Convolution with multiple step functions This is a question from Bertsekas' Data Networks. It is question 2.2 on page 141.
It is asking for the convolution of the following 2 functions.
Function 1: $ s(t) = 1 $, when $0 \leq t \leq T$. It is $0$ elsewhere.
Function 2: $h(t) = \alpha e^{-a t}$, when $t \geq 0$. It is $0$ elsewhere.
My work:
The output $r(t) = \int_{-\infty}^{\infty} s(\tau) h(t-\tau) d\tau$. Then, we can rewrite $s(t) = 1 \times u(t) \times u(T-t)$, where $u(t)$ is the standard step function. Also, we can rewrite $h(t) = \alpha e^{-a t} u(t)$.
Then, $$ r(t) = \int_{-\infty}^{\infty} u(\tau) u(T-\tau) \alpha e^{-a (t-\tau)} u(t-\tau) d\tau$$
I am confused at this point because there are three separate step functions in the integral. The first step function $u(\tau)$ says the lower limit of $\tau$ in the integral must be 0. 
The second step function $u(T-\tau)$ says $\tau \leq T$ or else the integral evalutes to zero. In the examples I am looking at this step function is not present, hence I have not seen this complication before.
The third step function $u(t-\tau)$ says the upper limit of $\tau$ in the integral must be $t$. I have seen this before in the examples.
Does the second step function have much influence on the result of the integral? I have two upper bounds, $\tau = T$ and $\tau = t$. I assume $\tau=t$ wins since it is more general. 
Note that: $r(t)$ is defined in the following time range: $[0+0, T+\infty]$. That is, $r(t)$ exists for all non-negative times.
So, $r(t) = \int_0^t \alpha e^{-a (t-\tau)} d\tau$
Thanks.
 A: Split into the cases $0 < t \leq T$ and $t > T$, to conclude that
$$
(s*h)(t) = \int_0^{\min \{ t,T\} } {1 \cdot h(t - \tau )d\tau } = \int_0^{\min \{ t,T\} } {\alpha e^{ - \alpha (t - \tau )} d\tau } .
$$
EDIT: It follows that
$$
(s*h)(t) = 1 - e^{-\alpha t}, \;\; 0 < t \leq T,
$$
$$
(s*h)(t) = e^{ - \alpha (t - T)}  - e^{ - \alpha t} ,\;\; t > T.
$$
(Since $(s*h)(t) = 0$ for $t \leq 0$, we see that $(s*h)(t)$ is continuous on $\mathbb{R}$.)
Relation to probability theory: With $s$ and $h$ as above, let $X$ be an exponential random variable with density function $h$, and $Y$ an independent uniform$[0,T]$ random variable, so that $Y$ has density function $\tilde s = s/T$.
By the law of total probability, conditioning on $Y$, we have
$$
{\rm P}(X + Y \le t) = \int_0^T {{\rm P}(X \le t - \tau )\frac{1}{T}d\tau }.
$$
It follows that if $0 < t \leq T$, then
$$
{\rm P}(X + Y \le t) = \int_0^t {{\rm P}(X \le t - \tau )\frac{1}{T}d\tau }  = \frac{1}{T}\int_0^t {(1 - e^{ - \alpha (t - \tau )} )d\tau }  = \frac{{t - (1 - e^{ - \alpha t} )/\alpha }}{T},
$$
while if $t > T$, then
$$
{\rm P}(X + Y \le t) = \int_0^T {{\rm P}(X \le t - \tau )\frac{1}{T}d\tau } = \frac{1}{T}\int_0^T {(1 - e^{ - \alpha (t - \tau )} )d\tau }  = \frac{{T - (e^{ - \alpha (t - T)}  - e^{ - \alpha t} )/\alpha }}{T}.
$$
Hence, the density function of $X+Y$ is given by
$$
f_{X+Y} (t) = \frac{{1 - e^{ - \alpha t} }}{T} ,\;\; 0 < t \leq T,
$$
$$
f_{X+Y} (t) = \frac{{e^{ - \alpha (t - T)}  - e^{ - \alpha t} }}{T}, \;\; t > T.
$$
On the other hand, since $X$ and $Y$ are independent with respective densities $h$ and $\tilde s \,(=s/T)$,
$$
f_{X+Y} (t) = (h*\tilde s)(t) = (\tilde s * h)(t) = \frac{{(s*h)(t)}}{T},
$$
from which it follows that
$$
(s*h)(t) = 1 - e^{ - \alpha t} ,\;\; 0 < t \leq T,
$$
$$
(s*h)(t) = e^{ - \alpha (t - T)}  - e^{ - \alpha t} ,\;\; t > T
$$
(as we have already seen above).
A: It seems a shame that you can get into so much mathematical detail on this question without noticing that it is the perhaps the very simplest non-trivial example of signal processing in electrical circuit theory, namely: the charging of a capacitor (through a resistor) when you close a switch.
