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I know that $ \int_{0}^{+ \infty} e^{- x^{2}} dx = \frac{\sqrt{\pi}}{2}. $

My question is: $ \int_{0}^{1} e^{- x^{2}} dx = ~? $

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    See here: http://en.wikipedia.org/wiki/Error_function for more information.2011-09-19

3 Answers 3

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$\mathrm{erf}(z)$ is the "error function" encountered in integrating the normal distribution (which is a normalized form of the Gaussian function). It is an entire function defined by $\mathrm{erf}(z)=\frac{2}{\sqrt{\pi}}\int_{0}^{z}e^{-t^2}\mathrm{dt}$ Note that some authors (e.g., Whittaker and Watson 1990, p. 341) define $\mathrm{erf}(z)$ without the leading factor of $2/\sqrt{\pi}$.

Your question's solution is $\int_{0}^{1}e^{-r^2}\mathrm{dr}=\frac{\sqrt{\pi}}{2}\mathrm{erf}(1)$

You can get approximate value by using series $\mathrm{erf}(x)=\frac{e^{-x^2}}{\sqrt{\pi}}\sum^{\infty}_{n=0}\frac{(2x)^{2n+1}}{(2n+1)!!}$

Hope this helped.

Added. approximate value of $\mathrm{erf}(1)$ with 20 digits is

$\mathrm{erf}(1)\approx0.84270079294971486934$.

so the value is

$\int_{0}^{1}e^{-r^2}\mathrm{dr}\approx0.74682413281242702540$

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    @J.M.: I show these two methods and an asymptotic expansion [here](http://www.whim.org/neb$u$la/math/cumnormdist.html).2012-12-24
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Since $ e^{-x^2}=\sum_{k=0}^\infty\frac{(-x^2)^k}{k!} $ we get by integrating that $ \int_0^te^{-x^2}\,\mathrm{d}x=\sum_{k=0}^\infty(-1)^k\frac{t^{2k+1}}{(2k+1)k!} $ Therefore, $ \int_0^1e^{-x^2}\,\mathrm{d}x=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)k!} $ which converges pretty quickly.

Unilateral Power Series

A bit more complicated, yet yielding a non-alternating series, is to note that if we define $ f(x)=e^{x^2}\int_0^xe^{-t^2}\,\mathrm{d}t $ Then $f'(x)=1+2xf(x)$ and $ f(x)=\sum_{k=0}^\infty a_kx^k $ yields $2a_{k-1}=(k+1)a_{k+1}$. Using $f(0)=0$ and $f'(0)=1$ gives $a_{2k+1}=\dfrac{2^k}{(2k+1)!!}$. Thus, $ \int_0^xe^{-t^2}\,\mathrm{d}t=e^{-x^2}\sum_{k=0}^\infty\frac{2^kx^{2k+1}}{(2k+1)!!} $ Therefore, $ \int_0^1e^{-t^2}\,\mathrm{d}t=e^{-1}\sum_{k=0}^\infty\frac{2^k}{(2k+1)!!} $

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    Nice to see the double factorial popping up every so often... :)2013-01-16
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According to Laplace, Legendre and the two masters of continued fractions, Jacobi and Ramanujan, $ \int_{0}^{1} e^{- x^{2}} \,d{x} = \frac{\sqrt{\pi}}{2} - \cfrac{\frac{1}{2} e^{-1}}{1 + \cfrac{1}{2 + \cfrac{2}{1 + \cfrac{3}{2 + \cfrac{4}{1 + \cfrac{5}{2 + \cdots}}}}}}. $