Antiderivative Of Error Function
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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 admit antiderivative − 2\operatorname − 1 (x)&={\frac − 0{\sqrt {\pi }}}\int _{-x}^ 9e^{-t^ 8}\,\mathrm 7 t\\&={\frac
Integral Complementary Error Function
6{\sqrt {\pi }}}\int _ 5^ 4e^{-t^ 3}\,\mathrm 2 t.\end 1}} The complementary error function, denoted erfc, is defined
Integral Of Error Function With Gaussian Density Function
as erfc ( x ) = 1 − erf ( x ) = 2 π ∫ x ∞ e − t 2 d t = e − x 2 erfcx ( x ) , {\displaystyle {\begin Φ 8\operatorname
Error Function Values
Φ 7 (x)&=1-\operatorname Φ 6 (x)\\&={\frac Φ 5{\sqrt {\pi }}}\int _ Φ 4^{\infty }e^{-t^ Φ 3}\,\mathrm Φ 2 t\\&=e^{-x^ Φ 1}\operatorname Φ 0 (x),\end 9}} 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 Φ 8 (x)} for non-negative x {\displaystyle x} is known as Craig's formula:[5] erfc ( x | x ≥ 0 ) = 2 π ∫ 0 error function integral table π / 2 exp ( − x 2 sin 2 θ ) d θ . {\displaystyle \operatorname Φ 6 (x|x\geq 0)={\frac Φ 5{\pi }}\int _ Φ 4^{\pi /2}\exp \left(-{\frac Φ 3}{\sin ^ Φ 2\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 − 6\operatorname − 5 (x)&=-i\operatorname − 4 (ix)\\&={\frac − 3{\sqrt {\pi }}}\int _ − 2^ − 1e^ − 0}\,\mathrm − 9 t\\&={\frac − 8{\sqrt {\pi }}}e^ − 7}D(x),\end − 6}} 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 4 (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^ 2}\operatorname 1 (-iz)=\operatorname 0 (-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 I
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack error function integral calculation Overflow the company Business Learn more about hiring developers or posting ads with us erf function calculator Mathematics Questions Tags Users Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site for people studying erf function table 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 https://en.wikipedia.org/wiki/Error_function up and rise to the top What is the antiderivative of $e^{-x^2}$ up vote 35 down vote favorite 13 I was wondering what the antiderivative of $e^{-x^2}$ was, and when I wolfram alpha'd it I got $$\displaystyle \int e^{-x^2} \textrm{d}x = \dfrac{1}{2} \sqrt{\pi} \space \text{erf} (x) + C$$ So, I of course didn't know what this $\text{erf}$ was and I looked it up on wikipedia, where it was defined as: $$ http://math.stackexchange.com/questions/523824/what-is-the-antiderivative-of-e-x2 \text{erf}(x) = \dfrac{2}{\sqrt{\pi}} \displaystyle \int_0^x e^{-t^2} \textrm{d}t $$ To my mathematically illiterate mind, this is a bit too circular to understand. Why can't we express $\int e^{-x^2} \textrm{d}x$ as a 'normal function'? Also, what is the use of the error function? error-function share|cite|improve this question edited Oct 14 '13 at 9:48 in_wolframAlpha_we_trust 2,175419 asked Oct 12 '13 at 19:34 Phaptitude 9071023 1 You should read this: en.wikipedia.org/wiki/Elementary_function –M Turgeon Oct 12 '13 at 19:37 9 This pretty much says that an antiderivative of $e^{-x^2}$ is an antiderivative of $e^{-x^2}$, and that's the best you can do. –1015 Oct 12 '13 at 19:38 You can read this paper on impossibility theorems for elementary integration: claymath.org/programs/outreach/academy/LectureNotes05/… –Kaa1el Oct 12 '13 at 21:30 add a comment| 6 Answers 6 active oldest votes up vote 73 down vote accepted When $f$ is a continuous function on the interval $[a,b]$, we can find a function $F$ defined on $[a,b]$ such that $F'(x)=f(x)$ for all $x\in[a,b]$. This is the “fundamental theorem of calculus”; just consider $$ F(x)=\int_{a}^{x} f(t)\,dt $$ There are other functions with the same property, precisely those of the form $F(x)+c$ where $c$ is a completely arbitrary constant. Sometimes this function can be expressed with the so-called “elementary functions
here for a quick overview of the site Help Center Detailed answers to any 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 http://math.stackexchange.com/questions/108109/steps-in-evaluating-the-integral-of-complementary-error-function hiring developers or posting ads with us Mathematics Questions Tags Users Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site 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 Steps in evaluating the integral of complementary error error function 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 $$ -\frac{2}{\sqrt\pi} e^{-t^2} $$ This was obtained of error function 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 migrated from meta.math.stackexchange.com Feb 11 '12 at 11:24 This question came from our