Integration is the basic operation inintegral calculus . Whiledifferentiation has straightforwardrules by which the derivative of a complicatedfunction can be found by differentiating its simpler component functions, integration does not, so tables of known integrals are often useful. This page lists some of the most commonantiderivatives .
Historical development of integrals [ edit ] A compilation of a list of integrals (Integraltafeln) and techniques of integral calculus was published by the German mathematicianMeier Hirsch [de ] (also spelled Meyer Hirsch) in 1810.[ 1] David Bierens de Haan for hisTables d'intégrales définies Supplément aux tables d'intégrales définies Nouvelles tables d'intégrales définies
These tables, which contain mainly integrals of elementary functions, remained in use until the middle of the 20th century. They were then replaced by the much more extensive tables ofGradshteyn and Ryzhik . In Gradshteyn and Ryzhik, integrals originating from the book by Bierens de Haan are denoted by BI.
Not allclosed-form expressions have closed-form antiderivatives; this study forms the subject ofdifferential Galois theory , which was initially developed byJoseph Liouville in the 1830s and 1840s, leading toLiouville's theorem which classifies which expressions have closed-form antiderivatives. A simple example of a function without a closed-form antiderivative ise −x 2 error function .
Since 1968 there is theRisch algorithm for determining indefinite integrals that can be expressed in term ofelementary functions , typically using acomputer algebra system . Integrals that cannot be expressed using elementary functions can be manipulated symbolically using general functions such as theMeijer G-function .
More detail may be found on the following pages for thelists of integrals :
Gradshteyn ,Ryzhik ,Geronimus ,Tseytlin , Jeffrey, Zwillinger, andMoll 's (GR)Table of Integrals, Series, and Products Integrals and Series byPrudnikov ,Brychkov , andMarichev (with volumes 1–3 listing integrals and series ofelementary andspecial functions , volume 4–5 are tables ofLaplace transforms ). More compact collections can be found in e.g. Brychkov, Marichev, Prudnikov'sTables of Indefinite Integrals , or as chapters in Zwillinger'sCRC Standard Mathematical Tables and Formulae orBronshtein and Semendyayev 'sGuide Book to Mathematics Handbook of Mathematics Users' Guide to Mathematics
Other useful resources includeAbramowitz and Stegun and theBateman Manuscript Project . Both works contain many identities concerning specific integrals, which are organized with the most relevant topic instead of being collected into a separate table. Two volumes of the Bateman Manuscript are specific to integral transforms.
There are several web sites which have tables of integrals and integrals on demand.Wolfram Alpha can show results, and for some simpler expressions, also the intermediate steps of the integration.Wolfram Research also operates another online service, the Mathematica Online Integrator.
Integrals of simple functions [ edit ] C is used for anarbitrary constant of integration that can only be determined if something about the value of the integral at some point is known. Thus, each function has an infinite number ofantiderivatives .
These formulas only state in another form the assertions in thetable of derivatives .
Integrals with a singularity [ edit ] When there is asingularity in the function being integrated such that the antiderivative becomes undefined at some point (the singularity), thenC does not need to be the same on both sides of the singularity. The forms below normally assume theCauchy principal value around a singularity in the value ofC , but this is not necessary in general. For instance, in∫ 1 x d x = ln | x | + C {\displaystyle \int {1 \over x}\,dx=\ln \left|x\right|+C} antiderivative becomes infinite there. If the integral above were to be used to compute a definite integral between −1 and 1, one would get the wrong answer 0. This however is the Cauchy principal value of the integral around the singularity. If the integration is done in the complex plane the result depends on the path around the origin, in this case the singularity contributes −i π when using a path above the origin andi π for a path below the origin. A function on the real line could use a completely different value ofC on either side of the origin as in:[ 2] ∫ 1 x d x = ln | x | + { A if x > 0 ; B if x < 0. {\displaystyle \int {1 \over x}\,dx=\ln |x|+{\begin{cases}A&{\text{if }}x>0;\\B&{\text{if }}x<0.\end{cases}}}
The following function has a non-integrable singularity at 0 forn ≤ −1
Exponential functions [ edit ] Trigonometric functions [ edit ] ∫ sin x d x = − cos x + C {\displaystyle \int \sin x\,dx=-\cos x+C} ∫ cos x d x = sin x + C {\displaystyle \int \cos x\,dx=\sin x+C} ∫ tan x d x = ln | sec x | + C = − ln | cos x | + C {\displaystyle \int \tan x\,dx=\ln \left|\sec x\right|+C=-\ln \left|\cos x\right|+C} ∫ cot x d x = − ln | csc x | + C = ln | sin x | + C {\displaystyle \int \cot x\,dx=-\ln \left|\csc x\right|+C=\ln \left|\sin x\right|+C} ∫ sec x d x = ln | sec x + tan x | + C = ln | tan ( x 2 + π 4 ) | + C {\displaystyle \int \sec x\,dx=\ln \left|\sec x+\tan x\right|+C=\ln \left|\tan \left({\dfrac {x}{2}}+{\dfrac {\pi }{4}}\right)\right|+C} ∫ csc x d x = − ln | csc x + cot x | + C = ln | csc x − cot x | + C = ln | tan x 2 | + C {\displaystyle \int \csc x\,dx=-\ln \left|\csc x+\cot x\right|+C=\ln \left|\csc x-\cot x\right|+C=\ln \left|\tan {\frac {x}{2}}\right|+C} ∫ sec 2 x d x = tan x + C {\displaystyle \int \sec ^{2}x\,dx=\tan x+C} ∫ csc 2 x d x = − cot x + C {\displaystyle \int \csc ^{2}x\,dx=-\cot x+C} ∫ sec x tan x d x = sec x + C {\displaystyle \int \sec x\,\tan x\,dx=\sec x+C} ∫ csc x cot x d x = − csc x + C {\displaystyle \int \csc x\,\cot x\,dx=-\csc x+C} ∫ sin 2 x d x = 1 2 ( x − sin 2 x 2 ) + C = 1 2 ( x − sin x cos x ) + C {\displaystyle \int \sin ^{2}x\,dx={\frac {1}{2}}\left(x-{\frac {\sin 2x}{2}}\right)+C={\frac {1}{2}}(x-\sin x\cos x)+C} ∫ cos 2 x d x = 1 2 ( x + sin 2 x 2 ) + C = 1 2 ( x + sin x cos x ) + C {\displaystyle \int \cos ^{2}x\,dx={\frac {1}{2}}\left(x+{\frac {\sin 2x}{2}}\right)+C={\frac {1}{2}}(x+\sin x\cos x)+C} ∫ tan 2 x d x = tan x − x + C {\displaystyle \int \tan ^{2}x\,dx=\tan x-x+C} ∫ cot 2 x d x = − cot x − x + C {\displaystyle \int \cot ^{2}x\,dx=-\cot x-x+C} ∫ sec 3 x d x = 1 2 ( sec x tan x + ln | sec x + tan x | ) + C {\displaystyle \int \sec ^{3}x\,dx={\frac {1}{2}}(\sec x\tan x+\ln \left|\sec x+\tan x\right|)+C} ∫ csc 3 x d x = 1 2 ( − csc x cot x + ln | csc x − cot x | ) + C = 1 2 ( ln | tan x 2 | − csc x cot x ) + C {\displaystyle \int \csc ^{3}x\,dx={\frac {1}{2}}(-\csc x\cot x+\ln \left|\csc x-\cot x\right|)+C={\frac {1}{2}}\left(\ln \left|\tan {\frac {x}{2}}\right|-\csc x\cot x\right)+C} ∫ sin n x d x = − sin n − 1 x cos x n + n − 1 n ∫ sin n − 2 x d x {\displaystyle \int \sin ^{n}x\,dx=-{\frac {\sin ^{n-1}x\cos x}{n}}+{\frac {n-1}{n}}\int \sin ^{n-2}x\,dx} ∫ cos n x d x = cos n − 1 x sin x n + n − 1 n ∫ cos n − 2 x d x {\displaystyle \int \cos ^{n}x\,dx={\frac {\cos ^{n-1}x\sin x}{n}}+{\frac {n-1}{n}}\int \cos ^{n-2}{x}\,dx} Inverse trigonometric functions [ edit ] ∫ arcsin x d x = x arcsin x + 1 − x 2 + C , for | x | ≤ 1 {\displaystyle \int \arcsin {x}\,dx=x\arcsin {x}+{\sqrt {1-x^{2}}}+C,{\text{ for }}\vert x\vert \leq 1} ∫ arccos x d x = x arccos x − 1 − x 2 + C , for | x | ≤ 1 {\displaystyle \int \arccos {x}\,dx=x\arccos {x}-{\sqrt {1-x^{2}}}+C,{\text{ for }}\vert x\vert \leq 1} ∫ arctan x d x = x arctan x − 1 2 ln | 1 + x 2 | + C , for all real x {\displaystyle \int \arctan {x}\,dx=x\arctan {x}-{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C,{\text{ for all real }}x} ∫ arccot x d x = x arccot x + 1 2 ln | 1 + x 2 | + C , for all real x {\displaystyle \int \operatorname {arccot} {x}\,dx=x\operatorname {arccot} {x}+{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C,{\text{ for all real }}x} ∫ arcsec x d x = x arcsec x − ln | x ( 1 + 1 − x − 2 ) | + C , for | x | ≥ 1 {\displaystyle \int \operatorname {arcsec} {x}\,dx=x\operatorname {arcsec} {x}-\ln \left\vert x\,\left(1+{\sqrt {1-x^{-2}}}\,\right)\right\vert +C,{\text{ for }}\vert x\vert \geq 1} ∫ arccsc x d x = x arccsc x + ln | x ( 1 + 1 − x − 2 ) | + C , for | x | ≥ 1 {\displaystyle \int \operatorname {arccsc} {x}\,dx=x\operatorname {arccsc} {x}+\ln \left\vert x\,\left(1+{\sqrt {1-x^{-2}}}\,\right)\right\vert +C,{\text{ for }}\vert x\vert \geq 1} Hyperbolic functions [ edit ] ∫ sinh x d x = cosh x + C {\displaystyle \int \sinh x\,dx=\cosh x+C} ∫ cosh x d x = sinh x + C {\displaystyle \int \cosh x\,dx=\sinh x+C} ∫ tanh x d x = ln ( cosh x ) + C {\displaystyle \int \tanh x\,dx=\ln(\cosh x)+C} ∫ coth x d x = ln | sinh x | + C , for x ≠ 0 {\displaystyle \int \coth x\,dx=\ln \left|\sinh x\right|+C,{\text{ for }}x\neq 0} ∫ sech x d x = arctan ( sinh x ) + C {\displaystyle \int \operatorname {sech} x\,dx=\arctan(\sinh x)+C} ∫ csch x d x = ln | coth x − csch x | + C = ln | tanh x 2 | + C , for x ≠ 0 {\displaystyle \int \operatorname {csch} \,x\,dx=\ln \left|\operatorname {coth} x-\operatorname {csch} x\right|+C=\ln \left|\tanh {x \over 2}\right|+C,{\text{ for }}x\neq 0} ∫ sech 2 x d x = tanh x + C {\displaystyle \int \operatorname {sech} ^{2}x\,dx=\tanh x+C} ∫ csch 2 x d x = − coth x + C {\displaystyle \int \operatorname {csch} ^{2}x\,dx=-\operatorname {coth} x+C} ∫ sech x tanh x d x = − sech x + C {\displaystyle \int \operatorname {sech} x\,\operatorname {tanh} x\,dx=-\operatorname {sech} x+C} ∫ csch x coth x d x = − csch x + C {\displaystyle \int \operatorname {csch} x\,\operatorname {coth} x\,dx=-\operatorname {csch} x+C} Inverse hyperbolic functions [ edit ] ∫ arcsinh x d x = x arcsinh x − x 2 + 1 + C , for all real x {\displaystyle \int \operatorname {arcsinh} \,x\,dx=x\,\operatorname {arcsinh} \,x-{\sqrt {x^{2}+1}}+C,{\text{ for all real }}x} ∫ arccosh x d x = x arccosh x − x 2 − 1 + C , for x ≥ 1 {\displaystyle \int \operatorname {arccosh} \,x\,dx=x\,\operatorname {arccosh} \,x-{\sqrt {x^{2}-1}}+C,{\text{ for }}x\geq 1} ∫ arctanh x d x = x arctanh x + ln ( 1 − x 2 ) 2 + C , for | x | < 1 {\displaystyle \int \operatorname {arctanh} \,x\,dx=x\,\operatorname {arctanh} \,x+{\frac {\ln \left(\,1-x^{2}\right)}{2}}+C,{\text{ for }}\vert x\vert <1} ∫ arccoth x d x = x arccoth x + ln ( x 2 − 1 ) 2 + C , for | x | > 1 {\displaystyle \int \operatorname {arccoth} \,x\,dx=x\,\operatorname {arccoth} \,x+{\frac {\ln \left(x^{2}-1\right)}{2}}+C,{\text{ for }}\vert x\vert >1} ∫ arcsech x d x = x arcsech x + arcsin x + C , for 0 < x ≤ 1 {\displaystyle \int \operatorname {arcsech} \,x\,dx=x\,\operatorname {arcsech} \,x+\arcsin x+C,{\text{ for }}0<x\leq 1} ∫ arccsch x d x = x arccsch x + | arcsinh x | + C , for x ≠ 0 {\displaystyle \int \operatorname {arccsch} \,x\,dx=x\,\operatorname {arccsch} \,x+\left|\operatorname {arcsinh} \,x\right|+C,{\text{ for }}x\neq 0} Products of functions proportional to their second derivatives [ edit ] ∫ cos a x e b x d x = e b x a 2 + b 2 ( a sin a x + b cos a x ) + C {\displaystyle \int \cos ax\,e^{bx}\,dx={\frac {e^{bx}}{a^{2}+b^{2}}}\left(a\sin ax+b\cos ax\right)+C} ∫ sin a x e b x d x = e b x a 2 + b 2 ( b sin a x − a cos a x ) + C {\displaystyle \int \sin ax\,e^{bx}\,dx={\frac {e^{bx}}{a^{2}+b^{2}}}\left(b\sin ax-a\cos ax\right)+C} ∫ cos a x cosh b x d x = 1 a 2 + b 2 ( a sin a x cosh b x + b cos a x sinh b x ) + C {\displaystyle \int \cos ax\,\cosh bx\,dx={\frac {1}{a^{2}+b^{2}}}\left(a\sin ax\,\cosh bx+b\cos ax\,\sinh bx\right)+C} ∫ sin a x cosh b x d x = 1 a 2 + b 2 ( b sin a x sinh b x − a cos a x cosh b x ) + C {\displaystyle \int \sin ax\,\cosh bx\,dx={\frac {1}{a^{2}+b^{2}}}\left(b\sin ax\,\sinh bx-a\cos ax\,\cosh bx\right)+C} Absolute-value functions [ edit ] Letf continuous function , that has at most onezero . Iff g f f g f ∫ | f ( x ) | d x = sgn ( f ( x ) ) g ( x ) + C , {\displaystyle \int \left|f(x)\right|\,dx=\operatorname {sgn}(f(x))g(x)+C,} sgn(x ) is thesign function , which takes the values −1, 0, 1 whenx
This can be proved by computing the derivative of the right-hand side of the formula, taking into account that the condition ong
This gives the following formulas (wherea ≠ 0f C must be replaced by apiecewise constant function):
∫ | ( a x + b ) n | d x = sgn ( a x + b ) ( a x + b ) n + 1 a ( n + 1 ) + C {\displaystyle \int \left|(ax+b)^{n}\right|\,dx=\operatorname {sgn}(ax+b){(ax+b)^{n+1} \over a(n+1)}+C}
whenn n ≠ − 1 {\displaystyle n\neq -1} ∫ | tan a x | d x = − 1 a sgn ( tan a x ) ln ( | cos a x | ) + C {\displaystyle \int \left|\tan {ax}\right|\,dx=-{\frac {1}{a}}\operatorname {sgn}(\tan {ax})\ln(\left|\cos {ax}\right|)+C}
whena x ∈ ( n π − π 2 , n π + π 2 ) {\textstyle ax\in \left(n\pi -{\frac {\pi }{2}},n\pi +{\frac {\pi }{2}}\right)} n ∫ | csc a x | d x = − 1 a sgn ( csc a x ) ln ( | csc a x + cot a x | ) + C {\displaystyle \int \left|\csc {ax}\right|\,dx=-{\frac {1}{a}}\operatorname {sgn}(\csc {ax})\ln(\left|\csc {ax}+\cot {ax}\right|)+C}
whena x ∈ ( n π , n π + π ) {\displaystyle ax\in \left(n\pi ,n\pi +\pi \right)} n ∫ | sec a x | d x = 1 a sgn ( sec a x ) ln ( | sec a x + tan a x | ) + C {\displaystyle \int \left|\sec {ax}\right|\,dx={\frac {1}{a}}\operatorname {sgn}(\sec {ax})\ln(\left|\sec {ax}+\tan {ax}\right|)+C}
whena x ∈ ( n π − π 2 , n π + π 2 ) {\textstyle ax\in \left(n\pi -{\frac {\pi }{2}},n\pi +{\frac {\pi }{2}}\right)} n ∫ | cot a x | d x = 1 a sgn ( cot a x ) ln ( | sin a x | ) + C {\displaystyle \int \left|\cot {ax}\right|\,dx={\frac {1}{a}}\operatorname {sgn}(\cot {ax})\ln(\left|\sin {ax}\right|)+C}
whena x ∈ ( n π , n π + π ) {\displaystyle ax\in \left(n\pi ,n\pi +\pi \right)} n If the functionf f sgn(f (x )) ∫f (x )dx is an antiderivative off interval on whichf f (x ) = 0step function . If we also use the fact that the absolute values of sine and cosine are periodic with periodπ , then we get:
Ci ,Si :Trigonometric integrals ,Ei :Exponential integral ,li :Logarithmic integral function ,erf :Error function
Definite integrals lacking closed-form antiderivatives [ edit ] There are some functions whose antiderivativescannot be expressed inclosed form . However, the values of the definite integrals of some of these functions over some common intervals can be calculated. A few useful integrals are given below.
∫ 0 ∞ x e − x d x = 1 2 π {\displaystyle \int _{0}^{\infty }{\sqrt {x}}\,e^{-x}\,dx={\frac {1}{2}}{\sqrt {\pi }}} Gamma function )∫ 0 ∞ e − a x 2 d x = 1 2 π a {\displaystyle \int _{0}^{\infty }e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}} a > 0Gaussian integral )∫ 0 ∞ x 2 e − a x 2 d x = 1 4 π a 3 {\displaystyle \int _{0}^{\infty }{x^{2}e^{-ax^{2}}\,dx}={\frac {1}{4}}{\sqrt {\frac {\pi }{a^{3}}}}} 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 {\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}}}}}
fora > 0n !! is thedouble factorial .∫ 0 ∞ x 3 e − a x 2 d x = 1 2 a 2 {\displaystyle \int _{0}^{\infty }{x^{3}e^{-ax^{2}}\,dx}={\frac {1}{2a^{2}}}} 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 {\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}}}}
fora > 0n = 0, 1, 2, ....∫ 0 ∞ x n e − a x b d x = 1 b a − n + 1 b Γ ( n + 1 b ) {\displaystyle \int _{0}^{\infty }x^{n}e^{-ax^{b}}dx={\frac {1}{b}}\ a^{-{\frac {n+1}{b}}}\Gamma \left({\frac {n+1}{b}}\right)} ∫ 0 ∞ x e x − 1 d x = π 2 6 {\displaystyle \int _{0}^{\infty }{\frac {x}{e^{x}-1}}\,dx={\frac {\pi ^{2}}{6}}} Bernoulli number )∫ 0 ∞ x 2 e x − 1 d x = 2 ζ ( 3 ) ≈ 2.40 {\displaystyle \int _{0}^{\infty }{\frac {x^{2}}{e^{x}-1}}\,dx=2\zeta (3)\approx 2.40} ∫ 0 ∞ x 3 e x − 1 d x = π 4 15 {\displaystyle \int _{0}^{\infty }{\frac {x^{3}}{e^{x}-1}}\,dx={\frac {\pi ^{4}}{15}}} Planck's law in physics)∫ 0 ∞ x n e x − 1 d x = Γ ( n + 1 ) ζ ( n + 1 ) {\displaystyle \int _{0}^{\infty }{\frac {x^{n}}{e^{x}-1}}\,dx=\Gamma (n+1)\zeta (n+1)} n > 0 {\displaystyle n>0} Riemann zeta function )∫ 0 ∞ sin x x d x = π 2 {\displaystyle \int _{0}^{\infty }{\frac {\sin {x}}{x}}\,dx={\frac {\pi }{2}}} sinc function and theDirichlet integral )∫ 0 ∞ sin 2 x x 2 d x = π 2 {\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}{x}}{x^{2}}}\,dx={\frac {\pi }{2}}} ∫ 0 π 2 sin n x d x = ∫ 0 π 2 cos n x d x = ( n − 1 ) ! ! n ! ! × { 1 if n is odd π 2 if n is even. {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}x\,dx={\frac {(n-1)!!}{n!!}}\times {\begin{cases}1&{\text{if }}n{\text{ is odd}}\\{\frac {\pi }{2}}&{\text{if }}n{\text{ is even.}}\end{cases}}}
(ifn double factorial ).∫ − π π cos ( α x ) cos n ( β x ) d x = { 2 π 2 n ( n m ) | α | = | β ( 2 m − n ) | 0 otherwise {\displaystyle \int _{-\pi }^{\pi }\cos(\alpha x)\cos ^{n}(\beta x)dx={\begin{cases}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&|\alpha |=|\beta (2m-n)|\\0&{\text{otherwise}}\end{cases}}}
(forα ,β ,m ,n β ≠ 0m ,n ≥ 0Binomial coefficient )∫ − t t sin m ( α x ) cos n ( β x ) d x = 0 {\displaystyle \int _{-t}^{t}\sin ^{m}(\alpha x)\cos ^{n}(\beta x)dx=0}
(forα ,β n m an odd, positive integer; since the integrand isodd )∫ − π π sin ( α x ) sin n ( β x ) d x = { ( − 1 ) ( n + 1 2 ) ( − 1 ) m 2 π 2 n ( n m ) n odd , α = β ( 2 m − n ) 0 otherwise {\displaystyle \int _{-\pi }^{\pi }\sin(\alpha x)\sin ^{n}(\beta x)dx={\begin{cases}(-1)^{\left({\frac {n+1}{2}}\right)}(-1)^{m}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&n{\text{ odd}},\ \alpha =\beta (2m-n)\\0&{\text{otherwise}}\end{cases}}}
(forα ,β ,m ,n β ≠ 0m ,n ≥ 0Binomial coefficient )∫ − π π cos ( α x ) sin n ( β x ) d x = { ( − 1 ) ( n 2 ) ( − 1 ) m 2 π 2 n ( n m ) n even , | α | = | β ( 2 m − n ) | 0 otherwise {\displaystyle \int _{-\pi }^{\pi }\cos(\alpha x)\sin ^{n}(\beta x)dx={\begin{cases}(-1)^{\left({\frac {n}{2}}\right)}(-1)^{m}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&n{\text{ even}},\ |\alpha |=|\beta (2m-n)|\\0&{\text{otherwise}}\end{cases}}}
(forα ,β ,m ,n β ≠ 0m ,n ≥ 0Binomial coefficient )∫ − ∞ ∞ e − ( a x 2 + b x + c ) d x = π a exp [ b 2 − 4 a c 4 a ] {\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\frac {\pi }{a}}}\exp \left[{\frac {b^{2}-4ac}{4a}}\right]}
(whereexp[u ] is theexponential function e u a > 0∫ 0 ∞ x z − 1 e − x d x = Γ ( z ) {\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-x}\,dx=\Gamma (z)}
(whereΓ ( z ) {\displaystyle \Gamma (z)} Gamma function )∫ 0 1 ( ln 1 x ) p d x = Γ ( p + 1 ) {\displaystyle \int _{0}^{1}\left(\ln {\frac {1}{x}}\right)^{p}\,dx=\Gamma (p+1)} ∫ 0 1 x α − 1 ( 1 − x ) β − 1 d x = Γ ( α ) Γ ( β ) Γ ( α + β ) {\displaystyle \int _{0}^{1}x^{\alpha -1}(1-x)^{\beta -1}dx={\frac {\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}}
(forRe(α ) > 0 andRe(β ) > 0 , seeBeta function )∫ 0 2 π e x cos θ d θ = 2 π I 0 ( x ) {\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)} I 0 (x )Bessel function of the first kind)∫ 0 2 π e x cos θ + y sin θ d θ = 2 π I 0 ( x 2 + y 2 ) {\displaystyle \int _{0}^{2\pi }e^{x\cos \theta +y\sin \theta }d\theta =2\pi I_{0}\left({\sqrt {x^{2}+y^{2}}}\right)} ∫ − ∞ ∞ ( 1 + x 2 ν ) − ν + 1 2 d x = ν π Γ ( ν 2 ) Γ ( ν + 1 2 ) {\displaystyle \int _{-\infty }^{\infty }\left(1+{\frac {x^{2}}{\nu }}\right)^{-{\frac {\nu +1}{2}}}\,dx={\frac {{\sqrt {\nu \pi }}\ \Gamma \left({\frac {\nu }{2}}\right)}{\Gamma \left({\frac {\nu +1}{2}}\right)}}}
(forν > 0probability density function ofStudent'st -distribution )If the functionf bounded variation on the interval[a ,b ] , then themethod of exhaustion provides a formula for the integral:∫ a b f ( x ) d x = ( b − a ) ∑ n = 1 ∞ ∑ m = 1 2 n − 1 ( − 1 ) m + 1 2 − n f ( a + m ( b − a ) 2 − n ) . {\displaystyle \int _{a}^{b}{f(x)\,dx}=(b-a)\sum \limits _{n=1}^{\infty }{\sum \limits _{m=1}^{2^{n}-1}{\left({-1}\right)^{m+1}}}2^{-n}f(a+m\left({b-a}\right)2^{-n}).}
The "sophomore's dream ":∫ 0 1 x − x d x = ∑ n = 1 ∞ n − n ( = 1.29128 59970 6266 … ) ∫ 0 1 x x d x = − ∑ n = 1 ∞ ( − n ) − n ( = 0.78343 05107 1213 … ) {\displaystyle {\begin{aligned}\int _{0}^{1}x^{-x}\,dx&=\sum _{n=1}^{\infty }n^{-n}&&(=1.29128\,59970\,6266\dots )\\[6pt]\int _{0}^{1}x^{x}\,dx&=-\sum _{n=1}^{\infty }(-n)^{-n}&&(=0.78343\,05107\,1213\dots )\end{aligned}}} Johann Bernoulli .
Abramowitz, Milton ;Stegun, Irene Ann , eds. (1983) [June 1964].Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables ISBN 978-0-486-61272-0 LCCN 64-60036 .MR 0167642 .LCCN 65-12253 .Bronstein, Ilja Nikolaevič; Semendjajew, Konstantin Adolfovič (1987) [1945]. Grosche, Günter; Ziegler, Viktor; Ziegler, Dorothea (eds.).Taschenbuch der Mathematik Verlag Harri Deutsch (andB. G. Teubner Verlagsgesellschaft , Leipzig).ISBN 3-87144-492-8 Gradshteyn, Izrail Solomonovich ;Ryzhik, Iosif Moiseevich ;Geronimus, Yuri Veniaminovich ;Tseytlin, Michail Yulyevich ; Jeffrey, Alan (2015) [October 2014]. Zwillinger, Daniel; Moll, Victor Hugo (eds.).Table of Integrals, Series, and Products Academic Press, Inc. ISBN 978-0-12-384933-5 LCCN 2014010276 .Prudnikov, Anatolii Platonovich (Прудников, Анатолий Платонович) ; Brychkov, Yuri A. (Брычков, Ю. А.);Marichev, Oleg Igorevich (Маричев, Олег Игоревич) (1988–1992) [1981−1986 (Russian)].Integrals and Series . Vol. 1– 5. Translated by Queen, N. M. (1 ed.). (Nauka ) Gordon & Breach Science Publishers/CRC Press .ISBN 2-88124-097-6 {{cite book }}: CS1 maint: multiple names: authors list (link )Yuri A. Brychkov (Ю. А. Брычков),Handbook of Special Functions: Derivatives, Integrals, Series and Other Formulas . Russian edition, Fiziko-Matematicheskaya Literatura, 2006. English edition, Chapman & Hall/CRC Press, 2008,ISBN 1-58488-956-X Daniel Zwillinger.CRC Standard Mathematical Tables and Formulae , 31st edition. Chapman & Hall/CRC Press, 2002.ISBN 1-58488-291-3 (Many earlier editions as well.) Meyer Hirsch [de ] ,Integraltafeln oder Sammlung von Integralformeln Meyer Hirsch [de ] ,Integral Tables Or A Collection of Integral Formulae Integraltafeln ]David Bierens de Haan ,Nouvelles Tables d'Intégrales définies (Engels, Leiden, 1862)Benjamin O. PierceA short table of integrals - revised edition (Ginn & co., Boston, 1899) Tables of integrals [ edit ] Open source programs [ edit ]
