# Kramers-kronig relations with an addtional constant

Suppose I have a complex function $$\chi(\omega)$$, using Kramers-Kronig relations, we have:

$$\chi_1(\omega) = \frac{1}{\pi}\mathcal{P}\int_{-\infty}^{\infty} \frac{\chi_2(\omega')}{\omega'-\omega}d\omega'$$

However, suppose I have another function, which is $$\varepsilon(\omega)=\chi(\omega)+1$$, using the Kramers-Kronig relations on $$\varepsilon(\omega)$$, we have:

$$\varepsilon_1(\omega) = \frac{1}{\pi}\mathcal{P}\int_{-\infty}^{\infty} \frac{\varepsilon_2(\omega')}{\omega'-\omega}d\omega'$$, which means that:

$$\chi_1(\omega) = \frac{1}{\pi}\mathcal{P}\int_{-\infty}^{\infty} \frac{\chi_2(\omega')}{\omega'-\omega}d\omega' -1$$ and the contradiction raises.

Why the contradiction, can anyone explain to me?

A condition for the Kramers-Kronig relations is that $$|\chi(\omega)|\to 0$$ as $$|\omega|\to\infty$$. But in that case $$|\varepsilon (\omega)|\to 1$$ in this same limit, so $$\varepsilon(\omega)$$ need not obey the Kramer-Kronig relations as such. Indeed, the validity of the Kramers-Kronig relations for $$\chi(\omega)$$ instead yields
$$\varepsilon_1(\omega) = 1+\frac{1}{\pi}\mathcal{P}\int_{-\infty}^{\infty} \frac{\varepsilon_2(\omega')}{\omega'-\omega}d\omega'$$ as the appropriate equivalent to Kramers-Kronig for $$\varepsilon(\omega)$$.