Local Maximum of the integral over an unknown function this problem is a part of my project but currently I don't have a very good idea to solve it. Here is the function:
$Z(t) = k\mathrm{e}^{-bt} \int_0^t I(\tau)\mathrm{e}^{b\tau}\mathrm{d}\tau$
where $k$, $b$ are positive constants and $t$ is a positive real number. $I(t)$ is an unknown, but monotonously decreasing function, and assume $I(0)>0$ and $\lim_{t\rightarrow\infty} I(t)\geq 0$. The objective is to see whether (and where, if possible) the function shows a local maximum. I have already proved that there will be at most one local maximum by Taylor expansion. And for $\lim_{t\rightarrow\infty}I(t)=0$ case, $Z(t)$ will definitely show a local maximum. But for more general case (i.e. $\lim_{t\rightarrow\infty}I(t)>0$), I haven't found any good idea to solve it.
Since this is homework-related, could you please give me some hint, such as towards which direction I should work on? Thank you very much!
 A: Just compute the derivative:
$$\tag1Z'(t)=k(-b)e^{-bt}\cdot\int_0^tI(\tau)e^{b\tau}\,\mathrm d\tau+ke^{-bt}\cdot I(t)e^{bt}=kI(t)-bZ(t).$$
(We may also note that $kI(t)=Z'(t)+bZ(t)$ is $e^{-bt}$ times the derivative of $Z(t)e^{bt}$).
Assume $Z$ has a local minimum at $t_0$. Then for suitable $t>t_0$ we have $Z(t)>Z(t_0)$ and $Z'(t)>0$, hence 
$$ kI(t_0)=bZ(t_0)<bZ(t)+Z'(z)=kI(t)\le kI(t_0),$$
contradiction. Therefore $Z$ has no local minimum.
(It does however have a minimum at the boundary at $t=0$ as $Z'(0)=kI(0)>0$.)
As between two local maxima there would be a local minimum, we conclude that $Z$ has at most one local maximum. 
Thus the following overall behaviours are possible:


*

*$Z$ strictly increases until it reaches a maximum and from then on strictly decreases, but remains bounded from below as otherwise the right hand side of (1) would become positive for large $t$.

*$Z$ strictly increases, but remains bounded as otherwise the right hand side of $(1)$ would become negative for big $t$. 


In both cases we see that $\lim_{t\to\infty} Z(t)$ exists, hence $|Z'(t)|$ becomes arbitrarily small for big $t$, hence 
$$\lim_{t\to\infty} Z(t)=\frac kb \lim_{t\to\infty}I(t).$$
However, i don't see that the second option (no maxmimum, but strictly increasing) can be ruled out.
