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I want to calculate this integral $$I:=\int dk^{0}\frac{e^{-ik^{0}(x^{0}-x'^{0})}}{\left(\left(k^{0}\right)^{2}-|\vec{k}|^{2}\right)} $$ for that I recall the Residue Theorem: $$I=2\pi i \left\{ \mathrm{Res}\left[\frac{e^{-ik^{0}(x^{0}-x'^{0})}}{\left(\left(k^{0}\right)^{2}-|\vec{k}|^{2}\right)},k^{0}=|\vec{k}|\right]+\mathrm{Res}\left[\frac{e^{-ik^{0}(x^{0}-x'^{0})}}{\left(\left(k^{0}\right)^{2}-|\vec{k}|^{2}\right)},k^{0}=-|\vec{k}|\right]\right\} $$ My question is, how do I calculate those residues?

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The two poles are simple (assuming $|\vec k| \ne 0$) and there is a simple formula for the residue at a simple pole – kiwi Feb 18 at 1:46
Yes you are right. But I'm missing a sign, I obtain $I=2\pi i \frac{1}{2|\vec{k}|}\left[e^{-i|\vec{k}|(x^{0}-x'^{0})}-e^{i|\vec{k}|(x^{0}-x'^{‌​0})}\right]$, but my book says that the result is with a plus sign instead of the minus. Do you think that sign is important? @kiwi – Anuar Feb 18 at 2:16
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Is $x^{0} - x^{'0}$ positive or negative? If positive, you need to complete your contour in lower half plane, your contour will circle around the poles clockwisely and pickup an extra minus sign. If negative, you need to complete your contour in upper half plane, your contour will circle around the poles counterclosewisely and no extra minus sign there. – achille hui Feb 18 at 3:16
@Anuar, what book is that, please? – DonAntonio Feb 18 at 4:34
"Electrodynamics and classical theory of fields and particles" by A. O. Barut p. 153. @DonAntonio – Anuar Feb 18 at 15:53

1 Answer

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Let us try first to write that integral in a little less horrible , more usual, way:

$$I:=\int\,dz\frac{e^{-iz(x_0-x'_0)}}{(z^2-t^2)}=\int f(z)\,dz$$

Now:

$$Res_{z=\pm t}=\lim_{z\to \pm t}(z\pm t)f(z)=\frac{e^{\mp it(x_0-x'_0)}}{\pm 2t}$$

so the integral equals

$$I=\frac{2\pi i}{2t}\left(e^{-it(x_0-x'_0)}-e^{it(x_0-x_0')}\right)=\frac{2\pi}{t}\sin\left(t(x_0-x_0')\right)$$

and this agrees with your result, so I think your book's wrong .

The above is, of course, assuming the integral is around a path enclosing both poles of $\,f(z)\,$ .

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OK. Thanks for your help. I'm going to study the consequences of that missing sign. – Anuar Feb 18 at 3:14
(-1) see my comment in question about sign of $x_0 - x_0'$. – achille hui Feb 18 at 3:32
First, I think you rush a little too much to downvote (even if the answer is wrong). Second, why do you think the sign of that exponent matters at all to calculate the residues of the integrand, which only are $\,z=\pm t\,$ , with my notation, or to determine the path which, most probably, is given? – DonAntonio Feb 18 at 3:36
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@Anuar there are 4 possible choices of assigning the small imaginary part $i\epsilon$ to move the poles away from the real axis. They correspond to different casual behavior of the Green's function/propagator. The most common one is the one you see, the causal propagator, which will be non-zero only when $x_0 > x_0'$. The wiki page of propagator have a little bit more info on this. – achille hui Feb 19 at 9:19
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@DonAntonio, done. – achille hui Feb 19 at 13:50
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