Question on UV and IR divergences

  • Thread starter mhill
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Main Question or Discussion Point

no matter what theory we use , are all UV and IR divergences of the form ??

[tex] \int_{0}^{\infty} dk. k^{m} [/tex] where 'm' is an integer

or there are another divergent integral different from a power-law or logarithmic divergence ?? , and another question , can be this true

[tex] i^{m+n}D^{m}\delta (w) D^{n}\delta (w)= Fourier. transform (\int_{-\infty}^{\infty}dt.t^{n}(x-t)^{m}) [/tex]

where i have used the fact that the Fourier transform of a convolution of two functions (f*g) is just the product of the Fourier transform F(w)G(w)
 

Answers and Replies

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UV divergences are of that form, but it depends on the type of regulator you use. I'm not sure about IR divergences.

I don't know what you mean by [tex]D^m\delta(\omega)[/tex]. A derivative?

[tex]D^m \delta(\omega) D^n \delta(\omega) \approx \int\!dt\,t^m e^{i \omega t}\int\!dx\,x^n e^{i \omega x}= \int\!dt\,\int\!dx\, t^m x^n e^{i \omega (x+ t)} = \int\!dx \left[\int\!dt\, t^m (x-t)^n \right] e^{i \omega x}[/tex]
You will have to put in factors of i and pi and such.
 

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