Difference between revisions of "Continued fraction for 2e^(z^2) integral from z to infinity e^(-t^2) dt for positive Re(z)"

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

Revision as of 02:01, 6 June 2016

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 $x_k^{(n)}$ and $H_k^{(n)}$ are the zeros and weight factors of the Hermite polynomials.

Proof

Refrences