Error Function Integral
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that occurs in probability, statistics, and partial differential equations describing diffusion. It is defined as:[1][2] erf ( x ) = 1 π ∫ − x x e − t 2 d t = 2 π ∫ 0 x e − t 2 d t . {\displaystyle {\begin gamma function integral − 6\operatorname − 5 (x)&={\frac − 4{\sqrt {\pi }}}\int _{-x}^ − 3e^{-t^ − 2}\,\mathrm − 1 t\\&={\frac normal distribution integral − 0{\sqrt {\pi }}}\int _ 9^ 8e^{-t^ 7}\,\mathrm 6 t.\end 5}} The complementary error function, denoted erfc, is defined gaussian integral as erfc ( x ) = 1 − erf ( x ) = 2 π ∫ x ∞ e − t 2 d t = e − x 2 erfcx ( x ) , {\displaystyle {\begin 2\operatorname error function integral table 1 (x)&=1-\operatorname 0 (x)\\&={\frac Φ 9{\sqrt {\pi }}}\int _ Φ 8^{\infty }e^{-t^ Φ 7}\,\mathrm Φ 6 t\\&=e^{-x^ Φ 5}\operatorname Φ 4 (x),\end Φ 3}} which also defines erfcx, the scaled complementary error function[3] (which can be used instead of erfc to avoid arithmetic underflow[3][4]). Another form of erfc ( x ) {\displaystyle \operatorname 2 (x)} for non-negative x {\displaystyle x} is known as Craig's formula:[5] erfc ( x | x ≥ 0 ) = 2 π ∫ 0
Error Function Integral Calculation
π / 2 exp ( − x 2 sin 2 θ ) d θ . {\displaystyle \operatorname 0 (x|x\geq 0)={\frac Φ 9{\pi }}\int _ Φ 8^{\pi /2}\exp \left(-{\frac Φ 7}{\sin ^ Φ 6\theta }}\right)d\theta \,.} The imaginary error function, denoted erfi, is defined as erfi ( x ) = − i erf ( i x ) = 2 π ∫ 0 x e t 2 d t = 2 π e x 2 D ( x ) , {\displaystyle {\begin Φ 0\operatorname − 9 (x)&=-i\operatorname − 8 (ix)\\&={\frac − 7{\sqrt {\pi }}}\int _ − 6^ − 5e^ − 4}\,\mathrm − 3 t\\&={\frac − 2{\sqrt {\pi }}}e^ − 1}D(x),\end − 0}} where D(x) is the Dawson function (which can be used instead of erfi to avoid arithmetic overflow[3]). Despite the name "imaginary error function", erfi ( x ) {\displaystyle \operatorname 8 (x)} is real when x is real. When the error function is evaluated for arbitrary complex arguments z, the resulting complex error function is usually discussed in scaled form as the Faddeeva function: w ( z ) = e − z 2 erfc ( − i z ) = erfcx ( − i z ) . {\displaystyle w(z)=e^{-z^ 6}\operatorname 5 (-iz)=\operatorname 4 (-iz).} Contents 1 The name "error function" 2 Properties 2.1 Taylor series 2.2 Derivative and integral 2.3 Bürmann series 2.4 Inverse functions 2.5 Asymptotic expansion 2.6 Continued fraction expansion 2.7 Integ
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Error Function Values
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here for a quick overview of the site Help Center Detailed answers to any http://math.stackexchange.com/questions/1210623/integral-involving-an-error-function questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Mathematics Questions Tags Users Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site error function for people studying math at any level and professionals in 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 Integral involving an error function up vote error function integral 5 down vote favorite 3 For $\sigma>0$, how can we prove that $$\frac{1}{2}\int_{-1}^1 \text{erf}\left(\frac{\sigma}{\sqrt{2}}+\text{erf}^{-1}(x)\right) \, \mathrm{d}x= \text{erf}\left(\frac{\sigma}{2}\right)$$ where erf is the error function, $$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}dt.$$ The result was obtained by tinkering, and I was wondering if there a concise derivation. integration definite-integrals error-function share|cite|improve this question edited Mar 28 '15 at 23:10 Eric Naslund 53.4k10119210 asked Mar 28 '15 at 21:49 Nero 1,492963 1 First thing that comes to mind is to change variable (y = invert of erf). Then look at the integral as a convolution of erf and its derivative, then use convolution theorem. Overall it requires around 5 steps. –user227136 Mar 28 '15 at 22:36 add a comment| 2 Answers 2 active oldest votes up vote 9 down vote accepted Let $u=\text{erf}^{-1}(x)$ so that $x=\text{erf}\left(u\right)$, and $dx=\frac{2}{\sqrt{\pi}}e^{-u^{2}}du.$ Then $$I(\sigma)=\frac{1}{2}\int_{-1}^{1}\text{erf}\left(\frac{\sigma}{\sqrt{2}}+\text{erf}^{-1}(x)\right)dx=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^{2}}\text{erf}\left(\frac{\sigma}{\sqrt{2}}+u\right)du.$$ Differentiating with respect to $\sigma$ we find that $$\frac{\partial I(\sigma)}{\partial\sigma}=\frac{\sqrt{2}}{\pi}\int_{-\infty}^{\infty}e^{-u^{2}}e^{-(u+\sigma/\sqrt{2})^{2}}du,$$ and so substituting $u=y/\sqrt{2}$ we have $$\frac{\partial I(\sigma)}{\partial\sigma}=\frac{1}{\pi}\int_{-\infty}^{\infty}e^{-y^{