# Question on UV and IR divergences

1. Apr 25, 2008

### mhill

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

$$\int_{0}^{\infty} dk. k^{m}$$ 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

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

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)

2. Apr 25, 2008

### lbrits

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 $$D^m\delta(\omega)$$. A derivative?

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