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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The following formula holds for $\mathrm{Re}(z)>0$:
 
The following formula holds for $\mathrm{Re}(z)>0$:
 
$$2e^{z^2}\displaystyle\int_z^{\infty} e^{-t^2} \mathrm{d}t = \dfrac{1}{z+\dfrac{\frac{1}{2}}{z+\dfrac{1}{z+\dfrac{\frac{3}{2}}{z+\dfrac{2}{z+\ldots}}}}},$$
 
$$2e^{z^2}\displaystyle\int_z^{\infty} e^{-t^2} \mathrm{d}t = \dfrac{1}{z+\dfrac{\frac{1}{2}}{z+\dfrac{1}{z+\dfrac{\frac{3}{2}}{z+\dfrac{2}{z+\ldots}}}}},$$
−
where $x_k^{(n)}$ and $H_k^{(n)}$ are the zeros and weight factors of the [[Hermite polynomials]].
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where $e^{z^2}$ denotes the [[exponential]] and the right hand side denotes a [[continued fraction]].
 
==Proof==
 
==Proof==
  

Latest revision as of 02:08, 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 $e^{z^2}$ denotes the exponential and the right hand side denotes a continued fraction.

Proof

Refrences

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