# Limit of a integral whose integrand diverges under the limit

I am trying to simplify the following limit of integral where $$\mu$$ is given:

$$p(y) = \lim_{\sigma \to 0} \int_{\mathbb R} |x| \cdot \frac{1}{\sqrt{2\pi\sigma^2} } e^{-\frac{1}{2\sigma^2} (xy - \mu)^2} f(x) dx,$$

however I am not sure if there is a way to simplify it, as the integrand does not converge under the limit $$\sigma \to 0$$ and the interchangeablity of limit and integral fails here:

$$\lim_{\sigma \to 0} |x| \cdot \frac{1}{\sqrt{2\pi\sigma^2} } e^{-\frac{1}{2\sigma^2} (xy - \mu)^2} f(x) = \begin{cases} \inf, & x \ne \mu /y; \\ 0, & x = \mu /y \end{cases}.$$

However, I DO know the answer of the limit of the integral, as naturally it is the probablity density function of $$Y = T/X$$, where $$X$$ has PDF $$f(x)$$ and $$T \sim N(\mu, \sigma^2)$$ is a normal distribution. The limit of integral at $$\sigma \to 0$$ then essentially means the PDF of $$\mu / X$$ and can be easily calculated.

So how do I calculate this limit of integral analytically, without the help of the intuition from probablity?

$$\newcommand\si\sigma\newcommand\R{\mathbb R}$$We have to find $$\lim_{\si\downarrow0}p_\si(y)$$ for all real $$y$$ such that the limit exists, where $$p_\si(y):=\int_\R\frac{|x|\,dx}{\si\sqrt{2\pi}}e^{-(xy-\mu)^2/(2\si^2)}\,f(x).$$ Assuming that $$f$$ is bounded, and using the substitution $$x=(\mu+\si z)/y$$ and dominated convergence, we get $$p_\si(y)=\frac1{y^2}\,\int_\R\frac{|\mu+\si z|\,dz}{\sqrt{2\pi}}e^{-z^2/2} f\Big(\frac{\mu+\si z}y\Big) \to p(y):=\frac{|\mu|}{y^2}\,f\Big(\frac\mu y\Big) \tag{1}\label{1}$$ (as $$\si\downarrow0$$) for all real $$y\ne0$$ such that $$\mu/y$$ is a point of continuity of $$f$$.

If $$y=0$$, then $$\lim_{\si\downarrow0}p_\si(y)=\infty$$. If $$y\ne0$$ and $$\mu/y$$ is not a point of continuity of $$f$$, then $$\lim_{\si\downarrow0}p_\si(y)$$ does not exist in general.

However, \eqref{1} holds for every real $$y\ne0$$ such $$\mu/y$$ is a Lebesgue point of continuity of $$f$$. So, \eqref{1} holds for almost all real $$y$$.

The Gaussian tends to a delta function in the $$\sigma\rightarrow 0$$ limit,

$$\lim_{\sigma \to 0} \frac{1}{\sqrt{2\pi\sigma^2} } e^{-\frac{1}{2\sigma^2} (xy - \mu)^2} =\delta(xy-\mu)=|y|^{-1}\delta(x-\mu/y),$$ so the integral evaluates to $$p(y) = |y|^{-1}\int_{-\infty}^\infty |x| f(x)\delta(x-\mu/y) \,dx=|y|^{-1}|\mu/y|f(\mu/y).$$

• This answer incorrect for any real $y<0$, because the limit (if exists) must be $\ge0$. Also, as noted in my answer, if $y\ne0$ and $\mu/y$ is not a point of continuity of $f$, then the limit does not exist in general. One also needs a condition such as the boundedness of $f$ to make the limit transition. Also, you need to specify in what sense and why the delta function is the limit. Commented Nov 1, 2023 at 21:04
• you're right, I missed an absolute value sign in the delta function, thank you for correcting me. Commented Nov 1, 2023 at 21:07
• There are a number of other points as well, listed above, missing in your answer. Commented Nov 1, 2023 at 21:08