Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 '13 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 – Ana Feb 18 '13 at 2:16
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 '13 at 3:16
@Anuar, what book is that, please? – DonAntonio Feb 18 '13 at 4:34
"Electrodynamics and classical theory of fields and particles" by A. O. Barut p. 153. @DonAntonio – Ana Feb 18 '13 at 15:53
up vote 0 down vote accepted

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

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


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

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 $\,f(z)\,$ .

share|cite|improve this answer
OK. Thanks for your help. I'm going to study the consequences of that missing sign. – Ana Feb 18 '13 at 3:14
(-1) see my comment in question about sign of $x_0 - x_0'$. – achille hui Feb 18 '13 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 '13 at 3:36
@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 '13 at 9:19
@DonAntonio, done. – achille hui Feb 19 '13 at 13:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.