How to compute $ \int_0^1 {e^{-x^2}} dx$ I know that
$$
\int_{0}^{+ \infty} e^{- x^{2}} dx = \frac{\sqrt{\pi}}{2}.
$$
My question is:
$$
\int_{0}^{1} e^{- x^{2}} dx = ~?
$$
 A: 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,
$$
\int_0^1e^{-t^2}\,\mathrm{d}t=e^{-1}\sum_{k=0}^\infty\frac{2^k}{(2k+1)!!}
$$
A: $\mathrm{erf}(z)$ 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
$$\int_{0}^{1}e^{-r^2}\mathrm{dr}\approx0.74682413281242702540$$
A: According to Laplace, Legendre and the two masters of continued fractions, Jacobi and Ramanujan,
$$
\int_{0}^{1} e^{- x^{2}} \,d{x} = \frac{\sqrt{\pi}}{2} - \cfrac{\frac{1}{2} e^{-1}}{1 + \cfrac{1}{2 + \cfrac{2}{1 + \cfrac{3}{2 + \cfrac{4}{1 + \cfrac{5}{2 + \cdots}}}}}}.
$$
