It is given $f \in C[0;1]$ is positive. Prove that $I(y)=\int_0^1{\frac{yf(x)}{x^2+y^2}}dx$ is discontinuous at point $y=0$ It is given $f \in C[0;1]$ is positive. Prove that
$I(y)=\int_0^1{\frac{yf(x)}{x^2+y^2}}dx$ is discontinuous at point $y=0$
So we need to show that $I(0) \neq lim_{y\to0}\int_0^1{\frac{yf(x)}{x^2+y^2}}dx$
$I(0)=c,c\in \mathbb{R}$
and limit equals $f(0)\frac{\pi}{2}$.  At this step I am unsure what I am doing wrong.Will be thankful for your help.
 A: Let $g_y(x)=\frac{yf(x)}{x^2+y^2}$ denote the integrand for $y\in\mathbb R$ and $x\in(0,1)$. Then we have $g_0\equiv 0$ and hence $I(0)=0$.
For $\varepsilon\in(0,1)$ let $I(y)=I_0(y)+I_1(y)$ with $I_0(y)=\int_\varepsilon^1g_y(x)\mathrm dx$ and
$I_1(y)=\int_0^\varepsilon g_y(x)\mathrm dx$.
Using the extreme value theorem let $f_+=\max\{f(x):x\in[0,1]\}$.
Hence, for $y>0$ we have $g_y(x)\le yf_+/x^2$ and further $0\le I_0(y)\le yf_+\int_\varepsilon^1 x^{-2}\mathrm dx=yf_+(\varepsilon^{-1}-1)$ using $f\ge 0$ and
monotonicity, so $\lim_{y\downarrow 0}I_0(y)=0$.
Now, let $f_-(\varepsilon)=\min\{f(x):x\in[0,\varepsilon]\}$ and $f_+(\varepsilon)=\max\{f(x):x\in[0,\varepsilon]\}$, then for $y>0$ we have
$$f_-(\varepsilon)\int_0^\varepsilon\frac{y}{x^2+y^2}\mathrm{d}x\le I_1(y)\le
f_+(\varepsilon)\int_0^\varepsilon\frac{y}{x^2+y^2}\mathrm{d}x.$$
Now, we substitute $x=yt$, yielding $\mathrm dx=y\mathrm dt$ and
$$\int_0^\varepsilon\frac{y}{x^2+y^2}\mathrm{d}x=
\int_0^{\varepsilon/y}\frac{y^2}{(yt)^2+y^2}\mathrm{d}x=
\int_0^{\varepsilon/y}\frac{1}{t^2+1}\mathrm{d}x=\tan^{-1}(\varepsilon/y),$$
where we solved the well-known integral. Using the well-known limit $\lim_{t\rightarrow\infty}\tan^{-1}(t)=\pi/2$ we obtain $f_-(\varepsilon)\frac{\pi}{2}\le \lim_{y\downarrow 0}I_1(y)\le f_+(\varepsilon)\frac{\pi}{2}$. Using that $f$ is continuous at $0$, taking the limit $\varepsilon\downarrow 0$ now yields $\lim_{y\downarrow 0}I_1(y)=f(0)\frac{\pi}{2}$ and thereby $\lim_{y\downarrow 0}I(y)=f(0)\frac{\pi}{2}$. Since the proof for $y<0$ is very similar, I will not repeat the entire argument.
