In mathematics, the definite integral

${\displaystyle \int _{a}^{b}f(x)\,dx}$

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:

${\displaystyle \int _{a}^{\infty }f(x)\,dx=\lim _{b\to \infty }\left[\int _{a}^{b}f(x)\,dx\right]}$

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

${\displaystyle \int _{0}^{\infty }{\frac {dx}{1+x^{p))}={\frac {\pi /p}{\sin(\pi /p)))\quad {\text{for ))\Re (p)>1}$
${\displaystyle \int _{0}^{\infty }{\frac {x^{p-1}dx}{1+x))={\frac {\pi }{\sin(p\pi )))\quad {\text{for ))0
${\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
${\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{1+2x\cos \beta +x^{2))}={\frac {\pi }{\sin(m\pi )))\cdot {\frac {\sin(m\beta )}{\sin(\beta )))}$
${\displaystyle \int _{0}^{a}{\frac {dx}{\sqrt {a^{2}-x^{2))))={\frac {\pi }{2))}$
${\displaystyle \int _{0}^{a}{\sqrt {a^{2}-x^{2))}dx={\frac {\pi a^{2)){4))}$
${\displaystyle \int _{0}^{a}x^{m}(a^{n}-x^{n})^{p}\,dx={\frac {a^{m+1+np}\Gamma \left({\dfrac {m+1}{n))\right)\Gamma (p+1)}{n\Gamma \left({\dfrac {m+1}{n))+p+1\right)))}$
${\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)

## Definite integrals involving trigonometric functions

${\displaystyle \int _{0}^{\pi }\sin(mx)\sin(nx)dx={\begin{cases}0&{\text{if ))m\neq n\\\\{\dfrac {\pi }{2))&{\text{if ))m=n\end{cases))\quad {\text{for ))m,n{\text{ positive integers))}$
${\displaystyle \int _{0}^{\pi }\cos(mx)\cos(nx)dx={\begin{cases}0&{\text{if ))m\neq n\\\\{\dfrac {\pi }{2))&{\text{if ))m=n\end{cases))\quad {\text{for ))m,n{\text{ positive integers))}$
${\displaystyle \int _{0}^{\pi }\sin(mx)\cos(nx)dx={\begin{cases}0&{\text{if ))m+n{\text{ even))\\\\{\dfrac {2m}{m^{2}-n^{2))}&{\text{if ))m+n{\text{ odd))\end{cases))\quad {\text{for ))m,n{\text{ integers)).}$
${\displaystyle \int _{0}^{\frac {\pi }{2))\sin ^{2}(x)dx=\int _{0}^{\frac {\pi }{2))\cos ^{2}(x)dx={\frac {\pi }{4))}$
${\displaystyle \int _{0}^{\frac {\pi }{2))\sin ^{2m}(x)dx=\int _{0}^{\frac {\pi }{2))\cos ^{2m}(x)dx={\frac {1\times 3\times 5\times \cdots \times (2m-1)}{2\times 4\times 6\times \cdots \times 2m))\cdot {\frac {\pi }{2))\quad {\text{for ))m=1,2,3\ldots }$
${\displaystyle \int _{0}^{\frac {\pi }{2))\sin ^{2m+1}(x)dx=\int _{0}^{\frac {\pi }{2))\cos ^{2m+1}(x)dx={\frac {2\times 4\times 6\times \cdots \times 2m}{1\times 3\times 5\times \cdots \times (2m+1)))\quad {\text{for ))m=1,2,3\ldots }$
${\displaystyle \int _{0}^{\frac {\pi }{2))\sin ^{2p-1}(x)\cos ^{2q-1}(x)dx={\frac {\Gamma (p)\Gamma (q)}{2\Gamma (p+q)))={\frac {1}{2)){\text{B))(p,q)}$
${\displaystyle \int _{0}^{\infty }{\frac {\sin(px)}{x))dx={\begin{cases}{\dfrac {\pi }{2))&{\text{if ))p>0\\\\0&{\text{if ))p=0\\\\-{\dfrac {\pi }{2))&{\text{if ))p<0\end{cases))}$ (see Dirichlet integral)
${\displaystyle \int _{0}^{\infty }{\frac {\sin px\cos qx}{x))\ dx={\begin{cases}0&{\text{ if ))q>p>0\\\\{\dfrac {\pi }{2))&{\text{ if ))00\end{cases))}$
${\displaystyle \int _{0}^{\infty }{\frac {\sin px\sin qx}{x^{2))}\ dx={\begin{cases}{\dfrac {\pi p}{2))&{\text{ if ))0
${\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}px}{x^{2))}\ dx={\frac {\pi p}{2))}$
${\displaystyle \int _{0}^{\infty }{\frac {1-\cos px}{x^{2))}\ dx={\frac {\pi p}{2))}$
${\displaystyle \int _{0}^{\infty }{\frac {\cos px-\cos qx}{x))\ dx=\ln {\frac {q}{p))}$
${\displaystyle \int _{0}^{\infty }{\frac {\cos px-\cos qx}{x^{2))}\ dx={\frac {\pi (q-p)}{2))}$
${\displaystyle \int _{0}^{\infty }{\frac {\cos mx}{x^{2}+a^{2))}\ dx={\frac {\pi }{2a))e^{-ma))$
${\displaystyle \int _{0}^{\infty }{\frac {x\sin mx}{x^{2}+a^{2))}\ dx={\frac {\pi }{2))e^{-ma))$
${\displaystyle \int _{0}^{\infty }{\frac {\sin mx}{x(x^{2}+a^{2})))\ dx={\frac {\pi }{2a^{2))}\left(1-e^{-ma}\right)}$
${\displaystyle \int _{0}^{2\pi }{\frac {dx}{a+b\sin x))={\frac {2\pi }{\sqrt {a^{2}-b^{2))))}$
${\displaystyle \int _{0}^{2\pi }{\frac {dx}{a+b\cos x))={\frac {2\pi }{\sqrt {a^{2}-b^{2))))}$
${\displaystyle \int _{0}^{\frac {\pi }{2)){\frac {dx}{a+b\cos x))={\frac {\cos ^{-1}\left({\dfrac {b}{a))\right)}{\sqrt {a^{2}-b^{2))))}$
${\displaystyle \int _{0}^{2\pi }{\frac {dx}{(a+b\sin x)^{2))}=\int _{0}^{2\pi }{\frac {dx}{(a+b\cos x)^{2))}={\frac {2\pi a}{(a^{2}-b^{2})^{3/2))))$
${\displaystyle \int _{0}^{2\pi }{\frac {dx}{1-2a\cos x+a^{2))}={\frac {2\pi }{1-a^{2))}\quad {\text{for ))0
${\displaystyle \int _{0}^{\pi }{\frac {x\sin x\ dx}{1-2a\cos x+a^{2))}={\begin{cases}{\dfrac {\pi }{a))\ln \left|1+a\right|&{\text{if ))|a|<1\\\\{\dfrac {\pi }{a))\ln \left|1+{\dfrac {1}{a))\right|&{\text{if ))|a|>1\end{cases))}$
${\displaystyle \int _{0}^{\pi }{\frac {\cos mx\ dx}{1-2a\cos x+a^{2))}={\frac {\pi a^{m)){1-a^{2))}\quad {\text{for ))a^{2}<1\ ,\ m=0,1,2,\dots }$
${\displaystyle \int _{0}^{\infty }\sin ax^{2}\ dx=\int _{0}^{\infty }\cos ax^{2}={\frac {1}{2)){\sqrt {\frac {\pi }{2a))))$
${\displaystyle \int _{0}^{\infty }\sin ax^{n}={\frac {1}{na^{1/n))}\Gamma \left({\frac {1}{n))\right)\sin {\frac {\pi }{2n))\quad {\text{for ))n>1}$
${\displaystyle \int _{0}^{\infty }\cos ax^{n}={\frac {1}{na^{1/n))}\Gamma \left({\frac {1}{n))\right)\cos {\frac {\pi }{2n))\quad {\text{for ))n>1}$
${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{\sqrt {x))}\ dx=\int _{0}^{\infty }{\frac {\cos x}{\sqrt {x))}\ dx={\sqrt {\frac {\pi }{2))))$
${\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
${\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
${\displaystyle \int _{0}^{\infty }\sin ax^{2}\cos 2bx\ dx={\frac {1}{2)){\sqrt {\frac {\pi }{2a))}\left(\cos {\frac {b^{2)){a))-\sin {\frac {b^{2)){a))\right)}$
${\displaystyle \int _{0}^{\infty }\cos ax^{2}\cos 2bx\ dx={\frac {1}{2)){\sqrt {\frac {\pi }{2a))}\left(\cos {\frac {b^{2)){a))+\sin {\frac {b^{2)){a))\right)}$

## Definite integrals involving exponential functions

${\displaystyle \int _{0}^{\infty }{\sqrt {x))\,e^{-x}\,dx={\frac {1}{2)){\sqrt {\pi ))}$ (see also Gamma function)
${\displaystyle \int _{0}^{\infty }e^{-ax}\cos bx\,dx={\frac {a}{a^{2}+b^{2))))$
${\displaystyle \int _{0}^{\infty }e^{-ax}\sin bx\,dx={\frac {b}{a^{2}+b^{2))))$
${\displaystyle \int _{0}^{\infty }{\frac ((}e^{-ax}\sin bx}{x))\,dx=\tan ^{-1}{\frac {b}{a))}$
${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx)){x))\,dx=\ln {\frac {b}{a))}$
${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-\cos(bx)}{x))\,dx=\ln {\frac {b}{a))}$
${\displaystyle \int _{0}^{\infty }e^{-ax^{2))\,dx={\frac {1}{2)){\sqrt {\frac {\pi }{a))}\quad {\text{for ))a>0}$ (the Gaussian integral)
${\displaystyle \int _{0}^{\infty }{e^{-ax^{2))}\cos bx\,dx={\frac {1}{2)){\sqrt {\frac {\pi }{a))}e^{\left({\frac {-b^{2)){4a))\right)))$
${\displaystyle \int _{0}^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\frac {1}{2)){\sqrt {\frac {\pi }{a))}e^{\left({\frac {b^{2}-4ac}{4a))\right)}\cdot \operatorname {erfc} {\frac {b}{2{\sqrt {a)))),{\text{ where ))\operatorname {erfc} (p)={\frac {2}{\sqrt {\pi ))}\int _{p}^{\infty }e^{-x^{2))\,dx}$
${\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\ dx={\sqrt {\frac {\pi }{a))}e^{\left({\frac {b^{2}-4ac}{4a))\right)))$
${\displaystyle \int _{0}^{\infty }x^{n}e^{-ax}\ dx={\frac {\Gamma (n+1)}{a^{n+1))))$
${\displaystyle \int _{0}^{\infty }{x^{2}e^{-ax^{2))\,dx}={\frac {1}{4)){\sqrt {\frac {\pi }{a^{3))))\quad {\text{for ))a>0}$
${\displaystyle \int _{0}^{\infty }x^{2n}e^{-ax^{2))\,dx={\frac {2n-1}{2a))\int _{0}^{\infty }x^{2(n-1)}e^{-ax^{2))\,dx={\frac {(2n-1)!!}{2^{n+1))}{\sqrt {\frac {\pi }{a^{2n+1))))={\frac {(2n)!}{n!2^{2n+1))}{\sqrt {\frac {\pi }{a^{2n+1))))\quad {\text{for ))a>0\ ,\ n=1,2,3\ldots }$ (where !! is the double factorial)
${\displaystyle \int _{0}^{\infty }{x^{3}e^{-ax^{2))\,dx}={\frac {1}{2a^{2))}\quad {\text{for ))a>0}$
${\displaystyle \int _{0}^{\infty }x^{2n+1}e^{-ax^{2))\,dx={\frac {n}{a))\int _{0}^{\infty }x^{2n-1}e^{-ax^{2))\,dx={\frac {n!}{2a^{n+1))}\quad {\text{for ))a>0\ ,\ n=0,1,2\ldots }$
${\displaystyle \int _{0}^{\infty }x^{m}e^{-ax^{2))\ dx={\frac {\Gamma \left({\dfrac {m+1}{2))\right)}{2a^{\left({\frac {m+1}{2))\right)))))$
${\displaystyle \int _{0}^{\infty }e^{\left(-ax^{2}-{\frac {b}{x^{2))}\right)}\ dx={\frac {1}{2)){\sqrt {\frac {\pi }{a))}e^{-2{\sqrt {ab))))$
${\displaystyle \int _{0}^{\infty }{\frac {x}{e^{x}-1))\ dx=\zeta (2)={\frac {\pi ^{2)){6))}$
${\displaystyle \int _{0}^{\infty }{\frac {x^{n-1)){e^{x}-1))\ dx=\Gamma (n)\zeta (n)}$
${\displaystyle \int _{0}^{\infty }{\frac {x}{e^{x}+1))\ dx={\frac {1}{1^{2))}-{\frac {1}{2^{2))}+{\frac {1}{3^{2))}-{\frac {1}{4^{2))}+\dots ={\frac {\pi ^{2)){12))}$
${\displaystyle \int _{0}^{\infty }{\frac {x^{n)){e^{x}+1))\ dx=n!\cdot \left({\frac {2^{n}-1}{2^{n))}\right)\zeta (n+1)}$
${\displaystyle \int _{0}^{\infty }{\frac {\sin mx}{e^{2\pi x}-1))\ dx={\frac {1}{4))\coth {\frac {m}{2))-{\frac {1}{2m))}$
${\displaystyle \int _{0}^{\infty }\left({\frac {1}{1+x))-e^{-x}\right)\ {\frac {dx}{x))=\gamma }$ (where ${\displaystyle \gamma }$ is Euler–Mascheroni constant)
${\displaystyle \int _{0}^{\infty }{\frac {e^{-x^{2))-e^{-x)){x))\ dx={\frac {\gamma }{2))}$
${\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1))-{\frac {e^{-x)){x))\right)\ dx=\gamma }$
${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx)){x\sec px))\ dx={\frac {1}{2))\ln {\frac {b^{2}+p^{2)){a^{2}+p^{2))))$
${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx)){x\csc px))\ dx=\tan ^{-1}{\frac {b}{p))-\tan ^{-1}{\frac {a}{p))}$
${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}(1-\cos x)}{x^{2))}\ dx=\cot ^{-1}a-{\frac {a}{2))\ln \left|{\frac {a^{2}+1}{a^{2))}\right|}$
${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2))\,dx={\sqrt {\pi ))}$
${\displaystyle \int _{-\infty }^{\infty }x^{2(n+1)}e^{-{\frac {1}{2))x^{2))\,dx={\frac {(2n+1)!}{2^{n}n!)){\sqrt {2\pi ))\quad {\text{for ))n=0,1,2,\ldots }$

## Definite integrals involving logarithmic functions

${\displaystyle \int _{0}^{1}x^{m}(\ln x)^{n}\,dx={\frac {(-1)^{n}n!}{(m+1)^{n+1))}\quad {\text{for ))m>-1,n=0,1,2,\ldots }$
${\displaystyle \int _{1}^{\infty }x^{m}(\ln x)^{n}\,dx={\frac {(-1)^{n+1}n!}{(m+1)^{n+1))}\quad {\text{for ))m<-1,n=0,1,2,\ldots }$
${\displaystyle \int _{0}^{1}{\frac {\ln x}{1+x))\,dx=-{\frac {\pi ^{2)){12))}$
${\displaystyle \int _{0}^{1}{\frac {\ln x}{1-x))\,dx=-{\frac {\pi ^{2)){6))}$
${\displaystyle \int _{0}^{1}{\frac {\ln(1+x)}{x))\,dx={\frac {\pi ^{2)){12))}$
${\displaystyle \int _{0}^{1}{\frac {\ln(1-x)}{x))\,dx=-{\frac {\pi ^{2)){6))}$
${\displaystyle \int _{0}^{\infty }{\frac {\ln(a^{2}+x^{2})}{b^{2}+x^{2))}\ dx={\frac {\pi }{b))\ln(a+b)\quad {\text{for ))a,b>0}$
${\displaystyle \int _{0}^{\infty }{\frac {\ln x}{x^{2}+a^{2))}\ dx={\frac {\pi \ln a}{2a))\quad {\text{for ))a>0}$

## Definite integrals involving hyperbolic functions

${\displaystyle \int _{0}^{\infty }{\frac {\sin ax}{\sinh bx))\ dx={\frac {\pi }{2b))\tanh {\frac {a\pi }{2b))}$

${\displaystyle \int _{0}^{\infty }{\frac {\cos ax}{\cosh bx))\ dx={\frac {\pi }{2b))\cdot {\frac {1}{\cosh {\frac {a\pi }{2b))))}$

${\displaystyle \int _{0}^{\infty }{\frac {x}{\sinh ax))\ dx={\frac {\pi ^{2)){4a^{2))))$

${\displaystyle \int _{0}^{\infty }{\frac {x^{2n+1)){\sinh ax))\ dx=c_{2n+1}\left({\frac {\pi }{a))\right)^{2(n+1)},\quad c_{2n+1}={\frac {(-1)^{n)){2))\left({\frac {1}{2))-\sum _{k=0}^{n-1}(-1)^{k}{2n+1 \choose 2k+1}c_{2k+1}\right),\quad c_{1}={\frac {1}{4))}$

${\displaystyle \int _{0}^{\infty }{\frac {1}{\cosh ax))\ dx={\frac {\pi }{2a))}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{2n)){\cosh ax))\ dx=d_{2n}\left({\frac {\pi }{a))\right)^{2n+1},\quad d_{2n}={\frac {(-1)^{n)){2))\left({\frac {1}{4^{n))}-\sum _{k=0}^{n-1}(-1)^{k}{2n \choose 2k}d_{2k}\right),\quad d_{0}={\frac {1}{2))}$

## Frullani integrals

${\displaystyle \int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x))\ dx=\left(\lim _{x\to 0}f(x)-\lim _{x\to \infty }f(x)\right)\ln \left({\frac {b}{a))\right)}$
holds if the integral exists and ${\displaystyle f'(x)}$ is continuous.