List of definite integrals

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In mathematics, the definite integral

$\underset{a}{\overset{b}{\int }}f(x)dx$

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:

$\underset{a}{\overset{\mathrm{\infty }}{\int }}f(x)dx=\underset{b\to \mathrm{\infty }}{\lim }[\underset{a}{\overset{b}{\int }}f(x)dx]$

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

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{a^p+x^p}=\frac{\pi }{p{a}^{p-1}\sin (\frac{\pi }{p})}\text{for }\mathrm{\Re }(p)>1\text{, }\mathrm{\Re }(a)>0$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{p-1}dx}{1+x}=\frac{\pi }{\sin (p\pi )}\text{for }0<p<1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{m}dx}{x^n+a^n}=\frac{\pi a^{m-n+1}}{n\sin \left(\frac{m+1}{n}\pi \right)}\text{for }0<m+1<n$

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

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

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

$\underset{0}{\overset{a}{\int }}{x}^m(a^n-x^n{)}^{p}dx=\frac{a^{m+1+np}\mathrm{\Gamma }\left(\frac{m+1}{n}\right)\mathrm{\Gamma }(p+1)}{n\mathrm{\Gamma }(\frac{m+1}{n}+p+1)}$

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

Definite integrals involving trigonometric functions

$\underset{0}{\overset{\pi }{\int }}\sin (mx)\sin (nx)dx=\{\begin{array}{ll}0& \text{if }m\ne n\\ \\ \frac{\pi }{2}& \text{if }m=n\end{array}\text{for }m,n\text{ positive integers}$

$\underset{0}{\overset{\pi }{\int }}\cos (mx)\cos (nx)dx=\{\begin{array}{ll}0& \text{if }m\ne n\\ \\ \frac{\pi }{2}& \text{if }m=n\end{array}\text{for }m,n\text{ positive integers}$

$\underset{0}{\overset{\pi }{\int }}\sin (mx)\cos (nx)dx=\{\begin{array}{ll}0& \text{if }m+n\text{ even}\\ \\ \frac{2m}{m^2-n^2}& \text{if }m+n\text{ odd}\end{array}\text{for }m,n\text{ integers}.$

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

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

$\underset{0}{\overset{x}{\int }}{\sin }^{2m}(t)dt=\frac{(2m-1)!!}{(2m)!!}(x-\sin (x)\cos (x)(1+\sum _{k=1}^{\mathrm{\infty }}\frac{{\sin }^{2k}(x)(2k)!!}{(2k+1)!!}\left)\right)\text{for }m=1,2,3\dots$

$\underset{0}{\overset{x}{\int }}{\cos }^{2m}(t)dt=\frac{(2m-1)!!}{(2m)!!}(x-\sin (x)\cos (x)(1+\sum _{k=1}^{\mathrm{\infty }}\frac{{\cos }^{2k}(x)(2k)!!}{(2k+1)!!}\left)\right)\text{for }m=1,2,3\dots$

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

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin (px)}{x}dx=\{\begin{array}{ll}\frac{\pi }{2}& \text{if }p>0\\ \\ 0& \text{if }p=0\\ \\ -\frac{\pi }{2}& \text{if }p<0\end{array}$ (see Dirichlet integral)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin px\cos qx}{x}dx=\{\begin{array}{ll}0& \text{ if }q>p>0\\ \\ \frac{\pi }{2}& \text{ if }0<q<p\\ \\ \frac{\pi }{4}& \text{ if }p=q>0\end{array}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin px\sin qx}{x^2}dx=\{\begin{array}{ll}\frac{\pi p}{2}& \text{ if }0<p\le q\\ \\ \frac{\pi q}{2}& \text{ if }0<q\le p\end{array}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\sin }^{2}px}{x^2}dx=\frac{\pi p}{2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1-\cos px}{x^2}dx=\frac{\pi p}{2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos px-\cos qx}{x}dx=\ln \frac{q}{p}$

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

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

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin mx}{x(x^2+a^2)}dx=\frac{\pi }{2a^2}(1-e^{-ma})$

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

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

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

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

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

$\underset{0}{\overset{\pi }{\int }}\frac{x\sin xdx}{1-2a\cos x+a^2}=\{\begin{array}{ll}\frac{\pi }{a}\ln |1+a|& \text{if }|a|<1\\ \\ \frac{\pi }{a}\ln |1+\frac{1}{a}|& \text{if }|a|>1\end{array}$

$\underset{0}{\overset{\pi }{\int }}\frac{\cos mxdx}{1-2a\cos x+a^2}=\frac{\pi a^m}{1-a^2}\text{for }{a}^2<1,m=0,1,2,\dots$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin a{x}^{2}dx=\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos a{x}^2=\frac{1}{2}\sqrt{\frac{\pi }{2a}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin a{x}^{n}dx=\frac{1}{n{a}^{1/n}}\mathrm{\Gamma }\left(\frac{1}{n}\right)\sin \frac{\pi }{2n}\text{for }n>1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos a{x}^{n}dx=\frac{1}{n{a}^{1/n}}\mathrm{\Gamma }\left(\frac{1}{n}\right)\cos \frac{\pi }{2n}\text{for }n>1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x}{\sqrt{x}}dx=\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos x}{\sqrt{x}}dx=\sqrt{\frac{\pi }{2}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x}{x^p}dx=\frac{\pi }{2\mathrm{\Gamma }(p)\sin \left(\frac{p\pi }{2}\right)}\text{for }0<p<1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos x}{x^p}dx=\frac{\pi }{2\mathrm{\Gamma }(p)\cos \left(\frac{p\pi }{2}\right)}\text{for }0<p<1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin a{x}^2\cos 2bxdx=\frac{1}{2}\sqrt{\frac{\pi }{2a}}(\cos \frac{b^2}{a}-\sin \frac{b^2}{a})$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos a{x}^2\cos 2bxdx=\frac{1}{2}\sqrt{\frac{\pi }{2a}}(\cos \frac{b^2}{a}+\sin \frac{b^2}{a})$

Definite integrals involving exponential functions

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sqrt{x}{e}^{-x}dx=\frac{1}{2}\sqrt{\pi }$ (see also Gamma function)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-ax}\cos bxdx=\frac{a}{a^2+b^2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-ax}\sin bxdx=\frac{b}{a^2+b^2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}\sin bx}{x}dx={\tan }^{-1}\frac{b}{a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-e^{-bx}}{x}dx=\ln \frac{b}{a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-\cos (bx)}{x}dx=\ln \frac{b}{a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}\text{for }a>0$ (the Gaussian integral)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}\cos bxdx=\frac{1}{2}\sqrt{\frac{\pi }{a}}{e}^{\left(\frac{-b^2}{4a}\right)}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-(a{x}^2+bx+c)}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}{e}^{\left(\frac{b^2-4ac}{4a}\right)}\cdot \mathrm{erfc}\frac{b}{2\sqrt{a}},\text{ where }\mathrm{erfc}(p)=\frac{2}{\sqrt{\pi }}\underset{p}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^2}dx$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-(a{x}^2+bx+c)}dx=\sqrt{\frac{\pi }{a}}{e}^{\left(\frac{b^2-4ac}{4a}\right)}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^n{e}^{-ax}dx=\frac{\mathrm{\Gamma }(n+1)}{a^{n+1}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-a{x}^2}dx=\frac{1}{4}\sqrt{\frac{\pi }{a^3}}\text{for }a>0$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{2n}{e}^{-a{x}^2}dx=\frac{2n-1}{2a}\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{2(n-1)}{e}^{-a{x}^2}dx=\frac{(2n-1)!!}{2^{n+1}}\sqrt{\frac{\pi }{a^{2n+1}}}=\frac{(2n)!}{n!2^{2n+1}}\sqrt{\frac{\pi }{a^{2n+1}}}\text{for }a>0,n=1,2,3\dots$ (where !! is the double factorial)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^3{e}^{-a{x}^2}dx=\frac{1}{2a^2}\text{for }a>0$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{2n+1}{e}^{-a{x}^2}dx=\frac{n}{a}\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{2n-1}{e}^{-a{x}^2}dx=\frac{n!}{2a^{n+1}}\text{for }a>0,n=0,1,2\dots$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^m{e}^{-a{x}^2}dx=\frac{\mathrm{\Gamma }\left(\frac{m+1}{2}\right)}{2a^{\left(\frac{m+1}{2}\right)}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^n{e}^{-a{x}^b}dx=\frac{1}{b}{a}^{-\frac{n+1}{b}}\mathrm{\Gamma }\left(\frac{n+1}{b}\right)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{(-a{x}^2-\frac{b}{x^2})}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}{e}^{-2\sqrt{ab}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{e^x-1}dx=\zeta (2)=\frac{{\pi }^2}{6}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{n-1}}{e^x-1}dx=\mathrm{\Gamma }(n)\zeta (n)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{e^x+1}dx=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\dots =\frac{{\pi }^2}{12}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^n}{e^x+1}dx=n!\cdot \left(\frac{2^n-1}{2^n}\right)\zeta (n+1)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin mx}{e^{2\pi x}-1}dx=\frac{1}{4}\coth \frac{m}{2}-\frac{1}{2m}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{1+x}-e^{-x})\frac{dx}{x}=\gamma$ (where $\gamma$ is Euler–Mascheroni constant)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-x^2}-e^{-x}}{x}dx=\frac{\gamma }{2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{e^x-1}-\frac{e^{-x}}{x})dx=\gamma$

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

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

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

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^2}dx=\sqrt{\pi }$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^{2(n+1)}{e}^{-\frac{1}{2}{x}^2}dx=\frac{(2n+1)!}{2^{n}n!}\sqrt{2\pi }\text{for }n=0,1,2,\dots$

Definite integrals involving logarithmic functions

$\underset{0}{\overset{1}{\int }}{x}^m(\ln x{)}^{n}dx=\frac{(-1{)}^{n}n!}{(m+1{)}^{n+1}}\text{for }m>-1,n=0,1,2,\dots$

$\underset{1}{\overset{\mathrm{\infty }}{\int }}{x}^m(\ln x{)}^{n}dx=\frac{(-1{)}^{n+1}n!}{(m+1{)}^{n+1}}\text{for }m<-1,n=0,1,2,\dots$

$\underset{0}{\overset{1}{\int }}\frac{\ln x}{1+x}dx=-\frac{{\pi }^2}{12}$

$\underset{0}{\overset{1}{\int }}\frac{\ln x}{1-x}dx=-\frac{{\pi }^2}{6}$

$\underset{0}{\overset{1}{\int }}\frac{\ln (1+x)}{x}dx=\frac{{\pi }^2}{12}$

$\underset{0}{\overset{1}{\int }}\frac{\ln (1-x)}{x}dx=-\frac{{\pi }^2}{6}$

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\ln x}{x^2+a^2}dx=\frac{\pi \ln a}{2a}\text{for }a>0$

Definite integrals involving hyperbolic functions

$$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin ax}{\sinh bx}dx=\frac{\pi }{2b}\tanh \frac{a\pi }{2b}$$

$$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos ax}{\cosh bx}dx=\frac{\pi }{2b}\cdot \frac{1}{\cosh \frac{a\pi }{2b}}$$

$$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{\sinh ax}dx=\frac{{\pi }^2}{4a^2}$$

$$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{2n+1}}{\sinh ax}dx=c_{2n+1}{\left(\frac{\pi }{a}\right)}^{2(n+1)},c_{2n+1}=\frac{(-1{)}^n}{2}(\frac{1}{2}-\sum _{k=0}^{n-1}(-1{)}^k(\genfrac{}{}{0ex}{}{2n+1}{2k+1})c_{2k+1}),c_1=\frac{1}{4}$$

$$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\cosh ax}dx=\frac{\pi }{2a}$$

$$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{2n}}{\cosh ax}dx=d_{2n}{\left(\frac{\pi }{a}\right)}^{2n+1},d_{2n}=\frac{(-1{)}^n}{2}(\frac{1}{4^n}-\sum _{k=0}^{n-1}(-1{)}^k(\genfrac{}{}{0ex}{}{2n}{2k})d_{2k}),d_0=\frac{1}{2}$$

Frullani integrals

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

See also

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

References

• "Derivation of Logarithmic and Logarithmic Hyperbolic Tangent Integrals Expressed in Terms of Special Functions". Mathematics 8 (687): 687. 2020. doi:10.3390/math8050687. 
• "A Definite Integral Involving the Logarithmic Function in Terms of the Lerch Function". Mathematics 7 (1148): 1148. 2019. doi:10.3390/math7121148. 
• "Definite Integral of Arctangent and Polylogarithmic Functions Expressed as a Series". Mathematics 7 (1099): 1099. 2019. doi:10.3390/math7111099. 
• "Eigenschaften Einiger Bestimmten Integrale". Hof, K.K., Ed.. 1861. 
• Mathematical handbook of formulas and tables (3rd ed.). McGraw-Hill. 2009. ISBN 978-0071548557. 
• CRC standard mathematical tables and formulae (32nd ed.). CRC Press. 2003. ISBN 978-143983548-7. 
• Abramowitz, Milton; Stegun, Irene Ann, eds (1983). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Applied Mathematics Series. 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. LCCN 65-12253. ISBN 978-0-486-61272-0.