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There is this question which ask me to calculate the principal integral $$ \mathcal{P} \int_{-\infty}^\infty \frac{e^{-ix}}{(x+1)(x^2+1)} dx \, . $$

I find out that the poles are $x=-1$, $x=-i$ and $x=i$. However from here, I'm actually stuck because so far normally I only deal with question where there is a single pole at $x=0$ but this time round I need to deal with 3 poles which I'm not sure how should I start with. Any help is appreciated. Thanks!

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  • $\begingroup$ anybody have any idea?? $\endgroup$ – john Oct 24 '17 at 14:30
  • $\begingroup$ The number of poles is not really relevant: integrating along a contour with the real axis (where the pole $z=-1$ is) closed by the lower semi-large circle is possible. The pole $z=-i$ is inside and the circular contribution vanishes. $\endgroup$ – Paul Enta Oct 25 '17 at 13:13
  • $\begingroup$ hello @PaulEnta , thx for your reply. So it is possible to loop a semicircle that avoids z= -1 ( by forming another infinitesimal semicircular loop that doesn't enclosed z= -1 and connects with the real axis) and with the loop containing z= - i only in the lower plane?? $\endgroup$ – john Oct 25 '17 at 13:18
  • $\begingroup$ This clearly explained in the robjohn answer... $\endgroup$ – Paul Enta Oct 25 '17 at 13:33
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Hint: Consider the following contour

enter image description here

where the red semicircle gets infinitesimally small and the black semicircle gets infinitely big.

Show that the total of the integrals along the red, green and black contours equals the negative of the residue of $\frac{e^{-iz}}{(z+1)(z^2+1)}$ at $-i$.

Show that the integral along the black contour tends to $0$.

Show that the integral along the red contour is half the residue of $\frac{e^{-iz}}{(z+1)(z^2+1)}$ at $-1$.

Compute the integral along the green contour.

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  • $\begingroup$ Thanks, I try to work it out $\endgroup$ – john Oct 25 '17 at 13:26

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