# the meaning of epsilon in this operator $\epsilon$

Consider the dimensional regularized integral

$$\int d^{d}k (k^{2}-m^{2}+i\epsilon)^{-\lambda}$$

For positive $\lambda$ this integral has a pole at $k=m$. Is this the reason we we insert the $i \epsilon$ part?

Using Shotkhotsky's theorem $$(k^{2}-m^{2}+i\epsilon)^{-\lambda}= -i\pi \delta ^{\lambda}(k^{2}-m^{2})+PVC(\lambda) (k^{2}-m^{2}+i\epsilon)^{-\lambda}$$ but how do we regularize the IR divergence at $k_0=m$ for every $\lambda$? Do we simply ignore this pole $k=m$ and compute the integral?

Can we compute the integral by defining a parameter $b^{2}=-m^{2}$ and compute the integral for every $b$ and then take the limit $b\to-im$?

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Without an indication of the variable of integration (such as $dx$), this "integral" is meaningless. And are all these variables a function of this variable of integration? A problem like this can't be solved as far as I know, since it is poorly defined. – fibonatic May 22 '13 at 14:49
@fibonatic What? The integration variable is a $d$-dimensional vector $k$... hence the $\mathrm{d}^d k$. It's a pretty standard QFT type integral. – Michael Brown May 22 '13 at 14:54
@Michael Brown: I am not familiar with QFT (so maybe adding a QFT tag to it would help) and usually the variable of integration is written at the end of the integral, so in a mathematical point of view this equation didn't make much sense to me. – fibonatic May 22 '13 at 15:30
@fibonatic QFT aside, it's a perfectly good integral. The measure $dx$ (or in this case $d^dk$) can be written anywhere - the place where it appears doesn't matter. $\int$ together with $dx$ is just a shorthand to mean taking the limit of some sequence of sums w.r.t. $x$. So $\int x^2~ dx$ is the same as $\int dx~ x^2$ is the same as $\int x~ (dx)~ x$. but of course for one's sanity we typically don't write it in the last form. – nervxxx May 22 '13 at 15:44
@fibonatic: Integrals are operators, and as such the integrative variable should be written immediately after the ∫ – the integral symbol and the d… are not a pair of parentheses enclosing some expression. I know it's taught in school that way, but it's ugly and in some cases even wrong. – datenwolf May 22 '13 at 18:05

## 1 Answer

The $i \epsilon$ part is introduced in order to define in what direction you go around the pole at $k^2 = m^2$. The choice of $+ i \epsilon$ corresponds to going above the pole at $k_0 = -\sqrt{\mathbf k^2 + m^2}$ and below the pole at $k_0 = +\sqrt{\mathbf k^2 + m^2}$ in the complex plane (where $\mathbf k$ is the trhee-vector part of $k_\mu$).

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and you might want to add that there is a physical meaning with this choice of contour integration, as opposed to say the retarded or advanced propagators.. – nervxxx May 22 '13 at 15:37
so if the integral is convergent then you use RESIDUE theorem to evaluate it am i right ?? – Jose Javier Garcia May 22 '13 at 17:17
@JoseJavierGarcia Yes, that's correct. – Neuneck May 23 '13 at 16:58
another question would this method be valid to obtain a finite value for $\int_{0}^{\infty}dx(x+i\epsilon )^{-1}$ – Jose Javier Garcia Jul 26 '13 at 20:07
No. I think 1/x does not fall off fast enough in order to argue that the closing contour in the complex plane does not contribute to the integral. I'm not an expert on general function theory though. – Neuneck Jul 27 '13 at 5:31