The Kramers-Kronig relations - part 1

[See part 2 for some technical aspects]
Very nice derivation of the Kramers-Kronig relations (on Wikipedia, of all places), exploiting the relation between the even and odd components of a function $\chi (t)$ and the real and imaginary parts of its Fourier transform $\chi (\omega) = \chi ' (\omega) + i \, \chi '' (\omega)$.

One usually invokes the analyticity of $\chi (\omega)$ in the upper half-plane, which must first be derived from the causality: $\chi (t) = 0$ for $t < 0$. Complex integration along a well-chosen contour then yields the Kramers-Kronig relations in their standard form [1]:

$$\begin{align} \label{eq:KK}   \chi (\omega) &= \frac{1}{i \, \pi} \mathcal{P} \int_{-\infty}^{\infty} \text{d} \omega ' \frac{\chi (\omega ')}{\omega ' - \omega} \quad \text{or, for the components:} \nonumber \\ \chi '(\omega) &= \frac{1}{\pi} \mathcal{P} \int_{-\infty}^{\infty} \text{d} \omega ' \frac{\chi ''(\omega ')}{\omega ' - \omega} \\ \chi ''(\omega) &= -\frac{1}{\pi} \mathcal{P} \int_{-\infty}^{\infty} \text{d} \omega ' \frac{\chi '(\omega ')}{\omega ' - \omega} \nonumber \end{align}$$
where $\mathcal{P}$ denotes Cauchy's principal value.

However, this requires two trips to the complex plane and the result is not very intuitive. It is better to split the reasoning into two steps: write the relation in the time domain and then go to the frequency domain by using general (and hopefully, already known) properties of the Fourier transform [2]. This can help separate the conceptual novelty from the technical difficulty.

Illustration of the Kramers-Kronig relations between a physical (i. e. causal and real) signal and its Fourier transform. Work by FDominec, released under a Creative Commons Attribution-Share Alike 3.0 Unported license.

The first step is writing the response function as the sum of its even and odd components:

$$\begin{align} \chi (t) &=  \chi _{e}(t) + \chi _{o}(t) \quad \text{where:} \nonumber\\  \chi _{e}(t) &=  \frac{1}{2} (\chi (t) + \chi (-t)) \quad \text{and}\\  \chi _{o}(t) &=  \frac{1}{2} (\chi (t) - \chi (-t)) \nonumber \end{align}$$
 something we can do for any real function. The causality relation ($\chi (t) = 0$ for $t < 0$) implies that:
$$\begin{align} \label{eq:parity} \chi _{e}(t) &= \operatorname{sgn}(t) \, \chi _{o}(t) \quad \text{and}\\ \chi _{o}(t) &= \operatorname{sgn}(t) \, \chi _{e}(t) \nonumber \end{align}$$
where $\operatorname{sgn}(t)$ gives the sign of its argument (-1 for negative and +1 for positive values.)

This is interesting because the real and imaginary parts of the Fourier transform $\mathcal{F}$ are given by the even and odd components of the function, respectively:

$$\begin{align} \label{eq:Fourier} \chi (\omega) &= \chi ' (\omega) + i \, \chi '' (\omega) = \mathcal{F} [\chi (t) ] \nonumber\\ \chi ' (\omega) &= \mathcal{F} [\chi _{e} (t) ] \\ i \, \chi '' (\omega) &= \mathcal{F} [\chi _{o} (t) ] \nonumber \end{align}$$

The first step is over; to obtain $\eqref{eq:KK}$ we must now combine $\eqref{eq:parity}$ with $\eqref{eq:Fourier}$ and go to the frequency domain. We need to remember that the Fourier transform of a product is the convolution of the Fourier transforms of the factors and we also need an expression for the Fourier transform of the $\operatorname{sgn}(t)$ function. We will do this in part 2.

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^{[1] P. M. Chaikin and T. C. Lubensky, Principles of condensed matter physics, Cambridge University Press (2000). Chapter 7.↩
[2] For a recent and readily available reference, see: C. Warwick, Understanding the Kramers-Kronig Relation Using A Pictorial Proof. The idea goes at least as far back as the (very terse) presentation of B. Y.-K. Hu, Kramers-Kronig in two lines, Am. J. Phys. 57, 821, (1989).  ↩}