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The integral I am trying to evaluate is:

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

The standard approach to this is to realise $\sin x$ as the complex part of $e^{ix}$ and take a contour integral over a semicircle on the upper half-plane. I have no trouble doing this and getting the (correct) answer of $I = \frac\pi e$.

However, I see no reason why one shouldn't be able to do the aforementioned contour integral directly, without switching $\sin x$ for the exponential. When, I do this, I get a different answer, and I'm not sure where I'm going wrong.

The integrand has a single simple pole on the upper half-plane at $z=i$. We can compute the residue:

$$\text{Res}(f, i) = \lim_{z\to i}(z-i)f(z) = \lim_{z\to i}\frac{z}{z+i}\sin z = \frac12i\sinh(1) = \frac i4(e-\frac1e)$$

By the residue theorem, we get an unexpected answer:

$$I = 2\pi i\ \text{Res}(f, i) = \frac\pi 2(\frac1e - e)$$

What's the issue here?

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    $\begingroup$ Have you checked that the contribution to the integral from the semicircle goes to 0 as $r \rightarrow \infty$? $\endgroup$ – user_of_math May 27 '16 at 16:33
  • $\begingroup$ Ah. I thought I had, but I see now that this is not in fact the case. Thank you! $\endgroup$ – Guy Paterson-Jones May 27 '16 at 18:23
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This interesting question reminds us that Residue theorem is correct only when $$ \lim_{z\to a+i\infty}f(z) = 0. $$

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