I am trying to prove that an exponential input to a linear system is also exponential with the same frequency but different amplitude and phase.
That is, assuming that $y(t)=L[x(t)]$, $y(t)$ can be computed using the convolution function$y(t)=x(t)*h(t)=\int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau$, where $x(t)$ is the input and $h(t)$ is the system impulse response.
Now, let assume that $x(t)=e^{jw_0t}$. Then, according to linear system theory the corresponding output should be $y(t)=H(w_0)e^{jw_0t}$, where $H(w)$ is the Fourier transform of $h(t)$. Also it is known that $x(t)*h(t)=h(t)*x(t)$, that is convolution function is commutative.
I am doing the following steps in the proof:
$$y(t)=\int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau=\int_{-\infty}^{\infty} h(\tau)x(t-\tau)d\tau$$ $$y(t)=\int_{-\infty}^{\infty} h(\tau)e^{jw_0(t-\tau)}d\tau$$ since $t$ is independent of $\tau$ $$y(t)=e^{jw_0t}\underbrace{\int_{-\infty}^{\infty} h(\tau)e^{-jw_0\tau}d\tau}_{H(w_0)}->y(t)=H(w_0)e^{jw_0t}$$
Up to now everything is fine. However, the problem starts when I change the limits of the convolution integral. Assuming that linear system is causal, i.e. $x(t)=0, h(t)=0$ for $t<0$, the convolution integral becomes:
$$y(t)=\int_{0}^{t} x(\tau)h(t-\tau)d\tau=\int_{0}^{t} h(\tau)x(t-\tau)d\tau$$
By applying the same logic above,
$$y(t)=\int_{0}^{t} h(\tau)e^{jw_0(t-\tau)}d\tau$$ since $t$ is independent of $\tau$ $$y(t)=e^{jw_0t}\underbrace{\int_{0}^{t} h(\tau)e^{-jw_0\tau}d\tau}_{\ne H(w_0)}$$
The integral does not equal to $H(w_0)$ since the integration depends on $t$.
I believe that the first derivation is correct but could not figure out what is wrong in my second derivation. Can you please help me to figure out the error I am making in the second derivation?
Thanks in advance.
Regards,