# Is this OK: $\int_a^b \!\mathrm{d}x \,\,f(x) =^? \int_{\mathrm{i}\,a}^{{\mathrm{i}\,b}} \!\mathrm{d} (\mathrm{-i}y)\,\,f(\mathrm{-i}y).$ Any proof?

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; $$y:=\mathrm{i}x\Rightarrow \mathrm{d}x = -\mathrm{id}y,$$

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

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. $q^2=q_0^2-q_1^2$ $$I:=\int_{-\infty}^{\infty} \frac{\mathrm{d}q_1}{2\pi}\int_{-\infty}^{\infty} \frac{\mathrm{d}q_0}{2\pi}\,\frac{1}{(q^2-\tilde{D})^m}.$$ (BTW $\tilde{D}$ is nice so don't bother its existence and $m≥2$)
Can we then using the change of variables $$q_E:=\mathrm{i}q_0, \text{with a Euclidean vector }\bar{q}:=(q_E,q_1)\text{ with the usual metric } \bar{q}^2=q_E^2+q_1^2,$$

$$I=^?(-1)^m(\mathrm{-i})\int_{-\infty}^{\infty} \frac{\mathrm{d}q_1}{2\pi}\int_{-\mathrm{i}\infty}^{\mathrm{i}\infty} \frac{\mathrm{d}q_E}{2\pi}\,\frac{1}{(\bar{q}^2+\tilde{D})^m}=^?(-1)^m(-\mathrm{i})\int_{\Omega}\frac{\mathrm{d}^2\bar{q}}{(2\pi)^2}\frac{1}{(\bar{q}^2+\tilde{D})^m}.$$ 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.

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In any event, every step up to the very last $=^?$ in $$I=^?(-1)^m(\mathrm{-i})\int_{-\infty}^{\infty} \frac{\mathrm{d}q_1}{2\pi}\int_{-\mathrm{i}\infty}^{\mathrm{i}\infty} \frac{\mathrm{d}q_E}{2\pi}\,\frac{1}{(\bar{q}^2+\tilde{D})^m}=^?(-1)^m(-\mathrm{i})\int_{\Omega}\frac{\mathrm{d}^2\bar{q}}{(2\pi)^2}\frac{1}{(\bar{q}^2+\tilde{D})^m}.$$ 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 $\pm\infty$.