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Ultravioletcutoff

\begin{align} \lim_{\omega\rightarrow \textrm{small}}\left\{ \Sigma'(k,\,\omega)-\Sigma'(k,\,\infty) \right\}&=\lim_{\omega\rightarrow \textrm{small}} P\int\frac{d\omega'}{\pi}\frac{\Sigma''(k,\,\omega')}{\omega'-\omega}\notag\\ &=P\int \frac{d\omega'}{\pi} \omega'+\omega P\int \frac{d\omega'}{\pi} \end{align} Therefore, $$\frac{\partial}{\partial \omega}\Sigma'(k,\,\omega)\bigg|_{\omega=0}=P\int \frac{d\omega'}{\pi}$$

Cut-off energy

My confusion when studying the marginal Fermi liquid. Because $\Sigma''(k,\,\omega)\sim \omega$, the authors state that $\Sigma'(k,\omega)\sim \omega \log (\omega/\omega_c)$, where $\omega_c$ is the ultraviolet cutoff. Repeating the above calculation for a linear relationship for $\Sigma''(k,\,\omega)$, we have

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For the regular Fermi liquid result, the limits were the infrared and ultraviolet cutoffs, while here the lower limit is $\omega$. If we assume this for the regular Fermi liquid, then $\Sigma'(k,\,\omega)\sim \omega^2$, which is incorrect. So why, for the marginal Fermi liquid, do we take different integration limits in the Kramers-Kronig relation than for the regular Fermi liquid? Any explanation why (or explanation of any mistake I made) would be greatly appreciated.

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Infrareddivergence

Infrared cutoffphysics

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I am going through Andre-Marie Tremblay's derivation of the real part of the self energy in his lecture notes on the many-body problem. On page 254, if we take the imaginary $\Sigma''(k,\,\omega)\sim \omega'^2$, then the real part $\Sigma'(k,\,\omega)$ can be found from the following:

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EDIT: Specifically, why can't we just take the infrared cutoff for the marginal Fermi liquid to be some $\omega_i>0$? Why is the ultraviolet cutoff some constant but the infrared $\omega$, and why in the regular Fermi liquid case are both cutoffs arbitrary?