Integral minus constant $f(0)$ passes under the integral

Question

Let $f\in C[0,\infty]$. Show that $\lim_{y\to 0}\frac{2}{\pi}\int_\limits{0}^{\infty}\frac{yf(x)}{x^2+y^2}dx=0$.

Defining $\varphi(y)=\frac{2}{\pi}\int_\limits{0}^{\infty}\frac{yf(x)}{x^2+y^2}dx$ We need to prove $\forall\epsilon>0,\exists\delta>0\:\:\:\:|y|<\delta$

$|\varphi(y)-f(0)|<\epsilon$

$\varphi(y)-f(0)= \frac{2}{\pi}\int_\limits{0}^{\infty}\frac{y (f(x)-f(0)}{x^2+y^2}dx$

Question:

Considering $\varphi(y)-f(0)= \frac{2}{\pi}\int_\limits{0}^{\infty}\frac{y (f(x)-f(0)}{x^2+y^2}dx$ How can $f(0)$ pass under the integral?

Thanks in advance!

Answer 1

Because $$ \frac{2}{\pi} \int_0^{\infty} \frac{y}{x^2+y^2} \, dx = 1, $$ and $f(0)$ is a constant, so $$ \frac{2}{\pi} \int_0^{\infty} \frac{y f(x)}{x^2+y^2} \, dx - f(0) = \frac{2}{\pi} \int_0^{\infty} \frac{yf(x)}{x^2+y^2} \, dx - f(0)\frac{2}{\pi} \int_0^{\infty} \frac{y}{x^2+y^2} \, dx \\ = \frac{2}{\pi} \int_0^{\infty} \frac{yf(x)}{x^2+y^2} \, dx - \frac{2}{\pi} \int_0^{\infty} \frac{yf(0)}{x^2+y^2} \, dx. $$