TITLE: How to compute $ \int_0^1 {e^{-x^2}} dx$
QUESTION [6 upvotes]: I know that
$$
\int_{0}^{+ \infty} e^{- x^{2}} dx = \frac{\sqrt{\pi}}{2}.
$$
My question is:
$$
\int_{0}^{1} e^{- x^{2}} dx = ~?
$$

REPLY [3 votes]: According to Laplace, Legendre and the two masters of continued fractions, Jacobi and Ramanujan,
$$
\int_{0}^{1} e^{- x^{2}} \,d{x} = \frac{\sqrt{\pi}}{2} - \cfrac{\frac{1}{2} e^{-1}}{1 + \cfrac{1}{2 + \cfrac{2}{1 + \cfrac{3}{2 + \cfrac{4}{1 + \cfrac{5}{2 + \cdots}}}}}}.
$$