Evaluate $\lim_{\mu \to \infty} \int_{\frac{\mu}{2}}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{(t-\mu)^2}{2}}dt$

Question

As in the title, the problem is to evaluate the integral \begin{equation} \lim_{\mu\to\infty}\int_{\frac{\mu}{2}}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{(t-\mu)^2}{2}}dt. \label{integral} \end{equation}

If one examines this graphically, then it appears that the integral should equate to 1 (see picture below). An accompanying intuitive argument is to note that the integrand is the density function of an $N(\mu,1)$ random variable, and at the limit, the distance between $\mu$ and $\mu/2$ approaches $\infty$. Since standard deviation is fixed, we should approach the situation where the area of the tail left of $\mu/2$ goes to zero, and hence the integral goes to 1.

However, when one examines the integral analytically, it's quite odd since a (naive) interpretation of the limit suggests that we are evaluating an improper integral of the form $\int_{\infty}^\infty \dots dt$, which I have not seen before.

Therefore:

1) Does this integral actually equal 1?

2) Can this be shown from the given expression?

Answer 1

If $X$ is normal with mean $0$ and variance $1$ then the integral equals $P\{X+\mu >\mu /2\}=P\{X >-\mu /2\} \to 1$ as $\mu \to \infty$.

Answer 2

My own solution based from Paul's suggestion of examining the error function:

Write the cumulative distribution of a $N(\mu,\sigma^2)$ random variable as

\begin{align} \Phi\left(\frac{x-\mu}{\sigma}\right) &= \frac{1}{2}\left[1 + \text{erf}\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)\right]. \end{align}Then the integral becomes \begin{align} 1 - \Phi\left(\frac{\mu}{2}\right) &= 1 - \frac{1}{2}\left(1 + \text{erf}\left(\frac{\mu/2 - \mu}{\sqrt{2}}\right)\right)\\ &= 1 - \frac{1}{2}\left(1 + \text{erf}\left(-\frac{\mu}{2\sqrt{2}}\right)\right)\\ &= 1 - \frac{1}{2}\left(1 - \text{erf}\left(\frac{\mu}{2\sqrt{2}}\right)\right)\\ &\to 1 - \frac{1}{2}\left(1 - 1\right),\\ &= 1, \end{align} as $\mu \to \infty$.