# Continued fraction for 2e^(z^2) integral from z to infinity e^(-t^2) dt for positive Re(z)

The following formula holds for $\mathrm{Re}(z)>0$: $$2e^{z^2}\displaystyle\int_z^{\infty} e^{-t^2} \mathrm{d}t = \dfrac{1}{z+\dfrac{\frac{1}{2}}{z+\dfrac{1}{z+\dfrac{\frac{3}{2}}{z+\dfrac{2}{z+\ldots}}}}},$$ where $e^{z^2}$ denotes the exponential and the right hand side denotes a continued fraction.