Prove $\lim_{y \to 0^+} \frac{y}{\pi} \int_{-\infty}^{+\infty} \frac{f(x) \, \mathrm{d}x}{x^2+y^2} = f(0)$ with $f$ continuous, bounded I have been trying to solve the following problem, given $f$ continuous and bounded prove that
$$\lim_{y \to 0^+} \frac{y}{\pi} \int_{-\infty}^{+\infty} \frac{f(x) \, \mathrm{d}x}{x^2+y^2} = f(0)$$
I tried first solving this by parts by integrating $\frac{1}{x^2+y^2}$ and differentiating $f(x)$ but this doesn't seem to work or bring me anywhere.
 A: $${y\over\pi}\int_{-\infty}^\infty {f(x)\, dx \over x^2 + y^2} = 
{y\over\pi}\int_{-\infty}^\infty {f(xy)y\,dx\over y^2x^2 + y^2 }
= {1\over \pi}\int_{-\infty}^\infty {f(xy)\,dx\over 1 + x^2}.$$
Now apply bounded convergence against the measure $dx/(1+x^2)$.
A: First note that for any $y>0$
$$\int_{-\infty}^\infty \frac{y}{\pi(x^2+y^2)}\,dx=1$$
So, it suffices to show that 
$$\lim_{y\to 0^+}\int_{-\infty}^\infty \frac{y(f(x)-f(0))}{\pi(x^2+y^2)}\,dx=0$$

Given $\epsilon>0$, fix a number $\delta>0$ such that $|f(x)-f(0)|<\epsilon$ for $|x|<\delta$.  
With this fixed $\delta$, we write for $y>0$
$$\begin{align}
\left|\int_{-\infty}^\infty \frac{y(f(x)-f(0))}{\pi(x^2+y^2)}\,dx\right|&\le\left|\int_{|x|\le \delta} \frac{y(f(x)-f(0))}{\pi(x^2+y^2)}\,dx\right|+\left|\int_{|x|\ge\delta} \frac{y(f(x)-f(0))}{\pi(x^2+y^2)}\,dx\right|\\\\
&\le \int_{|x|\le \delta} \frac{y\left|f(x)-f(0)\right|}{\pi(x^2+y^2)}\,dx+ \int_{|x|\ge \delta} \frac{y\left|f(x)-f(0)\right|}{\pi(x^2+y^2)}\,dx\\\\
&\le \frac{2\epsilon}\pi \arctan(\delta/y)+\frac{2\sup_{|x|\ge\delta }(f(x))}\pi \left(\pi/2-\arctan(\delta/y)\right)\tag 1
\end{align}$$
Letting $y\to 0^+$ in $(1)$, we find that for any given $\epsilon>0$, 
$$\lim_{y\to 0^+}\left|\int_{-\infty}^\infty \frac{y(f(x)-f(0))}{\pi(x^2+y^2)}\,dx\right|\le \epsilon$$
And we are done!
