weighted integral of a continuous function I am not sure about my answer and wanted to  doublecheck  my steps.
Suppose $ f \colon \mathbb{R} \to \mathbb{R} $ is a  continuous and bounded function. Prove that
$\lim_{y \to 0^+} \frac{y}{\pi}\int_{-\infty}^{\infty} \frac{f(x)}{x^2+y^2}\,dx =f(0)$
This is my attempt: Setting $t=arctg(\frac{x}{y})$ , the integral becomes:
$\lim_{y \to 0^+} \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(y\tan(t)) \,dt$.
Since $f$ is bounded  I can apply the Lebesgue dominated convergence theorem to bring the limit under the sign of integral and then,since f is continuous at zero, I bring the limit inside the argument of the f, which yelds to the desired result.  
 A: Let $ y\in\mathbb{R}_{+} $, we have :
\begin{aligned}\frac{y}{\pi}\int_{-\infty}^{+\infty}{\frac{f\left(x\right)}{x^{2}+y^{2}}\,\mathrm{d}x}&=\frac{1}{y\pi}\int_{-\infty}^{+\infty}{\frac{f\left(x\right)}{1+\left(\frac{x}{y}\right)^{2}}\,\mathrm{d}x}\\ &=\frac{1}{\pi}\int_{-\infty}^{+\infty}{\frac{f\left(xy\right)}{1+x^{2}}\,\mathrm{d}x}\end{aligned}
Since $ \frac{1}{\pi}\int_{-\infty}^{+\infty}{\frac{\mathrm{d}x}{1+x^{2}}}=1 $, and since $ f $ is continuous, given an $ \varepsilon>0 $, there exists some $ \eta>0 $ such that : $$ \left(\forall x\in\mathcal{B}\left(0,\eta\right)\right),\ \left|f\left(x\right)-f\left(0\right)\right|<\varepsilon $$ Thus : \begin{aligned} \small\left|\frac{1}{\pi}\int_{-\infty}^{+\infty}{\frac{f\left(xy\right)}{1+x^{2}}\,\mathrm{d}x}-f\left(0\right)\right|&\small=\left|\frac{1}{\pi}\int_{-\infty}^{+\infty}{\frac{f\left(xy\right)-f\left(0\right)}{1+x^{2}}\,\mathrm{d}x}\right|\\ &\small\leq\frac{1}{\pi}\int_{-\infty}^{+\infty}{\frac{\left|f\left(xy\right)-f\left(0\right)\right|}{1+x^{2}}\,\mathrm{d}x}\\ &\small\leq\frac{1}{\pi}\int_{-\infty}^{-\frac{\eta}{y}}{\frac{\left|f\left(xy\right)-f\left(0\right)\right|}{1+x^{2}}\,\mathrm{d}x}+\frac{1}{\pi}\int_{-\frac{\eta}{y}}^{\frac{\eta}{y}}{\frac{\left|f\left(xy\right)-f\left(0\right)\right|}{1+x^{2}}\,\mathrm{d}x}+\frac{1}{\pi}\int_{\frac{\eta}{y}}^{+\infty}{\frac{\left|f\left(xy\right)-f\left(0\right)\right|}{1+x^{2}}\,\mathrm{d}x}\\ &\small\leq\frac{1}{\pi}\int_{-\infty}^{-\frac{\eta}{y}}{\frac{\left|f\left(xy\right)-f\left(0\right)\right|}{1+x^{2}}\,\mathrm{d}x}+\frac{\varepsilon}{\pi}\int_{-\frac{\eta}{y}}^{\frac{\eta}{y}}{\frac{\mathrm{d}x}{1+x^{2}}}+\frac{1}{\pi}\int_{\frac{\eta}{y}}^{+\infty}{\frac{\left|f\left(xy\right)-f\left(0\right)\right|}{1+x^{2}}\,\mathrm{d}x} \end{aligned}
For the $ y $ that we fixed in the begining, let's denote $ g_{y}:x\mapsto\frac{\left|f\left(xy\right)-f\left(0\right)\right|}{1+x^{2}} \cdot $
For $ \left(a,b\right)\in\left(\mathbb{R}_{+},\mathbb{R}_{-}\right) $, $ \int_{a}^{+\infty}{g_{y}\left(x\right)\mathrm{d}x} $, and $ \int_{-\infty}^{b}{g_{y}\left(x\right)\mathrm{d}x} $ are respectively the remainders of $ \int_{0}^{+\infty}{f_{y}\left(x\right)\mathrm{d}x} $, and $ \int_{-\infty}^{0}{f_{y}\left(x\right)\mathrm{d}x} $ which both converge. Thus, we have : $$ \lim_{a\to +\infty}{\int_{a}^{+\infty}{g_{y}\left(x\right)\mathrm{d}x}}=0 \ \ \ \ \ \textrm{and} \ \ \ \ \ \lim_{b\to -\infty}{\int_{-\infty}^{b}{g_{y}\left(x\right)\mathrm{d}x}}=0 $$
Thus, there exists $ \eta_{1},\eta_{2}\in\mathbb{R}_{+}^{*}=\left(0,+\infty\right) $, such that : $$\small \left(\forall x\in\mathcal{B}\left(0,\eta_{1}\right)\right),\ \int_{-\infty}^{-\frac{\eta}{y}}{\frac{\left|f\left(xy\right)-f\left(0\right)\right|}{1+x^{2}}\,\mathrm{d}x}<\varepsilon \ \ \ \ \ \ \ \textrm{and} \ \ \ \ \ \ \ \left(\forall x\in\mathcal{B}\left(0,\eta_{2}\right)\right),\ \int_{\frac{\eta}{y}}^{+\infty}{\frac{\left|f\left(xy\right)-f\left(0\right)\right|}{1+x^{2}}\,\mathrm{d}x}<\varepsilon $$
Because they both approache $ 0 $ as $ y $ approaches $ 0^{+} \cdot $ We get that $ \left(\forall x\in\mathcal{B}\left(0,\min\left(\eta,\eta_{1},\eta_{2}\right)\right)\right) $ : $$ \left|\frac{1}{\pi}\int_{-\infty}^{+\infty}{\frac{f\left(xy\right)}{1+x^{2}}\,\mathrm{d}x}-f\left(0\right)\right|\leq\frac{2\varepsilon}{\pi}\left(1+\arctan{\left(\frac{\eta}{y}\right)}\right)<\left(\frac{2+\pi}{\pi}\right)\varepsilon $$
Hence : $$ \lim_{y\to 0^{+}}{\frac{y}{\pi}\int_{-\infty}^{+\infty}{\frac{f\left(x\right)}{x^{2}+y^{2}}\,\mathrm{d}x}}=\lim_{y\to 0^{+}}{\frac{1}{\pi}\int_{-\infty}^{+\infty}{\frac{f\left(xy\right)}{1+x^{2}}\,\mathrm{d}x}}=f\left(0\right) $$
