sci.physics.particle

Sorry and yes the denominator should be k^2+m^2+i*epsilon

now my Qs is why is there i*epsilon?

why can't I integrate it without i*epsilon in that case the poles of k4
would be on img axis
giving for fourth component k4 = +/- i*m and taking upper half semi circle
would give residue of i*m.
This is Euclidean space!

Why is it wrong? and why is i*epsilon added and why not subtracted?
I could not find a reasonable argument in Peskin and Zee books.
I see that integral without i*epsilon is not divergent and can be integrated
residue.

thanks.

"Jim Black" wrote in message
news:1ljnmz2x0itxn.1amureqqtl24j.dlg@40tude.net...
> On Fri, 16 Nov 2007 09:55:56 +0100, chilldown wrote:
>
>> Thanks Jim,
>>
>> The equation of the propagator is as follows:
>> Its in Euclidean space.
>>
>> Int(d^4k * exp(i*k*x)/ (k^4 +m^2 +i*epsilon).
>>
>> I have Peskin and Zee but cannot find the justification of adding
>> i-epsilon
>> except that its added to make it integrable.
>>
>> I would like to keep the poles on real axis and avoid the poles by
>> excluding
>> them through
>> closing the countor without them therefore avoiding the i-epsilon.
>
> I think you mean (k^2 +m^2 +i*epsilon).
>
> In either case, if this is in a 4-D Euclidean space, there are no poles on
> the real axis here unless m=0 [although there are poles at
> k_0 = +/- i*sqrt(m^2 + k_1^2 + k_2^2 + k_3^2)].
>
> --
> Jim E. Black