Arriving at the asymptotic $\int \limits_\lambda^\infty e^{-t^2/2}dt \sim \frac{e^{-\lambda^2/2}}{\lambda}$

Question

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.

However, it is not clear to me how to arrive at the expression $\frac{e^{-\lambda^2/2}}{\lambda}$ from scratch. Is there a method for that?

Answer 1

With $t=\dfrac u\lambda+\lambda$,

$$\int_\lambda^\infty e^{-t^2/2}dt=\int_0^\infty e^{-(u/\lambda+\lambda)^2/2}\frac{du}\lambda=\frac{e^{-\lambda^2/2}}\lambda\int_0^\infty e^{-u^2/2\lambda^2}e^{-u}dt.$$

For large $\lambda$, the second factor decreases faster and

$$\int_0^\infty e^{-u^2/2\lambda^2}e^{-u}du\approx\int_0^\infty e^{-u}du.$$

Answer 2

Almost from definition $$I=\int e^{-\frac{t^2}{2}}\,dt=\sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{t}{\sqrt{2}}\right)$$ So $$J=\int_\lambda^\infty e^{-\frac{t^2}{2}}\,dt=\sqrt{\frac{\pi }{2}} \text{erfc}\left(\frac{\lambda }{\sqrt{2}}\right)$$ Now, look here for the asymptotic expansion for large $\lambda$ to get $$J=e^{-\frac{\lambda ^2}{2}} \left(\frac{1}{\lambda }-\frac{1}{\lambda ^3}+\frac{3}{\lambda ^5}+O\left(\frac{1}{\lambda^7 }\right)\right)$$