Difference between revisions of "Two-sided inequality for e^(x^2) integral from x to infinity e^(-t^2) dt for non-negative real x"
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(Created page with "==Theorem== The following formula holds for $x\geq 0$: $$\dfrac{1}{x+\sqrt{x^2+2}}< e^{x^2} \displaystyle\int_x^{\infty} e^{-t^2} \mathrm{d}t \leq \dfrac{1}{x+\sqrt{x^2+\frac{...") |
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The following formula holds for $x\geq 0$: | The following formula holds for $x\geq 0$: | ||
$$\dfrac{1}{x+\sqrt{x^2+2}}< e^{x^2} \displaystyle\int_x^{\infty} e^{-t^2} \mathrm{d}t \leq \dfrac{1}{x+\sqrt{x^2+\frac{4}{\pi}}},$$ | $$\dfrac{1}{x+\sqrt{x^2+2}}< e^{x^2} \displaystyle\int_x^{\infty} e^{-t^2} \mathrm{d}t \leq \dfrac{1}{x+\sqrt{x^2+\frac{4}{\pi}}},$$ | ||
| − | where $e^{x^2}$ denotes the [[exponential]], and $\pi$ denotes [[ | + | where $e^{x^2}$ denotes the [[exponential]], and $\pi$ denotes [[pi]]. |
==Proof== | ==Proof== | ||
Latest revision as of 01:53, 6 June 2016
Theorem
The following formula holds for $x\geq 0$: $$\dfrac{1}{x+\sqrt{x^2+2}}< e^{x^2} \displaystyle\int_x^{\infty} e^{-t^2} \mathrm{d}t \leq \dfrac{1}{x+\sqrt{x^2+\frac{4}{\pi}}},$$ where $e^{x^2}$ denotes the exponential, and $\pi$ denotes pi.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): 7.1.13
