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

This is related to Wick rotation in QFT but it is not exactly it. I'll take a 2-dimensional spacetime to be brief but usually there are more.

I've checked with a few functions and with finite integration limits and it seems OK, but I have no proof and never seen it in any books:

Using a change of variable; \begin{equation} y:=\mathrm{i}x\Rightarrow \mathrm{d}x = -\mathrm{id}y, \end{equation}

\begin{equation} \int_a^b \!\mathrm{d}x \,\,f(x) =^? \int_{\mathrm{i}\,a}^{{\mathrm{i}\,b}} \!\mathrm{d} (\mathrm{-i}y)\,\,f(\mathrm{-i}y). \end{equation}

The reason is the following: Suppose you have a double integral and a $2-$dimensional spacetime vector $q=(q_0,q_1)$ with Minkowski metric $(+,-)$, i.e. $q^2=q_0^2-q_1^2$ \begin{equation} I:=\int_{-\infty}^{\infty} \frac{\mathrm{d}q_1}{2\pi}\int_{-\infty}^{\infty} \frac{\mathrm{d}q_0}{2\pi}\,\frac{1}{(q^2-\tilde{D})^m}. \end{equation} (BTW $\tilde{D}$ is nice so don't bother its existence and $m≥2$)
Can we then using the change of variables \begin{equation} q_E:=\mathrm{i}q_0, \text{with a Euclidean vector }\bar{q}:=(q_E,q_1)\text{ with the usual metric } \bar{q}^2=q_E^2+q_1^2, \end{equation}

\begin{equation} I=^?(-1)^m(\mathrm{-i})\int_{-\infty}^{\infty} \frac{\mathrm{d}q_1}{2\pi}\int_{-\mathrm{i}\infty}^{\mathrm{i}\infty} \frac{\mathrm{d}q_E}{2\pi}\,\frac{1}{(\bar{q}^2+\tilde{D})^m}=^?(-1)^m(-\mathrm{i})\int_{\Omega}\frac{\mathrm{d}^2\bar{q}}{(2\pi)^2}\frac{1}{(\bar{q}^2+\tilde{D})^m}. \end{equation} Now the last integral can be easily computed using spherical coordinates (while in the original we couldn't because of the Minkowski metric); given that we take for granted that $\Omega = \mathbb{R}^2$.

The question is wether all these step are allowed and wether anyone can prove or give me some refrences etc.

Thanks a lot,

Qazi Peshawa.

share|cite|improve this question

Your title didn't match up your question.

In any event, every step up to the very last $=^?$ in \begin{equation} I=^?(-1)^m(\mathrm{-i})\int_{-\infty}^{\infty} \frac{\mathrm{d}q_1}{2\pi}\int_{-\mathrm{i}\infty}^{\mathrm{i}\infty} \frac{\mathrm{d}q_E}{2\pi}\,\frac{1}{(\bar{q}^2+\tilde{D})^m}=^?(-1)^m(-\mathrm{i})\int_{\Omega}\frac{\mathrm{d}^2\bar{q}}{(2\pi)^2}\frac{1}{(\bar{q}^2+\tilde{D})^m}. \end{equation} is valid because you have only changed the integration variable. The last step is not. You have changed the contour of integration and changed the endpoints from $\pm \infty i$ to $\pm\infty$.

Once you changed the endpoints, there is no general theory about what could happen to your integral. You need to justify this step is valid on a case-by-case basis. (It is okay in your example because your integrand goes to zero fast enough at infinity).

The change of contour of integration is in a better shape. In physics, most of the time one only need to perform integration on rational functions or at worst meromorphic functions. Cauchy integral theorem tells us the integral remains unchanged under continuous deformation of the contour while keeping the endpoints fixed. This means if you can deform you contour without hitting any poles of your integrand, your integral remains the same. If not, your integral along the new contour will pickup some additional residues from the poles encountered during rotation.

In summary, to use Wick's rotation, you need to keep an eye on contributions from infinity and from the poles.

share|cite|improve this answer
Thanks for your elaborate answer, appreciated. – The Noob Feb 4 '13 at 9:11

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.