In mathematics, the definite integral
∫
a
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)\,dx}
is the area of the region in the xy -plane bounded by the graph of f , the x -axis, and the lines x = a and x = b , such that area above the x -axis adds to the total, and that below the x -axis subtracts from the total.
The fundamental theorem of calculus establishes the relationship between indefinite and definite integrals and introduces a technique for evaluating definite integrals.
If the interval is infinite the definite integral is called an improper integral and defined by using appropriate limiting procedures. for example:
∫
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]}
A constant, such pi, that may be defined by the integral of an algebraic function over an algebraic domain is known as a period .
The following is a list of some of the most common or interesting definite integrals. For a list of indefinite integrals see List of indefinite integrals .
Definite integrals involving rational or irrational expressions
∫
0
∞
d
x
1
+
x
p
=
π
/
p
sin
(
π
/
p
)
for
ℜ
(
p
)
>
1
{\displaystyle \int _{0}^{\infty }{\frac {dx}{1+x^{p}={\frac {\pi /p}{\sin(\pi /p)}\quad {\text{for }\Re (p)>1}
∫
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 {\text{for }0<p<1}
∫
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 {\text{for }0<m+1<n}
∫
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 {\text{for }n(r-2)<m+1<nr}
Definite integrals involving trigonometric functions
∫
0
π
sin
(
m
x
)
sin
(
n
x
)
d
x
=
{
0
if
m
≠
n
π
2
if
m
=
n
for
m
,
n
positive integers
{\displaystyle \int _{0}^{\pi }\sin(mx)\sin(nx)dx={\begin{cases}0&{\text{if }m\neq n\\\\{\dfrac {\pi }{2}&{\text{if }m=n\end{cases}\quad {\text{for }m,n{\text{ positive integers}
∫
0
π
cos
(
m
x
)
cos
(
n
x
)
d
x
=
{
0
if
m
≠
n
π
2
if
m
=
n
for
m
,
n
positive integers
{\displaystyle \int _{0}^{\pi }\cos(mx)\cos(nx)dx={\begin{cases}0&{\text{if }m\neq n\\\\{\dfrac {\pi }{2}&{\text{if }m=n\end{cases}\quad {\text{for }m,n{\text{ positive integers}
∫
0
π
sin
(
m
x
)
cos
(
n
x
)
d
x
=
{
0
if
m
+
n
even
2
m
m
2
−
n
2
if
m
+
n
odd
for
m
,
n
integers
.
{\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{if }m+n{\text{ odd}\end{cases}\quad {\text{for }m,n{\text{ integers}.}
∫
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 {\text{for }m=1,2,3\ldots }
∫
0
x
sin
2
m
(
t
)
d
t
=
(
2
m
−
1
)
!
!
(
2
m
)
!
!
(
x
−
sin
(
x
)
cos
(
x
)
(
1
+
∑
k
=
1
∞
sin
2
k
(
x
)
(
2
k
)
!
!
(
2
k
+
1
)
!
!
)
)
for
m
=
1
,
2
,
3
…
{\displaystyle \int _{0}^{x}\sin ^{2m}(t)dt={\frac {(2m-1)!!}{(2m)!!}{\Biggl (}x-\sin(x)\cos(x){\Biggl (}1+\sum _{k=1}^{\infty }{\frac {\sin ^{2k}(x)(2k)!!}{(2k+1)!!}{\Biggr )}{\Biggr )}\quad {\text{for }m=1,2,3\ldots }
∫
0
x
cos
2
m
(
t
)
d
t
=
(
2
m
−
1
)
!
!
(
2
m
)
!
!
(
x
−
sin
(
x
)
cos
(
x
)
(
1
+
∑
k
=
1
∞
cos
2
k
(
x
)
(
2
k
)
!
!
(
2
k
+
1
)
!
!
)
)
for
m
=
1
,
2
,
3
…
{\displaystyle \int _{0}^{x}\cos ^{2m}(t)dt={\frac {(2m-1)!!}{(2m)!!}{\Biggl (}x-\sin(x)\cos(x){\Biggl (}1+\sum _{k=1}^{\infty }{\frac {\cos ^{2k}(x)(2k)!!}{(2k+1)!!}{\Biggr )}{\Biggr )}\quad {\text{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 {\text{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
if
p
>
0
0
if
p
=
0
−
π
2
if
p
<
0
{\displaystyle \int _{0}^{\infty }{\frac {\sin(px)}{x}dx={\begin{cases}{\dfrac {\pi }{2}&{\text{if }p>0\\\\0&{\text{if }p=0\\\\-{\dfrac {\pi }{2}&{\text{if }p<0\end{cases}
(see Dirichlet integral )
∫
0
∞
sin
p
x
cos
q
x
x
d
x
=
{
0
if
q
>
p
>
0
π
2
if
0
<
q
<
p
π
4
if
p
=
q
>
0
{\displaystyle \int _{0}^{\infty }{\frac {\sin px\cos qx}{x}\ dx={\begin{cases}0&{\text{ if }q>p>0\\\\{\dfrac {\pi }{2}&{\text{ if }0<q<p\\\\{\dfrac {\pi }{4}&{\text{ if }p=q>0\end{cases}
∫
0
∞
sin
p
x
sin
q
x
x
2
d
x
=
{
π
p
2
if
0
<
p
≤
q
π
q
2
if
0
<
q
≤
p
{\displaystyle \int _{0}^{\infty }{\frac {\sin px\sin qx}{x^{2}\ dx={\begin{cases}{\dfrac {\pi p}{2}&{\text{ if }0<p\leq q\\\\{\dfrac {\pi q}{2}&{\text{ if }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 {\text{for }0<a<1}
∫
0
π
x
sin
x
d
x
1
−
2
a
cos
x
+
a
2
=
{
π
a
ln
|
1
+
a
|
if
|
a
|
<
1
π
a
ln
|
1
+
1
a
|
if
|
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{if }|a|<1\\\\{\dfrac {\pi }{a}\ln \left|1+{\dfrac {1}{a}\right|&{\text{if }|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 {\text{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 {\text{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 {\text{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 {\text{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 {\text{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)}
Definite integrals involving exponential functions
∫
0
∞
x
e
−
x
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{\sqrt {x}\,e^{-x}\,dx={\frac {1}{2}{\sqrt {\pi }
(see also Gamma function )
∫
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
−
cos
(
b
x
)
x
d
x
=
ln
b
a
{\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-\cos(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 {\text{for }a>0}
(the Gaussian integral )
∫
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
,
where
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{ where }\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 {\text{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 {\text{for }a>0\ ,\ n=1,2,3\ldots }
(where !! is the double factorial )
∫
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 {\text{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 {\text{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
∞
x
n
e
x
+
1
d
x
=
n
!
⋅
(
2
n
−
1
2
n
)
ζ
(
n
+
1
)
{\displaystyle \int _{0}^{\infty }{\frac {x^{n}{e^{x}+1}\ dx=n!\cdot \left({\frac {2^{n}-1}{2^{n}\right)\zeta (n+1)}
∫
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 }
(where
γ
{\displaystyle \gamma }
is Euler–Mascheroni constant )
∫
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 {\text{for }n=0,1,2,\ldots }
Definite integrals involving logarithmic functions
∫
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 {\text{for }m>-1,n=0,1,2,\ldots }
∫
1
∞
x
m
(
ln
x
)
n
d
x
=
(
−
1
)
n
+
1
n
!
(
m
+
1
)
n
+
1
for
m
<
−
1
,
n
=
0
,
1
,
2
,
…
{\displaystyle \int _{1}^{\infty }x^{m}(\ln x)^{n}\,dx={\frac {(-1)^{n+1}n!}{(m+1)^{n+1}\quad {\text{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 {\text{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 {\text{for }a>0}
Definite integrals involving hyperbolic functions
∫
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}
∫
0
∞
x
2
n
+
1
sinh
a
x
d
x
=
c
2
n
+
1
(
π
a
)
2
(
n
+
1
)
,
c
2
n
+
1
=
(
−
1
)
n
2
(
1
2
−
∑
k
=
0
n
−
1
(
−
1
)
k
(
2
n
+
1
2
k
+
1
)
c
2
k
+
1
)
,
c
1
=
1
4
{\displaystyle \int _{0}^{\infty }{\frac {x^{2n+1}{\sinh ax}\ dx=c_{2n+1}\left({\frac {\pi }{a}\right)^{2(n+1)},\quad c_{2n+1}={\frac {(-1)^{n}{2}\left({\frac {1}{2}-\sum _{k=0}^{n-1}(-1)^{k}{2n+1 \choose 2k+1}c_{2k+1}\right),\quad c_{1}={\frac {1}{4}
∫
0
∞
1
cosh
a
x
d
x
=
π
2
a
{\displaystyle \int _{0}^{\infty }{\frac {1}{\cosh ax}\ dx={\frac {\pi }{2a}
∫
0
∞
x
2
n
cosh
a
x
d
x
=
d
2
n
(
π
a
)
2
n
+
1
,
d
2
n
=
(
−
1
)
n
2
(
1
4
n
−
∑
k
=
0
n
−
1
(
−
1
)
k
(
2
n
2
k
)
d
2
k
)
,
d
0
=
1
2
{\displaystyle \int _{0}^{\infty }{\frac {x^{2n}{\cosh ax}\ dx=d_{2n}\left({\frac {\pi }{a}\right)^{2n+1},\quad d_{2n}={\frac {(-1)^{n}{2}\left({\frac {1}{4^{n}-\sum _{k=0}^{n-1}(-1)^{k}{2n \choose 2k}d_{2k}\right),\quad d_{0}={\frac {1}{2}
∫
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)}
holds if the integral exists and
f
′
(
x
)
{\displaystyle f'(x)}
is continuous.
See also
References
Reynolds, Robert; Stauffer, Allan (2020). "Derivation of Logarithmic and Logarithmic Hyperbolic Tangent Integrals Expressed in Terms of Special Functions" . Mathematics . 8 (687): 687. doi :10.3390/math8050687 .
Reynolds, Robert; Stauffer, Allan (2019). "A Definite Integral Involving the Logarithmic Function in Terms of the Lerch Function" . Mathematics . 7 (1148): 1148. doi :10.3390/math7121148 .
Reynolds, Robert; Stauffer, Allan (2019). "Definite Integral of Arctangent and Polylogarithmic Functions Expressed as a Series" . Mathematics . 7 (1099): 1099. doi :10.3390/math7111099 .
Winckler, Anton (1861). "Eigenschaften Einiger Bestimmten Integrale". Hof, K.K., Ed .
Spiegel, Murray R.; Lipschutz, Seymour; Liu, John (2009). Mathematical handbook of formulas and tables (3rd ed.). McGraw-Hill . ISBN 978-0071548557 .
Zwillinger, Daniel (2003). CRC standard mathematical tables and formulae (32nd ed.). CRC Press . ISBN 978-143983548-7 .
Abramowitz, Milton ; Stegun, Irene Ann , eds. (1983) [June 1964]. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables . Applied Mathematics Series. Vol. 55 (Ninth reprint with additional corrections of tenth original printing with corrections (December 1972); first ed.). Washington D.C.; New York: United States Department of Commerce, National Bureau of Standards; Dover Publications. ISBN 978-0-486-61272-0 . LCCN 64-60036 . MR 0167642 . LCCN 65-12253 .