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Consider the integral limit $$\lim_{n\to \infty}\int^\infty_{-\infty}\frac{n\sin(x/n)}{x(1+x^2)}dx$$

By dominated convergence theorem, the integral equals $$\int^\infty_{-\infty} \frac1{1+x^2}dx=\pi$$

If we use residue theorem to evaluate the integral, by taking a infinitely large semi-circle on the upper half plane(small indent at origin), the arc integral vanishes, then we get $$\int^\infty_{-\infty} \frac{n\sin(x/n)}{x(1+x^2)}dx=2\pi iRes_i=n\pi \sinh (\pi /n)$$ Then taking the limit on $n$, the integral equals $\pi^2$.

What mistake did I make?

Thanks in advance.

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  • $\begingroup$ @MarcoCantarini Can you please show how? $\endgroup$ – Szeto Apr 17 '18 at 14:41
  • $\begingroup$ @MarcoCantarini Oh sorry I found my mistake. I think this question should be deleted. Thank you for pointing out. $\endgroup$ – Szeto Apr 17 '18 at 14:46
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    $\begingroup$ There's a more fundamental error here - the integral should evaluate to $n \pi (1 - e^{-1/n})$, which has the correct limit. $\endgroup$ – B. Mehta Apr 17 '18 at 14:48
  • $\begingroup$ @B.Mehta Would you mind showing how your derived this expression? $\endgroup$ – Szeto Apr 17 '18 at 15:03
  • $\begingroup$ Sure, I'll write an answer now. $\endgroup$ – B. Mehta Apr 17 '18 at 15:06
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The first issue is that you assumed the large arc integral vanishes as $R \to \infty$, which is unfortunately not true for $\frac{\sin z}{z(1+z^2)}$, as it is unbounded there. Instead, observe that $\sin z$ is the imaginary part of $e^{iz}$, which is bounded for the relevant region. Specifically, if $z = x + iy$ and $y > 0$, then $|e^{iz}|=e^{-y}<1$. So, instead calculate $$\int_{-\infty}^\infty \frac{ne^{i\frac zn}}{z(1+z^2)}\,dz=\int_{-\infty}^\infty f(z)\,dz$$ and take its imaginary part. However, switching to this gives a (simple) pole at the origin, which needs to be dealt with. So, we take a small dent at the origin instead.

Use the contour $\gamma$ consisting of the four paths:

  • $[\epsilon, R]$,
  • a semicircle $|z| = R$ and $\operatorname{Im} z > 0$ (counterclockwise) - call this $C_R$,
  • $[-R, -\epsilon]$
  • a semicircle $|z| = \epsilon$ and $\operatorname{Im} z > 0$ (clockwise) - call this $C_\epsilon$.

Inside the contour, there is a simple pole at $i$ with residue $$\operatorname{Res}(f,i)= -\frac{ne^{-\frac 1n}}{2}$$ so we have $$\int_\gamma f(z) \, dz = -i\pi n e^{-\frac1n}.$$

On the outer semicircle, we have $f(z) < \frac{1}{z(1+z^2)}$, which is $\mathcal{O}(R^{-3})$. So, $\int_{C_R} f(z)\,dz$ is $\mathcal{O}(R^{-2})$, hence vanishes in the limit.

For the inner semicircle, observe that $f(z)$ has a simple pole at the origin so we 'pick up half the residue', as we are going around halfway. (For details of how this really works, see here.) Further, remember that this is traversed clockwise so we actually pick up $-\frac{1}{2}$ of the residue. $\operatorname{Res}(f,0) = n$, so $$\lim_{\epsilon \to 0} \int_{C_\epsilon} f(z)\,dz=-\frac{1}{2}\cdot 2\pi i \cdot n=-i\pi n.$$

Combining, we get $$\int_{-\infty}^\infty f(z)\,dz = n \pi i - n \pi i e^{-\frac{1}{n}}=i n \pi (1 - e^{-\frac{1}{n}})$$ and taking the imaginary part gives $$\int^\infty_{-\infty}\frac{n\sin(x/n)}{x(1+x^2)}dx=n\pi(1-e^{-\frac{1}{n}}).$$

Finally, notice that $\lim_{n \to \infty} n \pi (1 - e^{-\frac{1}{n}}) = \pi$, as expected.

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