The problem is related to the control area but I am unable to find it in the books related to control theory.
Let $a > 0$ and $t_0 < t$, if $w(t)$ is a continuous function in the interval $[t_0,\infty)$, if \begin{equation} \int_{t_0}^{t} e^{-a(t-\tau)}w(\tau)d\tau \to 0 \quad \text{when } t\to\infty \end{equation} does that implies that $w(t)\to0$ as $t\to\infty$?
Considering the control point of view is intuitive to see that this might be the case, but I was not able to find how to proof.
Any help is appreciated.
The answer to your question is no. A pair of counterexamples is given by the functions $\cos(e^t)$ and $\sin(e^t)$. To prove that$^{(*)}$ $$ \lim_{t\to\infty}\int_0^te^{-a(t-\tau)}\cos(e^{\tau})\,d\tau=0, \tag{1} $$ even though $\cos(e^t)$ does not vanish at infinity, let's first consider the case $0<a\leq 1$. In this case, integration by parts yields \begin{align} &\int_0^te^{-a(t-\tau)}\cos(e^{\tau})\,d\tau=e^{-at}\int_0^te^{(a-1)\tau} \cos(e^{\tau})\,e^{\tau}d\tau \\ &=e^{-at}\left\{e^{(a-1)t}\sin(e^t) -(a-1)\int_0^t e^{(a-1)\tau}\sin(e^{\tau})\,d\tau\right\}. \tag{2} \end{align} For the first term on the RHS of $(2)$ we have $$ \lim_{t\to\infty}e^{-at}e^{(a-1)t}\sin(e^t)=\lim_{t\to\infty}e^{-t}\sin(e^t)=0, \tag{3} $$ and for the second we have, since $a\leq 1$, $$ \left|\int_0^t e^{(a-1)\tau}\sin(e^{\tau})\,d\tau \right| \leq\int_0^t \left|e^{(a-1)\tau}\sin(e^{\tau})\right|\,d\tau \leq\int_0^t d\tau = t, \tag{4} $$ so also $$ \lim_{t\to\infty}e^{-at}(a-1)\int_0^t e^{(a-1)\tau}\sin(e^{\tau})\,d\tau = 0. \tag{5} $$ The case $a>1$ can be reduced to the case $0<a\leq 1$ in a finite number of integration by parts. Indeed, for any positive value of $a$ the first term in $(2)$ vanishes at infinity; the second term, on the other hand, is similar to the original integral $(1)$, with $a$ and $\cos(e^{\tau})$ replaced with $a-1$ and $\sin(e^{\tau})$, respectively. Integrating the second term by parts we obtain $$ -(a-1)e^{-at}\left\{-e^{(a-2)t}\cos(e^t)+\cos(1) +(a-2)\int_0^t e^{(a-2)\tau}\cos(e^{\tau})\,d\tau\right\}. \tag{6} $$ The first two terms of $(6)$ vanish in the limit $t\to\infty$, and the remaining integral is again similar to $(1)$, but now with $a$ replaced with $a-2$. Therefore, repeating this process $n$ times, until $a-n\leq 1$, we'll arrive at an integral that vanishes in the limit $t\to\infty$.
$^{(*)}$ The proof for $\sin(e^t)$ is completely similar.