Time domain and frequency domain analysis equivalence Let us imagine an LTI system with physically realizable input (ruling out fancy mathematical functions and the concomitant complexities and paradoxes) completely known from -$\infty$ to $\infty$.  We want to calculate the output. We can analyse it in time domain using the linear constant coefficient differential equation and in the frequency domain using the Fourier representation of the signal and frequency response of the system.  

The outputs calculated from both the methods should be identical. In time domain, we get (transients + steady state)  response of the system,  In frequency domain, the sinusoids are eternal and hence transients are generally neglected.  But since we are talking about the same system with the same input, the outputs calculated from both the methods should be the same.  

Will they be identical in spite of neglecting the transients in the frequency domain?
now, even a steady-state sinusoid can be thought of, in the limit, as a sum of weighted pulses, each with a beginning and some kinda end.  how do all of these "transient" signals add up to a steady-state sinusoid?  but they do:

$$\begin{align}
x(t) &amp;= \lim_{T \to 0} \sum_{n=-\infty}^{+\infty} x(nT) \cdot \operatorname{rect}\left( \frac{t-nT}{T} \right) \\
 &amp;= \lim_{T \to 0} \sum_{n=-\infty}^{+\infty} x(nT) \cdot \frac{1}{T}\operatorname{rect}\left( \frac{t-nT}{T} \right) \cdot T \\
 &amp;= \int\limits_{-\infty}^{\infty} x(u) \cdot \delta(t-u) \cdot du
 \end{align}$$

where $ \operatorname{rect}(u) \triangleq \begin{cases} 1 \quad |u|&lt;\tfrac12 \\ 0 \quad |u| &gt; \tfrac12  \end{cases} $

The Fourier integral models the entire time function as an infinitely dense and infinitely large set of sinusoids.  in a similar manner, those steady-state functions can add to something with a transient.

$$ x(t) = \int\limits_{-\infty}^{+\infty} X(f) \, e^{j2\pi ft} \ df $$