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I have to prove that if $f$ is a continuous function, then:

$$\lim\limits_{h \to 0^+}{\int\limits_{-1}^1{\frac{h}{h^2+x^2}}f(x)\:dx}=\pi f(0)$$

So far what I've tried is noting that $f$ is bounded on $[-1,1]$ so let $M=\sup\limits_{x\in{[-1,1]}}{f(x)}$, and $m=\inf\limits_{x\in{[-1,1]}}{f(x)}$. Then maybe we can use the squeeze theorem somehow to prove the result, but it didn't really lead me anywhere.

Does anyone have any hints? Are there any techniques to solving problems similar to this-- ones with generic $f$'s under the integrand?

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From this manipulation

$$\int_{-1}^1 \frac{h}{h^2+x^2}f(x)dx = \int_{-1}^1 \frac{1}{1+\frac{x^2}{h^2}}f(x)\frac{dx}{h}$$

the substitution $z = \frac{x}{h}$ becomes obvious.

$$ = \int_{-\frac{1}{h}}^{\frac{1}{h}} \frac{f(hz)}{1+z^2}dz$$

Then use dominated convergence to move the limit into the integral

$$ \lim_{h\to 0^+} \int_{-\frac{1}{h}}^{\frac{1}{h}} \frac{f(hz)}{1+z^2}dz = \int_{-\infty}^\infty \frac{f(0)}{1+z^2}dz = f(0)\arctan(z)\Bigr|_{-\infty}^\infty = \pi f(0)$$

Generally the trick is to absorb the limit variable into the $dx$ via some sort of substitution. For example in the integral

$$\lim_{n\to\infty} n\int_0^1 f(x)x^ndx$$

the trick is to use the substitution $u = x^{n+1}$ so that $du \sim x^ndx$

$$= \lim_{n\to \infty} \frac{n}{n+1} \int_0^1 f\left(u^{\frac{1}{n+1}}\right) du = \int_0^1 f(1)du = f(1)$$

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  • $\begingroup$ The upper limit of the integral of the first integral of the second last line should be $\frac{1}{h}$. Otherwise +1 for a very nice solution. $\endgroup$ – Severin Schraven Dec 13 '19 at 23:01
  • $\begingroup$ @SeverinSchraven Thank you! $\endgroup$ – Ninad Munshi Dec 13 '19 at 23:03
  • $\begingroup$ Why is it a necessary condition that f be continuous? It isn't clear to me how that's used in your solution. Unless it has something to do with taking the limit as $h\rightarrow 0$ of f(hz), but then wouldn't it only be necessary to say f is continuous in a neighborhood around x=0? $\endgroup$ – Connor Dec 14 '19 at 2:07
  • $\begingroup$ @Connor Since $f(z)$ was continuous on $[-1,1]$, $f(hz)$ was continuous on $\left[-\frac{1}{h},\frac{1}{h}\right]$. A continuous function on a compact interval must be bounded (extreme value theorem), and that is what allowed us to use the dominated convergence theorem in the first place. $\endgroup$ – Ninad Munshi Dec 14 '19 at 10:56
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Another possible way could be with Taylor series built around $x=0$ $$f(x)=f(0)+ f'(0)x+\frac{1}{2} f''(0)x^2+O\left(x^3\right)$$ which make $$I=\int{\frac{h}{h^2+x^2}}f(x)\:dx\sim\int\frac{h \left(f(0)+ f'(0)x+\frac{1}{2} f''(0)x^2\right)}{h^2+x^2}$$ that is to say $$I\sim h \left(\frac{\left(2 f(0)-h^2 f''(0)\right) \tan ^{-1}\left(\frac{x}{h}\right)}{2 h}+\frac{1}{2} x f''(0)+\frac{1}{2} f'(0) \log \left(h^2+x^2\right)\right)$$ which makes $$J=\int_{-a}^{+a}{\frac{h}{h^2+x^2}}f(x)\:dx\sim\tan ^{-1}\left(\frac{a}{h}\right) \left(2 f(0)-h^2 f''(0)\right)+a h f''(0)$$

Expanding again as a Taylor series around $h=0$ $$J\sim \pi f(0)+h \left(a f''(0)-\frac{2 f(0)}{a}\right)+O\left(h^2\right)$$ which shows the limit (it does not depend on $a$) and how it is approached.

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  • $\begingroup$ Wouldn't this require that f be differentiable at 0? $\endgroup$ – Connor Dec 14 '19 at 17:25
  • $\begingroup$ @Connor. For sure ! $\endgroup$ – Claude Leibovici Dec 15 '19 at 2:54
  • $\begingroup$ Ah ok, but the question just says f is continuous, not necessarily differentiable $\endgroup$ – Connor Dec 15 '19 at 6:13

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