Evaluating integral limit in two ways gives different limits

Question

Problem

Show that $$\lim\limits_{h \rightarrow 0^{+}} \int_{-1}^{1} \frac{h}{h^{2}+x^{2}} \, dx = \pi.$$

I can do this by evaluating the integral directly and showing that it is equal to $$2\lim\limits_{h \rightarrow 0^{+}} \arctan(1/h) = \pi.$$ The problem is that I have been asked to evaluate the limit in a more roundabout way using another method (presumably because I will learn something in the process). When I use the other way I get a different answer and I cannot figure out why.

Background

Let $\phi:\mathbb{R} \rightarrow \mathbb{R}$ be non-negative and integrable on $[-1,1],$ such that $\phi(x) = 0 \,\, \forall \,\, |x| \geq 1$ and $\int_{-1}^{1} \phi = 1.$ Then we may define $\phi_{h} = \frac{1}{h} \phi(x/h).$ I have shown that, for any function $f$ which is continuous at $0,$ we must have $$\lim\limits_{h\rightarrow 0^{+}} \int_{-1}^{1} \phi_{h}(x)f(x) \, dx = f(0)$$ I am confident that this result is correct.

My work

Using the above result, we define $$\phi(x) = \left\{ \begin{array}{ll} \frac{2}{\pi} \frac{1}{1+x^{2}}, \,\, x \in (-1,1) \\ 0, \,\, |x| \geq 1 \end{array} \right. $$ Then I believe $\phi$ satisfies all the criteria given in the background. What is more, if we define $f(x) = \frac{\pi}{2}$ we have $$ \lim\limits_{h\rightarrow 0^{+}} \int_{-1}^{1} \frac{h}{h^{2}+x^{2}} \, dx = \lim\limits_{h\rightarrow 0^{+}} \int_{-1}^{1} \phi_{h}(x) f(x) \, dx = f(0) = \frac{\pi}{2}$$

Where on earth is that pesky $\frac{1}{2}$ coming from???

Answer 1

The values $\phi(\hat x)=0$ for $|\hat x|>1$ correspond to $|x|>h$ and must be integrated too... You don't have $$ \lim\limits_{h\rightarrow 0^{+}} \int_{-1}^{1} \frac{h}{h^{2}+x^{2}} \, dx = \lim\limits_{h\rightarrow 0^{+}} \int_{-1}^{1} \phi_{h}(x) f(x) \, dx$$ as $\phi_{h}(x) f(x)$ is zero for $|x|>h$.

Answer 2

What you are doing effectively is using a pdf to evaluate the integral. The integral you have is related to the Cauchy Distribution which has pdf : $$\phi(x) = \frac1\pi\frac{1}{1+x^2}$$

The problem comes in the limit $h\to \infty$. You found $\phi(x)$ correctly as $$\int_{-1}^1\frac{1}{1+x^2} dx = \frac{\pi}{2}$$

However what you need is $\phi_h(x) = \frac1h\phi(x/h)$ so the interval you need to integrate over is now $[-1,1]/h\to[-\infty,\infty]$ as $h\to 1$ So try again and you obtain $$\int_{-\infty}^\infty\frac{1}{1+x^2} dx= \pi$$

Pesky $\frac12$ gone and the result follows.