List of definite integrals

${\displaystyle \int _{0}^{\infty }{\frac {dx}{1+x^{p}}}={\frac {\pi /p}{\sin(\pi /p)}}\quad {\text{for }}\Re (p)>1}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{p-1}dx}{1+x}}={\frac {\pi }{\sin(p\pi )}}\quad {\text{for }}0<p<1}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{x^{n}+a^{n}}}={\frac {\pi a^{m-n+1}}{n\sin \left({\dfrac {m+1}{n}}\pi \right)}}\quad {\text{for }}0<m+1<n}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{1+2x\cos \beta +x^{2}}}={\frac {\pi }{\sin(m\pi )}}\cdot {\frac {\sin(m\beta )}{\sin(\beta )}}}$

${\displaystyle \int _{0}^{a}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}={\frac {\pi }{2}}}$

${\displaystyle \int _{0}^{a}{\sqrt {a^{2}-x^{2}}}dx={\frac {\pi a^{2}}{4}}}$

${\displaystyle \int _{0}^{a}x^{m}(a^{n}-x^{n})^{p}\,dx={\frac {a^{m+1+np}\Gamma \left({\dfrac {m+1}{n}}\right)\Gamma (p+1)}{n\Gamma \left({\dfrac {m+1}{n}}+p+1\right)}}}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{({x^{n}+a^{n})}^{r}}}={\frac {(-1)^{r-1}\pi a^{m+1-nr}\Gamma \left({\dfrac {m+1}{n}}\right)}{n\sin \left({\dfrac {m+1}{n}}\pi \right)(r-1)!\,\Gamma \left({\dfrac {m+1}{n}}-r+1\right)}}\quad {\text{for }}n(r-2)<m+1<nr}$

${\displaystyle \int _{0}^{\pi }\sin(mx)\sin(nx)dx={\begin{cases}0&{\text{if }}m\neq n\\\\{\dfrac {\pi }{2}}&{\text{if }}m=n\end{cases}}\quad {\text{for }}m,n{\text{ positive integers}}}$

${\displaystyle \int _{0}^{\pi }\cos(mx)\cos(nx)dx={\begin{cases}0&{\text{if }}m\neq n\\\\{\dfrac {\pi }{2}}&{\text{if }}m=n\end{cases}}\quad {\text{for }}m,n{\text{ positive integers}}}$

${\displaystyle \int _{0}^{\pi }\sin(mx)\cos(nx)dx={\begin{cases}0&{\text{if }}m+n{\text{ even}}\\\\{\dfrac {2m}{m^{2}-n^{2}}}&{\text{if }}m+n{\text{ odd}}\end{cases}}\quad {\text{for }}m,n{\text{ integers}}.}$

${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2}(x)dx=\int _{0}^{\frac {\pi }{2}}\cos ^{2}(x)dx={\frac {\pi }{4}}}$

${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m}(x)dx=\int _{0}^{\frac {\pi }{2}}\cos ^{2m}(x)dx={\frac {1\times 3\times 5\times \cdots \times (2m-1)}{2\times 4\times 6\times \cdots \times 2m}}\cdot {\frac {\pi }{2}}\quad {\text{for }}m=1,2,3\ldots }$

${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}(x)dx=\int _{0}^{\frac {\pi }{2}}\cos ^{2m+1}(x)dx={\frac {2\times 4\times 6\times \cdots \times 2m}{1\times 3\times 5\times \cdots \times (2m+1)}}\quad {\text{for }}m=1,2,3\ldots }$

${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2p-1}(x)\cos ^{2q-1}(x)dx={\frac {\Gamma (p)\Gamma (q)}{2\Gamma (p+q)}}={\frac {1}{2}}{\text{B}}(p,q)}$

${\displaystyle \int _{0}^{\infty }{\frac {\sin(px)}{x}}dx={\begin{cases}{\dfrac {\pi }{2}}&{\text{if }}p>0\\\\0&{\text{if }}p=0\\\\-{\dfrac {\pi }{2}}&{\text{if }}p<0\end{cases}}}$ (see Dirichlet integral)

${\displaystyle \int _{0}^{\infty }{\frac {\sin px\cos qx}{x}}\ dx={\begin{cases}0&{\text{ if }}q>p>0\\\\{\dfrac {\pi }{2}}&{\text{ if }}0<q<p\\\\{\dfrac {\pi }{4}}&{\text{ if }}p=q>0\end{cases}}}$

${\displaystyle \int _{0}^{\infty }{\frac {\sin px\sin qx}{x^{2}}}\ dx={\begin{cases}{\dfrac {\pi p}{2}}&{\text{ if }}0<p\leq q\\\\{\dfrac {\pi q}{2}}&{\text{ if }}0<q\leq p\end{cases}}}$

${\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}px}{x^{2}}}\ dx={\frac {\pi p}{2}}}$

${\displaystyle \int _{0}^{\infty }{\frac {1-\cos px}{x^{2}}}\ dx={\frac {\pi p}{2}}}$

${\displaystyle \int _{0}^{\infty }{\frac {\cos px-\cos qx}{x}}\ dx=\ln {\frac {q}{p}}}$

${\displaystyle \int _{0}^{\infty }{\frac {\cos px-\cos qx}{x^{2}}}\ dx={\frac {\pi (q-p)}{2}}}$

${\displaystyle \int _{0}^{\infty }{\frac {\cos mx}{x^{2}+a^{2}}}\ dx={\frac {\pi }{2a}}e^{-ma}}$

${\displaystyle \int _{0}^{\infty }{\frac {x\sin mx}{x^{2}+a^{2}}}\ dx={\frac {\pi }{2}}e^{-ma}}$

${\displaystyle \int _{0}^{\infty }{\frac {\sin mx}{x(x^{2}+a^{2})}}\ dx={\frac {\pi }{2a^{2}}}\left(1-e^{-ma}\right)}$

${\displaystyle \int _{0}^{2\pi }{\frac {dx}{a+b\sin x}}={\frac {2\pi }{\sqrt {a^{2}-b^{2}}}}}$

${\displaystyle \int _{0}^{2\pi }{\frac {dx}{a+b\cos x}}={\frac {2\pi }{\sqrt {a^{2}-b^{2}}}}}$

${\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {dx}{a+b\cos x}}={\frac {\cos ^{-1}\left({\dfrac {b}{a}}\right)}{\sqrt {a^{2}-b^{2}}}}}$

${\displaystyle \int _{0}^{2\pi }{\frac {dx}{(a+b\sin x)^{2}}}=\int _{0}^{2\pi }{\frac {dx}{(a+b\cos x)^{2}}}={\frac {2\pi a}{(a^{2}-b^{2})^{3/2}}}}$

${\displaystyle \int _{0}^{2\pi }{\frac {dx}{1-2a\cos x+a^{2}}}={\frac {2\pi }{1-a^{2}}}\quad {\text{for }}0<a<1}$

${\displaystyle \int _{0}^{\pi }{\frac {x\sin x\ dx}{1-2a\cos x+a^{2}}}={\begin{cases}{\dfrac {\pi }{a}}\ln \left|1+a\right|&{\text{if }}|a|<1\\\\{\dfrac {\pi }{a}}\ln \left|1+{\dfrac {1}{a}}\right|&{\text{if }}|a|>1\end{cases}}}$

${\displaystyle \int _{0}^{\pi }{\frac {\cos mx\ dx}{1-2a\cos x+a^{2}}}={\frac {\pi a^{m}}{1-a^{2}}}\quad {\text{for }}a^{2}<1\ ,\ m=0,1,2,\dots }$

${\displaystyle \int _{0}^{\infty }\sin ax^{2}\ dx=\int _{0}^{\infty }\cos ax^{2}={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}}$

${\displaystyle \int _{0}^{\infty }\sin ax^{n}={\frac {1}{na^{1/n}}}\Gamma \left({\frac {1}{n}}\right)\sin {\frac {\pi }{2n}}\quad {\text{for }}n>1}$

${\displaystyle \int _{0}^{\infty }\cos ax^{n}={\frac {1}{na^{1/n}}}\Gamma \left({\frac {1}{n}}\right)\cos {\frac {\pi }{2n}}\quad {\text{for }}n>1}$

${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{\sqrt {x}}}\ dx=\int _{0}^{\infty }{\frac {\cos x}{\sqrt {x}}}\ dx={\sqrt {\frac {\pi }{2}}}}$

${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x^{p}}}\ dx={\frac {\pi }{2\Gamma (p)\sin \left({\dfrac {p\pi }{2}}\right)}}\quad {\text{for }}0<p<1}$

${\displaystyle \int _{0}^{\infty }{\frac {\cos x}{x^{p}}}\ dx={\frac {\pi }{2\Gamma (p)\cos \left({\dfrac {p\pi }{2}}\right)}}\quad {\text{for }}0<p<1}$

${\displaystyle \int _{0}^{\infty }\sin ax^{2}\cos 2bx\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}\left(\cos {\frac {b^{2}}{a}}-\sin {\frac {b^{2}}{a}}\right)}$

${\displaystyle \int _{0}^{\infty }\cos ax^{2}\cos 2bx\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}\left(\cos {\frac {b^{2}}{a}}+\sin {\frac {b^{2}}{a}}\right)}$

${\displaystyle \int _{0}^{\infty }{\sqrt {x}}\,e^{-x}\,dx={\frac {1}{2}}{\sqrt {\pi }}}$ (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)

${\displaystyle \int _{0}^{\infty }{\frac {e^{-x^{2}}-e^{-x}}{x}}\ dx={\frac {\gamma }{2}}}$

${\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {e^{-x}}{x}}\right)\ dx=\gamma }$

${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x\sec px}}\ dx={\frac {1}{2}}\ln {\frac {b^{2}+p^{2}}{a^{2}+p^{2}}}}$

${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x\csc px}}\ dx=\tan ^{-1}{\frac {b}{p}}-\tan ^{-1}{\frac {a}{p}}}$

${\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}(1-\cos x)}{x^{2}}}\ dx=\cot ^{-1}a-{\frac {a}{2}}\ln \left|{\frac {a^{2}+1}{a^{2}}}\right|}$

${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}$

${\displaystyle \int _{-\infty }^{\infty }x^{2(n+1)}e^{-{\frac {1}{2}}x^{2}}\,dx={\frac {(2n+1)!}{2^{n}n!}}{\sqrt {2\pi }}\quad {\text{for }}n=0,1,2,\ldots }$

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

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

${\displaystyle \int _{0}^{1}{\frac {\ln x}{1+x}}\,dx=-{\frac {\pi ^{2}}{12}}}$

${\displaystyle \int _{0}^{1}{\frac {\ln x}{1-x}}\,dx=-{\frac {\pi ^{2}}{6}}}$

${\displaystyle \int _{0}^{1}{\frac {\ln(1+x)}{x}}\,dx={\frac {\pi ^{2}}{12}}}$

${\displaystyle \int _{0}^{1}{\frac {\ln(1-x)}{x}}\,dx=-{\frac {\pi ^{2}}{6}}}$

${\displaystyle \int _{0}^{\infty }{\frac {\ln(a^{2}+x^{2})}{b^{2}+x^{2}}}\ dx={\frac {\pi }{b}}\ln(a+b)\quad {\text{for }}a,b>0}$

${\displaystyle \int _{0}^{\infty }{\frac {\ln x}{x^{2}+a^{2}}}\ dx={\frac {\pi \ln a}{2a}}\quad {\text{for }}a>0}$

$${\displaystyle \int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\ dx=\left(\lim _{x\to 0}f(x)-\lim _{x\to \infty }f(x)\right)\ln \left({\frac {b}{a}}\right)}$$

holds if the integral exists and ${\displaystyle f'(x)}$ is continuous.

• List of integrals
• Indefinite sum
• Gamma function
• List of limits

• 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.
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