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

==Theorem==
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]].
==Proof==

==Refrences==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Two-sided inequality for e^(x^2) integral from x to infinity e^(-t^2) dt for non-negative real x|next=Continued fraction for 1/sqrt(pi) integral from -infinity to infinity of e^(-t^2)/(z-t) dt}}: 7.1.14

[[Category:Theorem]]