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I want to calculate $$ p.v.\int_{-\infty}^{\infty}{\frac{dx}{x\sin{(\pi x)}}} $$ using residue calculus. The function $ f(z)=\frac{1}{z\sin{(\pi z)}} $ has the double pole $z=0$ and infinite simple poles at $z_n=n\in\mathbb{Z}$. According to residue theorem if we use the closed curve like the one shown in the picture enter image description here

then it can be proved that $$ \int_{-\infty}^{\infty}{f(x)dx}=\pi i\sum_{k=1}^{n}{\mathrm{Res}_{z=z_k}{f(z)}} $$

After calculating the residues the result is $$ \sum_{k=1}^{\infty}{\mathrm{Res}_{z=z_k}{f(z)}}=\sum_{k=1}^{\infty}{\frac{(-1)^n}{n\pi}}=-\frac{\ln{2}}{\pi} $$ Thus the integral will be $$ \int_{-\infty}^{\infty}{\frac{1}{x\sin{(\pi x)}}dx}=-i\ln{2}$$

Is there any mistake in the method I used? The $i$ in the last result doesn't seem like it is supposed to remain as the integral is real.

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  • $\begingroup$ I'd say there must be something wrong as the given one is a real integral, so how come the result is a complex non-real number? $\endgroup$ – DonAntonio Jun 8 at 14:46
  • $\begingroup$ That is my question exactly. A thought is that maybe the integral doen't converge but im not sure. $\endgroup$ – mac Jun 8 at 14:56
  • $\begingroup$ Yes: the integral doesn't seem to converge. $\endgroup$ – DonAntonio Jun 8 at 17:54
  • $\begingroup$ What the residue theorem says is that the integral around the loop is given by the residues for poles inside (picking up half-value for those on the loop itself). Note that your loop is not just the part along the real axis. It is also that half-circle. To find the integral along the real axis, you need the integral along the half circle to go to $0$ as the radius $R \to \infty$. But in this integral, that is not the case. $\endgroup$ – Paul Sinclair Jun 9 at 2:31

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