Prove that $$ \langle\alpha(t)\rangle=2 \mathrm{i} \int_{-\infty}^{t} \mathrm{~d} t^{\prime} \bar{K}^{\prime \prime}\left(t-t^{\prime}\right) \cdot \bar{F}\left(t^{\prime}\right) $$ where \(\bar{K}^{\prime \prime}(t)\) is the Fourier transform of \(\bar{X}^{\prime \prime}(\omega) .\) This form of the linear response is commonly seen in the literature. [Note that the identity $$ \lim _{\eta \rightarrow 0} \frac{1}{\omega^{\prime}-\omega \mp \mathrm{i} \eta}=P \frac{1}{\omega^{\prime}-\omega} \pm \mathrm{i} \pi \delta\left(\omega^{\prime}-\omega\right) $$ and the spectral representation of the Heaviside function $$ \theta\left(t-t^{\prime}\right)=-\lim _{\eta \rightarrow 0} \int_{-\infty}^{\infty} \frac{\mathrm{d} \omega}{(2 \pi \mathrm{i})} \frac{\mathrm{e}^{-\mathrm{i} \omega\left(t-t^{\prime}\right)}}{\omega+i \eta} $$ are useful.]

The Fourier transform is a powerful mathematical tool used to convert functions from the time domain to the frequency domain.
This is particularly useful to analyze signals and systems that vary with time.
The Fourier transform of a function \(f(t)\), denoted as \(F(\omega)\), is given by the integral:ewlineewline\[ F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} \mathrm{d} t \]ewlineewlineIn the context of the provided solution, the Fourier transform is used to represent the function \( \bar{X}''(\omega) \) in the frequency domain.
This transformation helps in simplifying the problem and applying the convolution theorem.
The inverse Fourier transform, which converts frequency domain functions back to the time domain, is given by:ewlineewline\[ f(t) = \int_{-\infty}^{\infty} F(\omega) e^{i \omega t} \frac{\mathrm{d} \omega}{2 \pi} \]ewlineWith this, we can understand how the frequency components contribute to the overall time-domain behavior of the system.

The Heaviside function, also known as the step function, is a piecewise function that takes the value 0 for negative arguments and 1 for positive arguments.
Mathematically, it is defined as:ewlineewline\[ \theta(t) = \begin{cases} 0, & t < 0 \ 1, & t \geq 0 \end{cases} \]ewlineIn the given problem, the Heaviside function \( \theta(t - t') \) is used to ensure the causality principle,
meaning that the effect (response) cannot occur before the cause (input).
The spectral representation of the Heaviside function is particularly important in the context of Fourier transforms and is given by:ewlineewline\[ \theta(t - t') = -\lim_{\eta \rightarrow 0} \int_{-\infty}^{\infty} \frac{\mathrm{d}\, \omega}{2\pi i} \frac{e^{-i \omega (t - t')}}{\omega + i \eta} \]ewlineThis representation helps in expressing the time-domain behavior in terms of frequency domain components.
It also introduces an infinitesimally small positive value \( \eta \) to ensure proper convergence and avoid singularities.

The spectral representation refers to expressing a function or signal in terms of its frequency components.
This is commonly done using the Fourier transform.
In the given problem, the spectral representation of both the Heaviside function and the function \( \bar{X}^{\prime \prime}(\omega) \) comes into play.
The spectral representation of the identity, \[ \lim_{\eta \rightarrow 0} \frac{1}{\omega' - \omega \mp i \eta} = P \frac{1}{\omega' - \omega} \pm i \pi \delta(\omega' - \omega) \] helps in breaking down complex time-domain interactions into manageable frequency-domain terms.ewlineThis representation is crucial in linear response theory where the system's response to an external input is analyzed.
It allows the decomposition of the linear response function into its spectral components, linking time-domain responses with their frequency counterparts.ewlineWhen combined with the Heaviside function's spectral form and Fourier transform, it aids in deriving the desired form of the linear response equation, making complex integrals more tractable.