Cauchy theorem and Wick rotation

[parton]

Homework Statement

Hi!

I have a little problem understanding a proof ('Wick rotation') which I found in a textbook on QFT.

Assume: f and g are polynomials with degree(g)-degree(f) >= 2

and f/g has no poles on the closed 1. and 3. quadrant.

proposition:

$$\int_{-\infty}^{+\infty} \mathrm{d}x \dfrac{f(x)}{g(x)} = i \int_{-\infty}^{+\infty} \mathrm{d}x \, \dfrac{f(ix)}{g(ix)}$$

proof (just until the part I don't undestand):

We rotate the real axis into the imaginary axis (counter-clockwise). Because the rotating real axis does not hit poles of f/g we can write

$$\int_{-\infty}^{+\infty} \mathrm{d}x \dfrac{f(x)}{g(x)} = \int_{-i\infty}^{+i\infty} \mathrm{d}x \dfrac{f(x)}{g(x)}$$

by the Cauchy theorem on the path-independence of integrals over holomorphic functions.


Could anyone explain this last step to me?

Homework Equations

The Attempt at a Solution

I heard about path-independence, but I thought we need two paths with identical initial and end point, but here it is obviously not the case... maybe I am missing something

 

[tiny-tim]

hi parton!

yes, you're right, we need two paths with identical initial and end point (or one continuous closed path), but we can take a finite closed path, apply the Cauchy theorem, and then let the path tend to infinity, checking that the bit "at infinity" has an integral that tends to zero

i think the idea here is that you integrate between ±r along one axis, then round a quadrant of a circle of radius r, back along the other axis, and then another quadrant to close the curve

the theorem assumes f(r)/g(r) is O(1/r²) or less, and the integral along the quadrants is proportional to πr times f(r)/g(r), and so is O(1/r)

so the difference between the integrals between ±r on each axis -> 0 as r -> ∞, ie the integrals between ∞ are the same