View source for Continued fraction for 1/sqrt(pi) integral from -infinity to infinity of e^(-t^2)/(z-t) dt

==Theorem==
The following formula holds:
$$\dfrac{1}{\sqrt{\pi}} \displaystyle\int_{-\infty}^{\infty} \dfrac{e^{-t^2}}{z-t} \mathrm{d}t = \dfrac{1}{z-\dfrac{\frac{1}{2}}{z-\dfrac{1}{z-\dfrac{\frac{3}{2}}{z-\dfrac{2}{z-\ldots}}}}}=\dfrac{1}{\sqrt{\pi}} \displaystyle\lim_{n \rightarrow \infty} \displaystyle\sum_{k=1}^n \dfrac{H_k^{(n)}}{z-x_k^{(n)}},$$
where $\pi$ denotes [[pi]], $e^{-t^2}$ denotes the [[exponential]] function, $x_k^{(n)}$ denotes the zeros of the [[Hermite polynomials]], and $H_k^{(n)}$ denote the [[weight function]] for the Hermite polynomials.
==Proof==

==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Continued fraction for 2e^(z^2) integral from z to infinity e^(-t^2) dt for positive Re(z)|next=Limit of erf when z approaches infinity and the modulus of arg(z) is less than pi/4}}: 7.1.15

[[Category:Theorem]]