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:

*

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


*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?
 A: 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 $\lim_{t \to 0} \int_{-1}^1 \frac{tf(x)}{t^2+x^2} dx = 0$.
