证明拉马努金等式

永远

\(\displaystyle\int_z^\infty  {{e^{ - \frac{{{x^2}}}{2}}}} dx = \frac{{{e^{ - \frac{{{z^2}}}{2}}}}}{{z + \frac{1}{{z + \frac{2}{{z + \frac{3}{{z + \frac{4}{{z +  \ddots }}}}}}}}}}\)

hbghlyj

limits of integration in second fundamental theorem of calculus

Try using Mill's ratio. This tells that if $x\geq 0$$$\int^{\infty}_x e^{-u^2/2}du\sim \frac{e^{-x^2/2}}{x}$$in fact, if you put $n=xe^{x^2/2}$ and $f(n)$ such that $f(n)=x$ the limit is constant. You can express also $f$ in terms of Lambert W function.

Arriving at the asymptotic

In the book "The Probabilistic Method", the integral $\int_\lambda^\infty e^{-t^2/2}dt$ is said to be "approximately equal" to $\frac{e^{-\lambda^2/2}}{\lambda}$ for large $\lambda$. I assume what is meant is $\int_\lambda^\infty e^{-t^2/2}dt \sim \frac{e^{-\lambda^2/2}}{\lambda}$, which can be verified using l'hopital, since the ratio of the derivatives is $\frac{\lambda^2}{\lambda^2+1}$, which tends to 1.

Prove that $P(\left|Z\right|\leq z) = \sqrt{\frac{2}{\pi}}\frac{e^{-z^2/2}}{z} \left(1 + \mathcal{O} \left(\frac{1}{z^2}\right)\right)$

As a by-product, if $z > 0$ then we have
$$ \int_{z}^{\infty} e^{-x^2/2} \, dx \leq \frac{e^{-z^2/2}}{z}. $$

Two similar integration about continued fractions

For the first one, we use
$$\cfrac{1}{x + \cfrac{1}{x + \cfrac{2}{x + \cdots}}} = e^{x^2/2} \int_x^\infty e^{-t^2/2} \, dt .$$

hbghlyj

\[\int_z^{\infty } \exp \left(-\frac{x^2}{2}\right) \, dx=\sqrt{\frac{\pi }{2}} \text{erfc}\left(\frac{z}{\sqrt{2}}\right)\]
DLMF §7.9
\[\sqrt{\pi}e^{z^{2}}\operatorname{erfc}z=\cfrac{z}{z^{2}+\cfrac{\frac{1}{2}}{1+%
\cfrac{1}{z^{2}+\cfrac{\frac{3}{2}}{1+\cfrac{2}{z^{2}+\cdots}}}}},\]
Wolfram Functions
\[{\mathrm{erf}(z)}=1-\cfrac{ⅇ^{-z^2}}{\sqrt\pi}\cfrac1{z+\cfrac{1/2}{z+\cfrac1{z+\cfrac{3/2}{z+\cfrac2{z+\cfrac{5/2}{z+\cfrac3{z+\dots}}}}}}}\operatorname{/;}{\mathrm{Re}(z)}>0\]