|
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 (see also Gamma function) \int_0^\infty e^{-a x^2}\,dx = \frac{1}{2} \sqrt \frac {\pi} {a} for a > 0 (the Gaussian integral) \int_0^\infty{x^2 e^{-a x^2}\,dx} = \frac{1}{4} \sqrt \frac {\pi} {a^3} for 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}}} for a > 0, n is 1, 2, 3, ... and !! is the double factorial. \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}} for a > 0, n = 0, 1, 2, .... \int_0^\infty \frac{x}{e^x-1}\,dx = \frac{\pi^2}{6} (see also Bernoulli number) \int_0^\infty \frac{x^2}{e^x-1}\,dx = 2\zeta(3) \simeq 2.40 \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} (see sinc function and Sine integral) \int_0^\infty\frac{\sin^2{x}}{x^2}\,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 n ≥ 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 n is an odd integer and n ≥ 3) \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} (for α , β, m, n integers with β ≠ 0 and m, n ≥ 0, see also Binomial coefficient) \int_{-\pi}^\pi \sin(\alpha x) \cos^n(\beta x) dx = 0 (for α , β real and n non-negative integer, see also Symmetry) \int_{-\pi}^\pi \sin(\alpha x) \sin^n(\beta x) dx = \begin{cases} (-1)^{(n+1)/2} (-1)^m \frac{2 \pi}{2^n} \binom{n}{m} & n \text{ odd},\ \alpha = \beta (2m-n) \\ 0 & \text{otherwise} \end{cases} (for α , β, m, n integers with β ≠ 0 and m, n ≥ 0, see also Binomial coefficient) \int_{-\pi}^{\pi} \cos(\alpha x) \sin^n(\beta x) dx = \begin{cases} (-1)^{n/2} (-1)^m \frac{2 \pi}{2^n} \binom{n}{m} & n \text{ even},\ |\alpha| = |\beta (2m-n)| \\ 0 & \text{otherwise} \end{cases} (for α , β, m, n integers with β ≠ 0 and m, n ≥ 0, see also Binomial coefficient) \int_{-\infty}^\infty e^{-(ax^2+bx+c)}\,dx = \sqrt{\frac{\pi}{a}}\exp\left[\frac{b^2-4ac}{4a}\right] (where exp[u] is the exponential function eu, and a > 0) \int_0^\infty x^{z-1}\,e^{-x}\,dx = \Gamma(z) (where \Gamma(z) is the Gamma function) \int_0^1 x^{\alpha-1}(1-x)^{\beta-1} dx = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} (for Re(α ) > 0 and Re(β) > 0, see Beta function) \int_0^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x) (where I0(x) is the modified Bessel function of the first kind) \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)} (for ν > 0 , this is related to the probability density function of the Student's t-distribution) If the function f has bounded variation on the interval [a,b], then the method of exhaustion provides a formula for the integral: \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!\; (Click "show" at right to see a proof or "hide" to hide it.)[show] 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 (-n)^{-n} &&(= 0.78343051071213\dots) \end{align} attributed to Johann Bernoulli. |
About us|Jobs|Help|Disclaimer|Advertising services|Contact us|Sign in|Website map|Search|
GMT+8, 2014-3-15 18:23 , Processed in 0.072386 second(s), 17 queries .
