Verify commutative law of convolution

The commutative law of convolution states $x(t) * h(t) = h(t) * x(t)$: the output of an LTI system is the same whether we convolve the input with the impulse response or the other way round.

Given: $x(t) = e^{-at}u(t)$, $a > 0$; $h(t) = u(t)$.

LHS: $y_1(t) = x(t) * h(t)$

$$y_1(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau = \int_{-\infty}^{\infty} e^{-a\tau}u(\tau)\,u(t-\tau)\,d\tau$$

Non-zero for $0 \le \tau \le t$ (needs $t \ge 0$):

$$y_1(t) = \int_0^{t} e^{-a\tau}d\tau = \left[\frac{e^{-a\tau}}{-a}\right]_0^t = \frac{1}{a}\left(1 - e^{-at}\right)u(t)$$

RHS: $y_2(t) = h(t) * x(t)$

$$y_2(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau = \int_{-\infty}^{\infty} u(\tau)\,e^{-a(t-\tau)}u(t-\tau)\,d\tau$$

Non-zero for $0 \le \tau \le t$ (needs $t \ge 0$):

$$\begin{aligned} y_2(t) &= e^{-at}\int_0^{t} e^{a\tau}d\tau \\ &= e^{-at}\cdot\frac{e^{at} - 1}{a} \\ &= \frac{1}{a}\left(1 - e^{-at}\right)u(t) \end{aligned}$$

Result

$$y_1(t) = y_2(t) = \frac{1}{a}\left(1 - e^{-at}\right)u(t)$$

so $x(t) * h(t) = h(t) * x(t)$ and the commutative law is verified.

Meaning: an input $e^{-at}u(t)$ applied to an integrator ($h = u(t)$) gives the same output as a unit step applied to a system with impulse response $e^{-at}u(t)$. In general the roles of input and impulse response can be interchanged.