Thursday, May 27, 2010

INTEGRATION FORMULAS

Integrals with a singularity

When there is a singularity in the function being integrated such that the integral becomes undefined, it is not , then C does not need to be the same on both sides of the singularity. The forms below normally assume the a singularity in the value of C but this is not in general necessary. For instance in

\int {1 \over x}\,dx = \ln \left|x \right| + C

There is a singularity at 0 and the integral becomes infinite there. If the integral above was used to give a definite integral between -1 and 1 the answer would be 0. This however is only the value assuming the Cauchy principal value for the integral around the singularity. If the integration was done in the complex plane the result would depend on the path round the origin, in this case the singularity contributes −iπ when using a path above the origin and iπ for a path below the origin. A function on the real line could use a completely different value of C on either side of the origin as Rational functions


These rational functions have a non-integrable singularity at 0 for a ≤ −1.

\int a\,dx = ax + C Exponential functions
more integrals:exponitial funmtion
\int e^x\,dx = e^x + C
\int a^x\,dx = \frac{a^x}{\ln a} + C

Logarithms

more integrals:logrthmic funtion
\int \ln x\,dx = x \ln x - x + C
\int \log_a x\,dx = x\log_a x - \frac{x}{\ln a} + C

Trigonometric functions

more integrals:trigonomertic funtion
\int \sin{x}\, dx = -\cos{x} + C
\int \cos{x}\, dx = \sin{x} + C
\int \tan{x} \, dx = -\ln{\left| \cos {x} \right|} + C = \ln{\left| \sec{x} \right|} + C
\int \cot{x} \, dx = \ln{\left| \sin{x} \right|} + C
\int \sec{x} \, dx = \ln{\left| \sec{x} + \tan{x}\right|} + C
\int \csc{x} \, dx = \ln{\left| \csc{x} - \cot{x}\right|} + C
\int \sec^2 x \, dx = \tan x + C
\int \csc^2 x \, dx = -\cot x + C
\int \sec{x} \, \tan{x} \, dx = \sec{x} + C
\int \csc{x} \, \cot{x} \, dx = -\csc{x} + C
\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
\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
\int \sec^3 x \, dx = \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C

\int \sin^n x \, dx = - \frac{\sin^{n-1} {x} \cos {x}}{n} + \frac{n-1}{n} \int \sin^{n-2}{x} \, dx
\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


\int \arcsin{x} \, dx = x \, \arcsin{x} + \sqrt{1 - x^2} + C
\int \arccos{x} \, dx = x \, \arccos{x} - \sqrt{1 - x^2} + C
\int \arctan{x} \, dx = x \, \arctan{x} - \frac{1}{2} \ln{\left| 1 + x^2\right|} + C
\int \arccot{x} \, dx = x \, \arccot{x} + \frac{1}{2} \ln{\left| 1 + x^2\right|} + C
\int \arcsec{x} \, dx = x \, \arcsec{x} - \operatorname{arcosh} \, x + C
\int \arccsc{x} \, dx = x \, \arccsc{x} + \operatorname{arcosh} \, x + C

Hyperbolic functions


\int \sinh x \, dx = \cosh x + C
\int \cosh x \, dx = \sinh x + C
\int \tanh x \, dx = \ln| \cosh x | + C
\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C
\int \mbox{sech}\,x \, dx = \arcsin\,(\tanh x) + C
\int \coth x \, dx = \ln| \sinh x | + C

Inverse hyperbolic functions


\int \operatorname{arsinh}\, x \, dx = x\, \operatorname{arsinh}\, x - \sqrt{x^2+1} + C
\int \operatorname{arcosh}\, x \, dx = x\, \operatorname{arcosh}\, x - \sqrt{x^2-1} + C
\int \operatorname{artanh}\, x \, dx = x\, \operatorname{artanh}\, x + \frac{1}{2}\ln{(1-x^2)} + C
\int \operatorname{arcsch}\,x \, dx = x\, \operatorname{arcsch}\, x+ \ln{\left[x\left(\sqrt{1+\frac{1}{x^2}} + 1\right)\right]} + C
\int \operatorname{arsech}\,x \, dx = x\, \operatorname{arsech}\, x- \arctan{\left(\frac{x}{x-1}\sqrt{\frac{1-x}{1+x}}\right)} + C
\int \operatorname{arcoth}\,x \, dx = x\, \operatorname{arcoth}\, x+ \frac{1}{2}\ln{(x^2-1)} + C

Composed functions

\int \cos ax\, e^{bx}\, dx = \frac{e^{bx}}{a^2+b^2}\left( a\sin ax + b\cos ax \right) + C
\int \sin ax\, e^{bx}\, dx = \frac{e^{bx}}{a^2+b^2}\left( b\sin ax - a\cos ax \right) + C
\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
\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

\int \left| (ax + b)^n \right|\,dx = {(ax + b)^{n+2} \over a(n+1) \left| ax + b \right|} + C \,\, [\,n\text{ is odd, and } n \neq -1\,]
\int \left| \sin{ax} \right|\,dx = {-1 \over a} \left| \sin{ax} \right| \cot{ax} + C
\int \left| \cos{ax} \right|\,dx = {1 \over a} \left| \cos{ax} \right| \tan{ax} + C
\int \left| \tan{ax} \right|\,dx = {\tan(ax)[-\ln\left|\cos{ax}\right|] \over a \left| \tan{ax} \right|} + C
\int \left| \csc{ax} \right|\,dx = {-\ln \left| \csc{ax} + \cot{ax} \right|\sin{ax} \over a \left| \sin{ax} \right|} + C
\int \left| \sec{ax} \right|\,dx = {\ln \left| \sec{ax} + \tan{ax} \right| \cos{ax} \over a \left| \cos{ax} \right|} + C
\int \left| \cot{ax} \right|\,dx = {\tan(ax)[\ln\left|\sin{ax}\right|] \over a \left| \tan{ax} \right|} + C

Special functions

Ci, Si: Trignometry integrals, Ei: Exponential integrals, li: Logarithmic integral function, erf: Error function

\int \operatorname{Ci}(x) dx = x\,\operatorname{Ci}(x) - \sin x
\int \operatorname{Si}(x) dx = x\,\operatorname{Si}(x) + \cos x
\int \operatorname{Ei}(x) dx = x\,\operatorname{Ei}(x) - e^x
\int \operatorname{li}(x)dx = x\, \operatorname{li}(x)-\operatorname{Ei}(2 \ln x)
\int \frac{\operatorname{li}(x)}{x}\,dx = \ln x\, \operatorname{li}(x) -x
\int \operatorname{erf}(x)\, dx = \frac{e^{-x^2}}{\sqrt{\pi }}+x\, \text{erf}(x)

Definite integrals lacking closed-form antiderivatives

There are some functions whose antiderivatives cannot be expressed in closed 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.

\int_0^\infty{\sqrt{x}\,e^{-x}\,dx} = \frac{1}{2}\sqrt \pi
\int_0^\infty{e^{-a x^2}\,dx} = \frac{1}{2} \sqrt \frac {\pi} {a}
\int_0^\infty{x^2 e^{-a x^2}\,dx} = \frac{1}{4} \sqrt \frac {\pi} {a^3} when a > 0
\int_0^\infty{x^{2n} e^{-a x^2}\,dx} = \frac{2n-1}{2a} \int_0^\infty{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}}} when a > 0, n is 1,2,3,... and !! is the double factional
\int_0^\infty{x^3 e^{-a x^2}\,dx} = \frac{1}{2 a^2} when a > 0
\int_0^\infty{x^{2n+1} e^{-a x^2}\,dx} = \frac {n} {a} \int_0^\infty{x^{2n-1} e^{-a x^2}\,dx} = \frac{n!}{2 a^{n+1}} when a > 0, n is 0, 1, 2, ....
\int_0^\infty{\frac{x}{e^x-1}\,dx} = \frac{\pi^2}{6} (
\int_0^\infty{\frac{x^3}{e^x-1}\,dx} = \frac{\pi^4}{15}
\int_0^\infty\frac{\sin(x)}{x}\,dx=\frac{\pi}{2} (
\int_0^\frac{\pi}{2}\sin^n{x}\,dx=\int_0^\frac{\pi}{2}\cos^n{x}\,dx=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot n}\frac{\pi}{2} (if n is an even integer and \scriptstyle{n \ge 2})
\int_0^\frac{\pi}{2}\sin^n{x}\,dx=\int_0^\frac{\pi}{2}\cos^n{x}\,dx=\frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (n-1)}{3 \cdot 5 \cdot 7 \cdot \cdots \cdot n} (if \scriptstyle{n} is an odd integer and \scriptstyle{n \ge 3} )
\int_{-\pi}^{\pi} \cos(\alpha x)\cos^n(\beta x) dx = \left \{ \begin{array}{cc} \frac{2 \pi}{2^n} \binom{n}{m} & |\alpha|= |\beta (2m-n)| \\ 0 & \mbox{otherwise} \\ \end{array} \right . (for \scriptstyle \alpha, \beta, m, n integers with \scriptstyle \beta \neq 0 and \scriptstyle m, n \geq 0
\int_0^1 x^{m-1}(1-x)^{n-1} dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}
\int_{-\pi}^{\pi} \sin(\alpha x) \cos^n(\beta x) dx = 0 (for \scriptstyle \alpha,\beta real and \scriptstyle n non-negative integer,
\int_{-\pi}^{\pi} \sin(\alpha x) \sin^n(\beta x) dx = \left \{ \begin{array}{cc} (-1)^{(n+1)/2} (-1)^m \frac{2 \pi}{2^n} \binom{n}{m} & n \mbox{ odd},\ \alpha = \beta (2m-n) \\ 0 & \mbox{otherwise} \\ \end{array} \right . (for \scriptstyle \alpha, \beta, m, n integers with \scriptstyle \beta \neq 0 and \scriptstyle m, n \geq 0
\int_{-\pi}^{\pi} \cos(\alpha x) \sin^n(\beta x) dx = \left \{ \begin{array}{cc} (-1)^{n/2} (-1)^m \frac{2 \pi}{2^n} \binom{n}{m} & n \mbox{ even},\ |\alpha| = |\beta (2m-n)| \\ 0 & \mbox{otherwise} \\ \end{array} \right . (for \scriptstyle \alpha, \beta, m, n integers with \scriptstyle \beta \neq 0 and \scriptstyle m,n \geq 0
\int_0^\infty\frac{\sin^2{x}}{x^2}\,dx=\frac{\pi}{2}
\int_0^\infty x^{z-1}\,e^{-x}\,dx = \Gamma(z) (where Γ(z) is the gamma funtion
\int_{-\infty}^\infty e^{-(ax^2+bx+c)}\,dx=\sqrt{\frac{\pi}{a}}\exp\left[\frac{b^2-4ac}{4a}\right] ( eu, and a > 0)
\int_{0}^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x) (
\int_{0}^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_{0} \left(\sqrt{x^2 + y^2}\right)
\int_{-\infty}^{\infty}{(1 + x^2/\nu)^{-(\nu + 1)/2}dx} = \frac { \sqrt{\nu \pi} \ \Gamma(\nu/2)} {\Gamma((\nu + 1)/2))}\,, 0\," src="http://upload.wikimedia.org/math/8/7/e/87ecc6d98450874d2b57cbc8e56ffdd1.png">,


\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} ).
\int_0^1 [\ln(1/x)]^p\,dx = p!

The "sophomore's dream" "

\begin{align} \int_0^1 x^{-x}\,dx &= \sum_{n=1}^\infty n^{-n} &&(= 1.29128599706266\dots)\\ \int_0^1 x^x \,dx &= -\sum_{n=1}^\infty (-1)^nn^{-n} &&(= 0.783430510712\dots) \end{align}
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