Find $ \lim_{\sigma \rightarrow \infty} \int_{-\infty}^{\infty} q(x) \frac{\exp \big(-\frac{(x-\mu)^2}{2 \sigma^2}\big)}{\sigma} dx $, where $ q(x) $ is "erf(x).
I need to compute the expression shown above. I believed the dominated convergence theorem could be used to solve this. But based on the provided answer, it does not seem possible. I would appreciate that would help me in sketching the general steps I need to follow. Or is there a simpler way to solve this? Thank you.
We can calculate the limes explicitly as follows. First, recall that the error function is given by $$q(x) = \frac{2}{\sqrt{\pi}} \int_x^\infty \exp(-x^2) \, \mathrm{d} x$$ and therefore $q(x) \le 2$ because of $$\int_{-\infty}^\infty \exp(-y^2) \, dy = \sqrt{\pi}.$$ We start with the change of variables $ y= (x-\mu)/\sqrt{2 \sigma^2}$ to get $$\int_{-\infty}^\infty q(x) \frac{\exp \Big( - \frac{(x-\mu)^2}{2 \sigma^2} \Big)}{\sigma} \, dx = \sqrt{2} \int_{-\infty}^\infty q(\sqrt{2} \sigma y + \mu) \exp(-y^2) \, dy.$$ If $y >0$, then $q(\sqrt{2} \sigma y + \mu) \rightarrow 0$ for $\sigma \rightarrow \infty$ and if $y <0$, then $q(\sqrt{2} \sigma y + \mu) \rightarrow 2$ for $\sigma \rightarrow \infty$. Thus, since the integrand is bounded by $2 \exp(-y^2)$, we get after an application of Lebesgue's dominated convergence theorem that $$\lim_{\sigma \rightarrow \infty} \int_{-\infty}^\infty q(x) \frac{\exp \Big( - \frac{(x-\mu)^2}{2 \sigma^2} \Big)}{\sigma} \, dx = 2 \sqrt{2} \int_{-\infty}^0 \exp(-y^2) \, dy = \sqrt{2\pi},$$ where we used the Gaussian integral formula again.