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 $\mathrm{Re}(z)>0$:
 
The following formula holds for $\mathrm{Re}(z)>0$:
$$2e^{z^2}\displaystyle\int_z^{\infty} e^{-t^2} \mathrm{d}t = \dfrac{1}{z+\dfrac{\frac{1}{2}}{z+\dfrac{1}{z+\dfrac{\frac{3}{2}}{z+\dfrac{2}{z+\ldots}}}}}$$
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$$2e^{z^2}\displaystyle\int_z^{\infty} e^{-t^2} \mathrm{d}t = \dfrac{1}{z+\dfrac{\frac{1}{2}}{z+\dfrac{1}{z+\dfrac{\frac{3}{2}}{z+\dfrac{2}{z+\ldots}}}}},$$
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where $e^{z^2}$ denotes the [[exponential]] and the right hand side denotes a [[continued fraction]].
 
==Proof==
 
==Proof==
  
 
==Refrences==
 
==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
 
* {{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
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[[Category:Theorem]]

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