Difference between revisions of "Continued fraction for 2e^(z^2) integral from z to infinity e^(-t^2) dt for positive Re(z)"
From specialfunctionswiki
(Created page with "==Theorem== The following formula holds for $\mathrm{Re}(z)>0$: $$2e^{z^2}\displaystyle\int_z^{\infty} e^{-t^2} \mathrm{d}t = \dfrac{1}{z+\dfrac{\frac{1}{2}}{z+\dfrac{1}{z+\df...") |
|||
| Line 1: | Line 1: | ||
==Theorem== | ==Theorem== | ||
| − | The following formula holds for $\mathrm{Re}(z)>0$: | + | The following formula holds for $\mathrm{Re}(z)>0$ (and the final formula holds with the additional requirement on $z$: $\mathrm{Im}(z)\neq 0$): |
| − | $$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}}}}}$$ | + | $$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)}},$$ |
| + | where $x_k^{(n)}$ and $H_k^{(n)}$ are the zeros and weight factors of the [[Hermite polynomials]]. | ||
==Proof== | ==Proof== | ||
==Refrences== | ==Refrences== | ||
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Two-sided inequality for e^(x^2) integral from x to infinity e^(-t^2) dt for non-negative real x|next=Continued fraction for 1/sqrt(pi) integral from -infinity to infinity of e^(-t^2)/(z-t) dt}}: 7.1.14 | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Two-sided inequality for e^(x^2) integral from x to infinity e^(-t^2) dt for non-negative real x|next=Continued fraction for 1/sqrt(pi) integral from -infinity to infinity of e^(-t^2)/(z-t) dt}}: 7.1.14 | ||
Revision as of 02:00, 6 June 2016
Theorem
The following formula holds for $\mathrm{Re}(z)>0$ (and the final formula holds with the additional requirement on $z$: $\mathrm{Im}(z)\neq 0$): $$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.
Proof
Refrences
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): 7.1.14
