Error Function Integration Parts
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Integral Complementary Error Function
Erf is the "error function" encountered in integrating the normal distribution (which is a normalized form of the Gaussian function). It integral of error function with gaussian density function is an entire function defined by (1) Note that some authors (e.g., Whittaker and Watson 1990, p.341) define without the leading factor of . Erf is implemented in the Wolfram Language as Erf[z]. A two-argument error function integral table form giving is also implemented as Erf[z0, z1]. Erf satisfies the identities (2) (3) (4) where is erfc, the complementary error function, and is a confluent hypergeometric function of the first kind. For , (5) where is the incomplete gamma function. Erf can also be defined as a Maclaurin series (6) (7) (OEIS A007680). Similarly, (8) (OEIS A103979 and A103980). For , may be computed from (9) (10) (OEIS A000079
Error Function Integral Calculation
and A001147; Acton 1990). For , (11) (12) Using integration by parts gives (13) (14) (15) (16) so (17) and continuing the procedure gives the asymptotic series (18) (19) (20) (OEIS A001147 and A000079). Erf has the values (21) (22) It is an odd function (23) and satisfies (24) Erf may be expressed in terms of a confluent hypergeometric function of the first kind as (25) (26) Its derivative is (27) where is a Hermite polynomial. The first derivative is (28) and the integral is (29) Min Max Re Im Erf can also be extended to the complex plane, as illustrated above. A simple integral involving erf that Wolfram Language cannot do is given by (30) (M.R.D'Orsogna, pers. comm., May 9, 2004). More complicated integrals include (31) (M.R.D'Orsogna, pers. comm., Dec.15, 2005). Erf has the continued fraction (32) (33) (Wall 1948, p.357), first stated by Laplace in 1805 and Legendre in 1826 (Olds 1963, p.139), proved by Jacobi, and rediscovered by Ramanujan (Watson 1928; Hardy 1999, pp.8-9). Definite integrals involving include Definite integrals involving include (34) (35) (36) (37) (38) The first two of these appear in Prudnikov et al. (1990, p.123, eqns. 2.8.19.8 and 2.8.19.11), with , . A complex generalization of is defined as (39) (40) Integral
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Error Function Values
related fields. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Steps in evaluating http://mathworld.wolfram.com/Erf.html the integral of complementary error function? up vote 5 down vote favorite 2 Could you please check the below and show me any errors? $$ \int_ x^ \infty {\rm erfc} ~(t) ~dt ~=\int_ x^ \infty \left[\frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du \right]\ dt $$ If I let dv=dt and u equal the term inside the bracket, and do integration by parts, $$ \int u ~dv ~=uv - \int v~ du $$ v=t and du becomes http://math.stackexchange.com/questions/108109/steps-in-evaluating-the-integral-of-complementary-error-function $$ -\frac{2}{\sqrt\pi} e^{-t^2} $$ This was obtained from using the Leibniz rule below, $$ \frac {d} {dt} \left[ \int_ a^ b f(u)du \right]\ = \int_ a^ b \frac {d} {dt} f(u) du + f \frac {db} {dt} - f \frac {da} {dt} $$ Then, $$ \frac {d} {dt} \left[\frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du \right]\ = \frac{2}{\sqrt\pi} \left[ \int_ t^ \infty \frac {d} {dt} \left( e^{-u^2} \right) du + e^{-\infty ^2} * 0 - e^{-t^2}*1 \right]= \frac{2}{\sqrt\pi} \left[0~+~0~- e^{-t^2} \right]$$ Is the first and second term going to zero correct? The upper limit b=infinity, and is db/dt=0 in the second term correct? The integral becomes $$ \left[~ t~ \frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du ~\right] _{x}^\infty + \int_ x^ \infty t \left[\frac{2}{\sqrt\pi} e^{-t^2} \right]\ dt =$$ $$ \left[~ t~ \frac{2}{\sqrt\pi} \int_ t^ \infty e^{-u^2} du ~\right] _{x}^\infty - \left[\frac{1}{\sqrt\pi} e^{-t^2} \right] _{x}^\infty =$$ $$ \left[ 0 - ~ x~ \frac{2}{\sqrt\pi} \int_ x^ \infty e^{-u^2} du ~\right] - \left[ 0 - \frac{1}{\sqrt\pi} e^{-x^2} \right] = $$ (Is the first limit going to zero OK? infinity times 0 = 0). The above becomes $$ -x~ {\rm erfc}~(x) + \frac{1}{\sqrt\pi} e^{-x^2} $$ Is everything correct here? Could you please give explanation to the questions I listed? integration special-functions share|cite|improve this question edited Feb 11 '12 at 14:12 asked Feb 11 '12 at 10:46 Tony 187311 migra
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