My function is $f(z)$ is analytic in Re$(s)>1$, defined by meromorphic continuation elsewhere. It has a simple pole only at $s=0$. I want to integrate $$\int_{-i\infty}^{i\infty}f(z)dz$$ knowing that $f(z)$ has sufficient decay at infinity. My problem is integrating over the pole: I believe that it should be interpreted as a Cauchy principal value, or equivalently, as a deformed contour $$\lim_{R\to\infty}\lim_{\epsilon\to0}(\int_{-iR}^{-i\epsilon}+\int_{C_\epsilon}+\int_{i\epsilon}^{iR}+\int_{C_R}f(z)dz)$$ where $C_R$ is a contour to the right connecting $-iR$ to $iR$ (using a Jordan lemma type argument), and $C_\epsilon$ is a semicircle around the pole $s=0$.

I am told that the contribution of the pole should be half of the residue, as the angle is $\pi$, but I am confused about the sign the residue should take, as my integral is oriented clockwise.

Question: should the contribution of the pole be $-i\pi\text{Res}(f(z),0)$?

  • $\begingroup$ in the end it shouldn't matter in which way you enclose the pole $\endgroup$ – tired Oct 22 '15 at 14:41
  • $\begingroup$ i would suggest you concince yourself with an example. integrate $f(z)=1/z$ around a small semicircle at the origin which encloses 0 one time to the right and one time to the left... $\endgroup$ – tired Oct 22 '15 at 14:45
  • $\begingroup$ @tired I think I'm convinced of that, that it doesn't matter which way one circles the pole, but here I am asking about the orientation coming from the direction I am integrating in. For example, there should be a change in sign depending on whether I integrate $1/z$ from down to up or up to down, is that right? $\endgroup$ – TA Wong Oct 22 '15 at 14:51
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    $\begingroup$ Yes, if the contour is negatively oriented, the residue of the simple pole at $0$ gets the factor $-\pi i$. As a rule of thumb, when the contour passes through a simple pole and is smooth there, then "half of the pole is inside the contour, half outside". (The fraction changes when the contour has a corner at the pole.) The sign is the same as for poles inside the contour. $\endgroup$ – Daniel Fischer Oct 22 '15 at 14:59
  • $\begingroup$ Sorry i missunderstood you... as u are always going clockwise the sign should be correct..:) $\endgroup$ – tired Oct 22 '15 at 15:21

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