- #1
zetafunction
- 391
- 0
the idea is let us supppose we have a well behaved theory where only logarithmic divergences of the form
[tex] \int_{0}^{\infty} \frac{dx}{x+a}=I(a) [/tex] for several values of 'a' occur ,
then would this theory be renormalizable ?? , i think QED works because only logarithmic divergencies appear , in fact the integral above I(a) can be regularized in the sense of Hadamard (or either differentiating respect to 'a' ) in the form
[tex] I(a)= -log(a)+c_a [/tex] here c_a is a free parameter to be fixed by experiments... Hadamard finite part integral says that
[tex] \int_{0}^{\infty} \frac{dx}{x+a}=I(a)= \int_{-\infty}^{\infty}dx \frac{H(x-a)}{x} [/tex]
so in the sense of distribution theory the integral I(a) exits and is equal to log(a)
[tex] \int_{0}^{\infty} \frac{dx}{x+a}=I(a) [/tex] for several values of 'a' occur ,
then would this theory be renormalizable ?? , i think QED works because only logarithmic divergencies appear , in fact the integral above I(a) can be regularized in the sense of Hadamard (or either differentiating respect to 'a' ) in the form
[tex] I(a)= -log(a)+c_a [/tex] here c_a is a free parameter to be fixed by experiments... Hadamard finite part integral says that
[tex] \int_{0}^{\infty} \frac{dx}{x+a}=I(a)= \int_{-\infty}^{\infty}dx \frac{H(x-a)}{x} [/tex]
so in the sense of distribution theory the integral I(a) exits and is equal to log(a)