
1
{\displaystyle \int _{0}^{\infty }x^{n}e^{-ax}\ dx={\frac {\Gamma (n+1)}{a^{n+1}}}}
∫
0
∞
x
2
e
−
a
x
2
d
x
=
1
4
π
a
3
for
a
>
0
{\displaystyle \int _{0}^{\infty }{x^{2}e^{-ax^{2}}\,dx}={\frac {1}{4}}{\sqrt {\frac {\pi }{a^{3}}}}\quad {\mbox{for }}a>0}
∫
0
∞
x
2
n
e
−
a
x
2
d
x
=
2
n
−
1
2
a
∫
0
∞
x
2
(
n
−
1
)
e
−
a
x
2
d
x
=
(
2
n
−
1
)
!
!
2
n
+
1
π
a
2
n
+
1
=
(
2
n
)
!
n
!
2
2
n
+
1
π
a
2
n
+
1
for
a
>
0
,
n
=
1
,
2
,
3
…
{\displaystyle \int _{0}^{\infty }x^{2n}e^{-ax^{2}}\,dx={\frac {2n-1}{2a}}\int _{0}^{\infty }x^{2(n-1)}e^{-ax^{2}}\,dx={\frac {(2n-1)!!}{2^{n+1}}}{\sqrt {\frac {\pi }{a^{2n+1}}}}={\frac {(2n)!}{n!2^{2n+1}}}{\sqrt {\frac {\pi }{a^{2n+1}}}}\quad {\mbox{for }}a>0\ ,\ n=1,2,3\ldots }
двоен факториел )
∫
0
∞
x
3
e
−
a
x
2
d
x
=
1
2
a
2
for
a
>
0
{\displaystyle \int _{0}^{\infty }{x^{3}e^{-ax^{2}}\,dx}={\frac {1}{2a^{2}}}\quad {\mbox{for }}a>0}
∫
0
∞
x
2
n
+
1
e
−
a
x
2
d
x
=
n
a
∫
0
∞
x
2
n
−
1
e
−
a
x
2
d
x
=
n
!
2
a
n
+
1
for
a
>
0
,
n
=
0
,
1
,
2
…
{\displaystyle \int _{0}^{\infty }x^{2n+1}e^{-ax^{2}}\,dx={\frac {n}{a}}\int _{0}^{\infty }x^{2n-1}e^{-ax^{2}}\,dx={\frac {n!}{2a^{n+1}}}\quad {\mbox{for }}a>0\ ,\ n=0,1,2\ldots }
∫
0
∞
x
m
e
−
a
x
2
d
x
=
Γ
(
m
+
1
2
)
2
a
(
m
+
1
2
)
{\displaystyle \int _{0}^{\infty }x^{m}e^{-ax^{2}}\ dx={\frac {\Gamma \left({\dfrac {m+1}{2}}\right)}{2a^{\left({\frac {m+1}{2}}\right)}}}}
∫
0
∞
e
(
−
a
x
2
−
b
x
2
)
d
x
=
1
2
π
a
e
−
2
a
b
{\displaystyle \int _{0}^{\infty }e^{\left(-ax^{2}-{\frac {b}{x^{2}}}\right)}\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}e^{-2{\sqrt {ab}}}}
∫
0
∞
x
e
x
−
1
d
x
=
ζ
(
2
)
=
π
2
6
{\displaystyle \int _{0}^{\infty }{\frac {x}{e^{x}-1}}\ dx=\zeta (2)={\frac {\pi ^{2}}{6}}}
∫
0
∞
x
n
−
1
e
x
−
1
d
x
=
Γ
(
n
)
ζ
(
n
)
{\displaystyle \int _{0}^{\infty }{\frac {x^{n-1}}{e^{x}-1}}\ dx=\Gamma (n)\zeta (n)}
∫
0
∞
x
e
x
+
1
d
x
=
1
1
2
−
1
2
2
+
1
3
2
−
1
4
2
+
⋯
=
π
2
12
{\displaystyle \int _{0}^{\infty }{\frac {x}{e^{x}+1}}\ dx={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\dots ={\frac {\pi ^{2}}{12}}}
∫
0
∞
sin
m
x
e
2
π
x
−
1
d
x
=
1
4
coth
m
2
−
1
2
m
{\displaystyle \int _{0}^{\infty }{\frac {\sin mx}{e^{2\pi x}-1}}\ dx={\frac {1}{4}}\coth {\frac {m}{2}}-{\frac {1}{2m}}}
∫
0
∞
(
1
1
+
x
−
e
−
x
)
d
x
x
=
γ
{\displaystyle \int _{0}^{\infty }\left({\frac {1}{1+x}}-e^{-x}\right)\ {\frac {dx}{x}}=\gamma }
γ
{\displaystyle \gamma }
Ојлерова-Маскерониева константа )
∫
0
∞
e
−
x
2
−
e
−
x
x
d
x
=
γ
2
{\displaystyle \int _{0}^{\infty }{\frac {e^{-x^{2}}-e^{-x}}{x}}\ dx={\frac {\gamma }{2}}}
∫
0
∞
(
1
e
x
−
1
−
e
−
x
x
)
d
x
=
γ
{\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {e^{-x}}{x}}\right)\ dx=\gamma }
∫
0
∞
e
−
a
x
−
e
−
b
x
x
sec
p
x
d
x
=
1
2
ln
b
2
+
p
2
a
2
+
p
2
{\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x\sec px}}\ dx={\frac {1}{2}}\ln {\frac {b^{2}+p^{2}}{a^{2}+p^{2}}}}
∫
0
∞
e
−
a
x
−
e
−
b
x
x
csc
p
x
d
x
=
tan
−
1
b
p
−
tan
−
1
a
p
{\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x\csc px}}\ dx=\tan ^{-1}{\frac {b}{p}}-\tan ^{-1}{\frac {a}{p}}}
∫
0
∞
e
−
a
x
(
1
−
cos
x
)
x
2
d
x
=
cot
−
1
a
−
a
2
ln
|
a
2
+
1
a
2
|
{\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}(1-\cos x)}{x^{2}}}\ dx=\cot ^{-1}a-{\frac {a}{2}}\ln \left|{\frac {a^{2}+1}{a^{2}}}\right|}
∫
−
∞
∞
e
−
x
2
d
x
=
π
{\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}
∫
−
∞
∞
x
2
(
n
+
1
)
e
−
1
2
x
2
d
x
=
(
2
n
+
1
)
!
2
n
n
!
2
π
for
n
=
0
,
1
,
2
,
…
{\displaystyle \int _{-\infty }^{\infty }x^{2(n+1)}e^{-{\frac {1}{2}}x^{2}}\,dx={\frac {(2n+1)!}{2^{n}n!}}{\sqrt {2\pi }}\quad {\mbox{for }}n=0,1,2,\ldots }
уреди
∫
0
1
x
m
(
ln
x
)
n
d
x
=
(
−
1
)
n
n
!
(
m
+
1
)
n
+
1
for
m
>
−
1
,
n
=
0
,
1
,
2
,
…
{\displaystyle \int _{0}^{1}x^{m}(\ln x)^{n}\,dx={\frac {(-1)^{n}n!}{(m+1)^{n+1}}}\quad {\mbox{for }}m>-1,n=0,1,2,\ldots }
∫
0
1
ln
x
1
+
x
d
x
=
−
π
2
12
{\displaystyle \int _{0}^{1}{\frac {\ln x}{1+x}}\,dx=-{\frac {\pi ^{2}}{12}}}
∫
0
1
ln
x
1
−
x
d
x
=
−
π
2
6
{\displaystyle \int _{0}^{1}{\frac {\ln x}{1-x}}\,dx=-{\frac {\pi ^{2}}{6}}}
∫
0
1
ln
(
1
+
x
)
x
d
x
=
π
2
12
{\displaystyle \int _{0}^{1}{\frac {\ln(1+x)}{x}}\,dx={\frac {\pi ^{2}}{12}}}
∫
0
1
ln
(
1
−
x
)
x
d
x
=
−
π
2
6
{\displaystyle \int _{0}^{1}{\frac {\ln(1-x)}{x}}\,dx=-{\frac {\pi ^{2}}{6}}}
∫
0
∞
ln
(
a
2
+
x
2
)
b
2
+
x
2
d
x
=
π
b
ln
(
a
+
b
)
for
a
,
b
>
0
{\displaystyle \int _{0}^{\infty }{\frac {\ln(a^{2}+x^{2})}{b^{2}+x^{2}}}\ dx={\frac {\pi }{b}}\ln(a+b)\quad {\mbox{for }}a,b>0}
∫
0
∞
ln
x
x
2
+
a
2
d
x
=
π
ln
a
2
a
for
a
>
0
{\displaystyle \int _{0}^{\infty }{\frac {\ln x}{x^{2}+a^{2}}}\ dx={\frac {\pi \ln a}{2a}}\quad {\mbox{for }}a>0}
уреди
∫
0
∞
sin
a
x
sinh
b
x
d
x
=
π
2
b
tanh
a
π
2
b
{\displaystyle \int _{0}^{\infty }{\frac {\sin ax}{\sinh bx}}\ dx={\frac {\pi }{2b}}\tanh {\frac {a\pi }{2b}}}
∫
0
∞
cos
a
x
cosh
b
x
d
x
=
π
2
b
⋅
1
cosh
a
π
2
b
{\displaystyle \int _{0}^{\infty }{\frac {\cos ax}{\cosh bx}}\ dx={\frac {\pi }{2b}}\cdot {\frac {1}{\cosh {\frac {a\pi }{2b}}}}}
∫
0
∞
x
sinh
a
x
d
x
=
π
2
4
a
2
{\displaystyle \int _{0}^{\infty }{\frac {x}{\sinh ax}}\ dx={\frac {\pi ^{2}}{4a^{2}}}}
∫
−
∞
∞
1
cosh
x
d
x
=
π
{\displaystyle \int _{-\infty }^{\infty }{\frac {1}{\cosh x}}\ dx=\pi }
∫
0
∞
f
(
a
x
)
−
f
(
b
x
)
x
d
x
=
(
lim
x
→
0
f
(
x
)
−
lim
x
→
∞
f
(
x
)
)
ln
(
b
a
)
{\displaystyle \int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\ dx=\left(\lim _{x\to 0}f(x)-\lim _{x\to \infty }f(x)\right)\ln \left({\frac {b}{a}}\right)}
f
′
(
x
)
{\displaystyle f'(x)}
![{\displaystyle \int _{a}^{\infty }f(x)\,dx=\lim _{b\to \infty }\left[\int _{a}^{b}f(x)\,dx\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a095fbd3f49faf63702fb3b314abd0e76c40f8a)
