Help with $ \lim\limits_{t \to 0} \int_{-1}^1 \frac{t}{t^2+x^2} f(x)\ dx$

Question

Show that if $f$ is continuous on $[-1,1]$, then $$ \lim_{t \to 0} \int_{-1}^{1}\frac{t}{t^2+x^2}f(x)\,dx=\pi f(0) $$ Any hints?

Answer 1

Let us extend (by preserving continuity) the domain of $f$ by setting $f(x)=f(1)$ for any $x\geq 1$ and $f(x)=f(-1)$ for any $x\leq -1$. By the substitution $x=tz,\,dx=t\,dz$ we get:

$$ \int_{-1}^{1}\frac{t\,f(x)}{t^2+x^2}\,dx = \int_{-1/t}^{1/t}\frac{f(zt)}{1+z^2}\,dx =O(t)+\int_{-\infty}^{+\infty}\frac{f(zt)}{1+z^2}\,dz$$ and by the dominated convergence theorem: $$ \lim_{t\to 0^+}\int_{-1}^{1}\frac{t\,f(x)}{t^2+x^2}\,dx = \lim_{t\to 0^+}\int_{-\infty}^{+\infty}\frac{f(zt)}{1+z^2}\,dz = \int_{-\infty}^{+\infty}\frac{f(0)}{1+z^2}\,dz = \pi\,f(0) $$ as wanted. The $O(t)$ term comes from $$ \int_{1/t}^{+\infty}\frac{1}{1+z^2}\,dz = \arctan(t) = O(t).$$

Answer 2

$\boldsymbol{\delta}$-$\boldsymbol{\epsilon}$ Approach

Choose an $\epsilon\gt0$. Pick a $0\lt\delta\lt1$ so that $|x|\le\delta\implies|f(x)-f(0)|\le\epsilon$. Pick an $M$ so that $\int_{|x|\ge M}\frac1{1+x^2}\,\mathrm{d}x\le\frac{\epsilon}{\max\limits_{|x|\le1}|f(x)|}$. Then for $t\lt \delta/M$, $$ \begin{align} &\left|\,\int_{-1}^1\frac{t}{t^2+x^2}f(x)\,\mathrm{d}x-\pi f(0)\,\right|\\ &=\left|\,\int_{-1/t}^{1/t}\frac1{1+x^2}f(xt)\,\mathrm{d}x-\int_{-\infty}^\infty\frac1{1+x^2}f(0)\,\mathrm{d}x\,\right|\\ &\le\int_{|x|\ge1/t}\frac1{1+x^2}|f(0)|\,\mathrm{d}x+\int_{|x|\lt1/t}\frac1{1+x^2}|f(xt)-f(0)|\,\mathrm{d}x\\ &\le\color{#C00}{\int_{|x|\ge1/t}\frac1{1+x^2}|f(0)|\,\mathrm{d}x}+\color{#090}{\int_{\delta/t\le|x|\lt1/t}\frac1{1+x^2}|f(xt)-f(0)|\,\mathrm{d}x}\\ &+\color{#00F}{\int_{|x|\lt\delta/t}\frac1{1+x^2}|f(xt)-f(0)|\,\mathrm{d}x}\\[6pt] &\le\color{#C00}{\epsilon}+\color{#090}{2\epsilon}+\color{#00F}{\pi\epsilon} \end{align} $$ Thus, for any $\epsilon\gt0$ $$ \lim_{t\to0}\left|\,\int_{-1}^1\frac{t}{t^2+x^2}f(x)\,\mathrm{d}x-\pi f(0)\,\right|\le(3+\pi)\epsilon $$ which means that $$ \lim_{t\to0}\left|\,\int_{-1}^1\frac{t}{t^2+x^2}f(x)\,\mathrm{d}x-\pi f(0)\,\right|=0 $$

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Dominated Convergence Approach

$[|xt|\le1]$, where $[\cdots]$ are Iverson brackets, is the characteristic function of $\left[-\frac1t,\frac1t\right]$. Then

$\left|\color{#C00}{\frac{[|xt|\le1]}{1+x^2}f(xt)}\right|\le\overbrace{\max\limits_{|u|\le1}|f(u)|\frac1{1+x^2}}^{L^1}$ and for all $x\in\mathbb{R}$, $\lim\limits_{t\to0}\color{#C00}{\frac{[|xt|\le1]}{1+x^2}f(xt)}=\color{#090}{\frac{f(0)}{1+x^2}}$. Therefore, $$ \begin{align} \lim_{t\to0}\int_{-1}^1\frac{t}{t^2+x^2}f(x)\,\mathrm{d}x &=\lim_{t\to0}\int_{-1/t}^{1/t}\frac1{1+x^2}f(xt)\,\mathrm{d}x\\ &=\lim_{t\to0}\int_{-\infty}^\infty\color{#C00}{\frac{[|xt|\le1]}{1+x^2}f(xt)}\,\mathrm{d}x\\[3pt] &=\int_{-\infty}^\infty\color{#090}{\frac{f(0)}{1+x^2}}\,\mathrm{d}x\\[9pt] &=\pi f(0) \end{align} $$