# 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

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 $k_\mu$).
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