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 $\mathrm{Re}(z)>0$ (and the final formula holds with the additional requirement on $z$: $\mathrm{Im}(z)\neq 0$): $$2e^{z^2}\displaystyle\int_z^{\infty} e^{-t^2} \mathrm{d}t = \dfrac{1}{z+\dfrac{\frac{1}{2}}{z+\dfrac{1}{z+\dfrac{\frac{3}{2}}{z+\dfrac{2}{z+\ldots}}}}}=\dfrac{1}{\sqrt{\pi}} \displaystyle\lim_{n \rightarrow \infty} \displaystyle\sum_{k=1}^n \dfrac{H_k^{(n)}}{z-x_k^{(n)}},$$ where $x_k^{(n)}$ and $H_k^{(n)}$ are the zeros and weight factors of the Hermite polynomials.

Refrences

• 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): 7.1.14