# Sign of Wick rotation [closed]

Suppose you have the integral

$$i \int^\infty_{-\infty} L_M(t) dt$$

and that $L_M$ contains two poles: when $t>0$ the pole lies above the t-axis and when $t<0$ the poles lies below the t-axis. Therefore you can rotate the contour from the real axis to the contour going from $i\infty$ to $-i\infty$. This path can be parameterized as $z=i\tau$ where $\tau$ is from $\infty$ to $-\infty$:

$$i \int^\infty_{-\infty} L_M(t) dt=i \int L_M(z) dz= i\int^{-\infty}_{\infty} L_M(i\tau) id\tau= \int^{\infty}_{-\infty} L_M(i\tau) d\tau \\\equiv-\int^{\infty}_{-\infty} L_E(\tau) d\tau$$

$$\int^{\infty}_{-\infty} L_M(-i\tau) d\tau \equiv-\int^{\infty}_{-\infty} L_E(\tau) d\tau$$

so they get the sign wrong (or I got the sign wrong).

## closed as off-topic by ACuriousMind♦, Kyle Kanos, Floris, Daniel Griscom, Sebastian RieseFeb 6 '16 at 20:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Floris, Daniel Griscom, Sebastian Riese
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• Why did you switch the integral limits in your second step? Also, if the pole lies above the x-axis for positive $t$, shouldn't you be doing $t\mapsto -\mathrm{i}\tau$ (i.e. rotate clockwise) to not hit it? In any case, hunting sign conventions and errors in your (or the textbook's) calculation seems to me to be off-topic as homework-like. – ACuriousMind Feb 6 '16 at 1:18
• Which textbooks? – Qmechanic Feb 6 '16 at 3:30
• related (or duplicate?): Performing Wick Rotation to get Euclidean action of scalar field – AccidentalFourierTransform Feb 6 '16 at 18:01

$$\int^\infty_{-\infty}f(x)dx=\int^\infty_{-\infty}f(-x)dx$$