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Let $P_n(x) = \frac{n}{1+n^2x^2}$. Prove that for every $n\in\mathbb{N}$ $$\int_{-\infty}^\infty P_n(x) \, dx = \frac{\pi}{n}$$

And for every $\delta > 0$:

$$\lim_{n\to\infty} \int_\delta^\infty P_n(x) \, dx = \lim_{n\to\infty} \int_{-\infty}^{-\delta} P_n(x) = 0$$

This question is Fourier-series-oriented and maybe has something with Fejér's theorem

I'd be glad to get a guidance.

Thanks.

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    $\begingroup$ For the first part, the form of the integrand suggests the substitution $nx = \tan \theta$. $\endgroup$ – Travis Willse Mar 4 '15 at 9:52
  • $\begingroup$ When $m>n$, $P_m(x)<P_n(x)$ except at $0$. So it's not plausible that the integral is independent of $n$. $\endgroup$ – user99914 Mar 4 '15 at 9:53
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Hint:

$$\int\frac1{1+(nx)^2}dx=\frac1n\int\frac{n\;dx}{1+(nx)^2}=\frac1n\arctan nx+C$$

BTW, the above shows that apparently you forgot a factor $\;\frac1n\;$ in the result of your integral

Addition: for every $\;\delta >0\;$ :

$$\int_\delta^\infty P_n(x) dx=\lim_{b\to\infty}\left.\frac1n\arctan nx\right|_\delta^b=\frac1n\left(\frac\pi2-\arctan\delta\right)\xrightarrow[\delta\to 0]{}\frac\pi{2n}$$

and now you can take the limit when $\;n\to\infty\;$ and get zero indeed.

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  • $\begingroup$ Thanks. You're absolutely right. I've corrected it $\endgroup$ – AlonAlon Mar 4 '15 at 9:57
  • $\begingroup$ For the second part:$$\lim_{n\to\infty} \int_\delta^\infty \frac{n}{1+(nx)^2} \, dx = [t= nx] = \lim_{t\to\infty} \int_{n\delta}^\infty \frac{1}{1+t^2} dt = \arctan(\infty)-\lim_{t\to\infty}\arctan(\delta n) \\= \arctan(\infty)-\arctan(\infty) = 0$$ $\endgroup$ – AlonAlon Mar 4 '15 at 10:04
  • $\begingroup$ Do you concur? :) $\endgroup$ – AlonAlon Mar 4 '15 at 10:04
  • $\begingroup$ @Alon, I can't understand the very first expression on the left (and neither the next one). Did you forget to add an integral sign there? $\endgroup$ – Timbuc Mar 4 '15 at 10:05
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    $\begingroup$ @AlonAlon Check what I added to my answer. $\endgroup$ – Timbuc Mar 4 '15 at 10:19
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Recall from calculus that $$ \frac{d}{d\theta} \arctan \theta = \frac{1}{1+\theta^2}. $$

Therefore $$ \frac{d}{d\theta} \left(\frac{1}{n}\arctan n\theta \right)= \frac{1}{1+n^2\theta^2}. $$ The fundamental theorem of calculus then informs us that for any $t_1$ and $t_2$ $$ \int_{t_1}^{t_2}\frac{1}{1+n^2\theta^2} \; d \theta = \frac{1}{n}(\arctan n t_1 - \arctan n t_2). $$ Note that $\lim_{\theta \rightarrow \pm \infty} \arctan \theta = \pm \pi/2$. So upon taking the limits $t_1 \rightarrow - \infty$ and $t_2 \rightarrow \infty$, we find that that $$ \int_{- \infty}^{\infty}\frac{1}{1+n^2\theta^2} = \frac{1}{n}(\pi/2 - (-\pi/2)) = \frac{\pi}{n}. $$

Using the above formula for the integral on any interval $[t_1,t_2]$ we may also readily compute the two latter two integrals. These will be equal to $$ \frac{\pi}{2n} - \frac{1}{n}\delta $$ and $$ - \frac{1}{n}\delta - \frac{\pi}{2n}, $$ respectively. Taking $n \rightarrow \infty$ then gives the desired result.

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