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==Definite integrals lacking closed-form antiderivatives== There are some functions whose antiderivatives ''cannot'' be expressed in [[Closed-form expression|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. *<math>\int_0^\infty \sqrt{x}\,e^{-x}\,dx = \frac{1}{2}\sqrt \pi</math> (see also [[Gamma function]]) *<math>\int_0^\infty e^{-a x^2}\,dx = \frac{1}{2} \sqrt \frac {\pi} {a} </math> for {{math|''a'' > 0}} (the [[Gaussian integral]]) *<math>\int_0^\infty{x^2 e^{-a x^2}\,dx} = \frac{1}{4} \sqrt \frac {\pi} {a^3} </math> for {{math|''a'' > 0}} *<math>\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}}} </math>{{pb}}for {{math|''a'' > 0}}, {{math|''n''}} is a positive integer and {{math|!!}} is the [[double factorial]]. *<math>\int_0^\infty{x^3 e^{-a x^2}\,dx} = \frac{1}{2 a^2} </math> when {{math|''a'' > 0}} *<math>\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}} </math>{{pb}}for {{math|''a'' > 0}}, {{math|1=''n'' = 0, 1, 2, ....}} *<math>\int_0^\infty \frac{x}{e^x-1}\,dx = \frac{\pi^2}{6}</math> (see also [[Bernoulli number]]) *<math>\int_0^\infty \frac{x^2}{e^x-1}\,dx = 2\zeta(3) \approx 2.40</math> *<math>\int_0^\infty \frac{x^3}{e^x-1}\,dx = \frac{\pi^4}{15}</math> *<math>\int_0^\infty \frac{\sin{x}}{x}\,dx = \frac{\pi}{2}</math> (see [[sinc function]] and the [[Dirichlet integral]]) *<math>\int_0^\infty\frac{\sin^2{x}}{x^2}\,dx = \frac{\pi}{2}</math> *<math>\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}</math>{{pb}}(if {{math|''n''}} is a positive integer and !! is the [[double factorial]]). *<math>\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} </math>{{pb}}(for {{math|''α'', ''β'', ''m'', ''n''}} integers with {{math|''β'' ≠ 0}} and {{math|''m'', ''n'' ≥ 0}}, see also [[Binomial coefficient]]) *<math>\int_{-t}^t \sin^m(\alpha x) \cos^n(\beta x) dx = 0</math>{{pb}}(for {{math|''α'', ''β''}} real, {{math|''n''}} a non-negative integer, and {{mvar|m}} an odd, positive integer; since the integrand is [[Odd function|odd]]) *<math>\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} </math>{{pb}}(for {{math|''α'', ''β'', ''m'', ''n''}} integers with {{math|''β'' ≠ 0}} and {{math|''m'', ''n'' ≥ 0}}, see also [[Binomial coefficient]]) *<math>\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} </math>{{pb}}(for {{math|''α'', ''β'', ''m'', ''n''}} integers with {{math|''β'' ≠ 0}} and {{math|''m'', ''n'' ≥ 0}}, see also [[Binomial coefficient]]) *<math>\int_{-\infty}^\infty e^{-(ax^2+bx+c)}\,dx = \sqrt{\frac{\pi}{a}}\exp\left[\frac{b^2-4ac}{4a}\right]</math>{{pb}}(where {{math|exp[''u'']}} is the [[exponential function]] {{math|''e''<sup>''u''</sup>}}, and {{math|''a'' > 0}}.) *<math>\int_0^\infty x^{z-1}\,e^{-x}\,dx = \Gamma(z)</math>{{pb}}(where <math>\Gamma(z)</math> is the [[Gamma function]]) *<math>\int_0^1 \left(\ln\frac{1}{x}\right)^p\,dx = \Gamma(p+1)</math> *<math>\int_0^1 x^{\alpha-1}(1-x)^{\beta-1} dx = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} </math>{{pb}}(for {{math|Re(''α'') > 0}} and {{math|Re(''β'') > 0}}, see [[Beta function]]) *<math>\int_0^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x)</math> (where {{math|''I''<sub>0</sub>(''x'')}} is the modified [[Bessel function]] of the first kind) *<math>\int_0^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_{0} \left(\sqrt{x^2 + y^2}\right) </math> *<math>\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)}</math>{{pb}}(for {{math|''ν'' > 0}} , this is related to the [[probability density function]] of [[Student's t-distribution|Student's ''t''-distribution]]) If the function {{math|''f''}} has [[bounded variation]] on the interval {{closed-closed|''a'',''b''}}, then the [[method of exhaustion]] provides a formula for the integral: <math display="block">\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} ).</math> The "[[sophomore's dream]]": <math display="block">\begin{align} \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{align}</math> attributed to [[Johann Bernoulli]].
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