# Integral challenge

The challenge is: if $$f:\mathbb{R}\to\mathbb{R}$$ is a continuous function, then $$\lim_{t\to0+}\frac{1}{\pi}\int_{-1}^1\frac{tf(x)}{t^2+(x-a)^2}\,\mathrm{d}x=f(a)$$ for every $$a\in[-1,1]$$

My attemps:

1. Under certainly conditions, i put the limit symbol under the integral sign, but the limit would be $$0$$. So it is a bad idea.

2. Therefore, i think use Weierstrass theorem for $$f$$ in $$[-1,1]$$, but i can't solve the problem in the case of $$f(x)=x^n$$.

What do you recommendme?

• It could be insightful to start proving the result when $f(x)$ is a constant function. In any case, that $1/ \pi$ makes me think we need Fourier series. Mar 2 at 22:14
• To add to @Crostul's suggestion, another possibly useful thing could be to assume that $f$ is periodic with a period of $2$ since the equation is only concerned with $a \in [-1, 1]$. (However, this may lead to $f$ being discontinuous at odd integers.) Mar 2 at 22:18

I consider the case $$a=0$$.
Note that $$\lim_{t \to 0 } \int_{-1}^1 \frac{tf(0)}{t^2+x^2} dx = f(0) \lim_{t \to 0} \left( \arctan(1/t) - \arctan(-1/t) \right) = \pi f(0).$$
This essentially reduces the problem to the case where $$f(0)=0$$.
If $$f(0)=0$$, by continuity for every $$\epsilon > 0$$ there exists a $$\delta > 0$$ such that $$|f(x)| \le \epsilon$$ for $$|x| \le \delta$$. It then follows that $$\lim_{t \to 0} \int_{|x| > \delta} \frac{tf(x)}{t^2+x^2} dx = \int_{|x| > \delta} \lim_{t \to0} \frac{tf(x)}{t^2+x^2} dx = \int_{|x|>\delta} 0 \, dx = 0.$$ On the other hand, we have $$\left| \lim_{t \to 0} \int_{-\delta}^\delta \frac{tf(x)}{t^2+x^2} dx \right| \le \lim_{t \to 0} \int_{-\delta}^\delta \left| \frac{tf(x)}{t^2+x^2} \right| dx \le 2\epsilon \lim_{t \to 0} \int_{0}^\delta \frac{t}{t^2+x^2} dx = \lim_{t \to 0} 2\epsilon \arctan(\delta/t) = 2\pi \epsilon.$$ It follows that $$\left| \lim_{t \to 0} \int_{-1}^1 \frac{tf(x)}{t^2+x^2} dx \right| \le 2\pi \epsilon$$ for all $$\epsilon > 0$$, so $$\lim_{t \to 0} \int_{-1}^1 \frac{tf(x)}{t^2+x^2} dx = 0$$.
• At first I didn't understand how your reduction to the $f(0)=0$ case worked, so to clarify it here: we write $f(x)=f(0)+g(x)$, so that $g(0)=0$, and integrate the resulting two terms separately.