Why Is the Divergence of a Diagram ln(lambda) When D=0?

[ndung200790]

Please teach me this:
Why the naively divergence of a diagram is ln(lambda) (where lambda is ultraviolet cutoff) when the superficial degree of divergence D=0(the divergence of lambda^D when D=0)).I am reading the renormalization theory in Schroeder&Peskin and I do not understand this.I do know that sometimes this naive divergence is failed because some reasons.But I really do not understand this''naive'' divergence.
Thank you very much in advanced.

 

[ndung200790]

It seem to me that it is consequence of the regulation of Feymann integrals(I have just read in
Critical Properties of Phi4 Theory by Kleinert.H).

 

[vanhees71]

Take a diagram of superficial degree of divergence of D=0 with all subdivergences removed. The remaining divergent integral goes like (with the external momenta fixed at a finite value):

\int \mathrm{d}^4 l/l^4 \sim \int \mathrm{d} |l|/|l| \sim \ln (l)

and is renormalized by subtracting the integral for external momenta fixed at the renormalization scale.

Perhaps, my manuscript on QFT, which has a lot about renormalization, can help you:

http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdf