# 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.

• In the second formula there is not $\frac{1}{y}$ in front of the integral. A part this, I think the argument is correct Apr 9 '20 at 23:12
• yes ,my mistake Apr 9 '20 at 23:16

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 : $$\lim_{y\to 0^{+}}{\frac{y}{\pi}\int_{-\infty}^{+\infty}{\frac{f\left(x\right)}{x^{2}+y^{2}}\,\mathrm{d}x}}=\lim_{y\to 0^{+}}{\frac{1}{\pi}\int_{-\infty}^{+\infty}{\frac{f\left(xy\right)}{1+x^{2}}\,\mathrm{d}x}}=f\left(0\right)$$

• thanks for the nice solution, could you please also have a look to my attempt and see if there is any mistake? Apr 9 '20 at 23:41