Limit of an integral $\lim_{r\to 0^+} \int_0^1 \left(\frac{f(x)r}{x^2+r^2}\right )~dx$ Suppose $f$ is a real-valued continuous function on the unit interval $[0,1]$. How can we compute $$\lim_{r\to 0^+} \int_0^1 \left(\dfrac{f(x)r}{x^2+r^2} \right)~dx?$$
If we let $\min f(x)=m$ and $\max f(x)=M$, then we have $$m\tan^{-1}\left(\frac{1}{r}\right)\leq \int_0^1 \left(\dfrac{f(x)r}{x^2+r^2} \right)~dx \leq M\tan^{-1}\left(\frac{1}{r}\right), $$ since $\int_0^1 r/(x^2+r^2)~dx=\tan^{-1}(1/r)$.
 A: Define $M$ such that $|f(x)|<M$ for $x\in [0,1]$. First, note that for all fixed $\gamma>0$ and all $r>0$ we have
$$r\int_\gamma^1\frac{f(x)}{x^2+r^2}dx\leq rM\int_\gamma^1 \frac{1}{x^2+r^2}dx<rM\int_\gamma^1 \frac{1}{x^2}dx=rM\left(\frac{1}{\gamma}-1\right)$$
$$r\int_\gamma^1\frac{f(x)}{x^2+r^2}dx\geq -rM\int_\gamma^1 \frac{1}{x^2+r^2}dx>-rM\int_\gamma^1 \frac{1}{x^2}dx=-rM\left(\frac{1}{\gamma}-1\right)$$
This implies that for all fixed $\gamma>0$
$$\lim_{r\to 0^{+}}r\int_\gamma^1\frac{f(x)}{x^2+r^2}dx=0$$
Second, since $f(x)$ is continuous at $0$, for all $\epsilon>0$ there exists $\delta>0$ such that $0\leq x\leq\delta$ implies
$$|f(0)-f(x)|<\epsilon$$
$$f(0)-\epsilon<f(x)<f(0)+\epsilon$$
Finally, let $\epsilon>0$ be arbitrary. Split the integral up at $\delta$ (from above):
$$r\int_0^1\frac{f(x)}{x^2+r^2}dx=r\int_0^\delta\frac{f(x)}{x^2+r^2}dx+r\int_\delta^1\frac{f(x)}{x^2+r^2}dx$$
From the first step, we know
$$\lim_{r\to 0^{+}}r\int_0^1\frac{f(x)}{x^2+r^2}dx=\lim_{r\to 0^{+}}\left[r\int_0^\delta\frac{f(x)}{x^2+r^2}dx+r\int_\delta^1\frac{f(x)}{x^2+r^2}dx\right]=\lim_{r\to 0^{+}}r\int_0^\delta\frac{f(x)}{x^2+r^2}dx$$
This integral can be bounded by
$$r\int_0^\delta\frac{f(0)-\epsilon}{x^2+r^2}dx<r\int_0^\delta\frac{f(x)}{x^2+r^2}dx<r\int_0^\delta\frac{f(0)+\epsilon}{x^2+r^2}dx$$
$$r(f(0)-\epsilon)\int_0^\delta\frac{1}{x^2+r^2}dx<r\int_0^\delta\frac{f(x)}{x^2+r^2}dx<r(f(0)+\epsilon)\int_0^\delta\frac{1}{x^2+r^2}dx$$
But we know
$$\int_0^\delta\frac{1}{x^2+r^2}dx=\frac{1}{r}\left[\arctan(\delta/r)-\arctan(0/r)\right]=\frac{1}{r}\arctan(\delta/r)$$
Thus
$$\lim_{r\to 0^{+}}r(f(0)\pm\epsilon)\int_0^\delta\frac{1}{x^2+r^2}dx=\lim_{r\to 0^{+}}(f(0)\pm\epsilon)\arctan(\delta/r)=(f(0)\pm\epsilon)\frac{\pi}{2}$$
This implies
$$f(0)\frac{\pi}{2}-\epsilon\frac{\pi}{2}\leq \lim_{r\to 0^{+}}r\int_0^\delta\frac{f(x)}{x^2+r^2}dx\leq f(0)\frac{\pi}{2}+\epsilon\frac{\pi}{2}$$
However, since $\epsilon$ was arbitrary, this simplifies to
$$f(0)\frac{\pi}{2}\leq \lim_{r\to 0^{+}}r\int_0^\delta\frac{f(x)}{x^2+r^2}dx\leq f(0)\frac{\pi}{2}$$
We conclude
$$\lim_{r\to 0^{+}}\int_0^\delta\frac{f(x)r}{x^2+r^2}dx= f(0)\frac{\pi}{2}$$
