I am wondering, why we integrate functions of the time in the following manner:
$$\int_0^t q({\tau})p(t - {\tau}) d{\tau} $$
it seems the $ p(t - {\tau} ) $ is integrated in the "other direction" of time flow than $q({\tau})$, simply because as ${\tau}$ progresses through time, the $q({\tau})$ it being incremented, while for $ p(t - {\tau} )$ is being decremented.
WHY is that ? What are implications to real-world application ? (By The Way, I have lectures from Control Theory and it seems like this is the way to do stuff) WHY ?
What is the difference if both $q()$ and $p()$ are integrated in the same "direction" of time, such that
$$\int_0^t q({\tau})p({\tau}) d{\tau} $$
it seems like the result would be completely different.
This type of relation occurs if one studies the linear response of a system to some cause. The standard example is the response of a dielectric to an electric field. There $q$ is the electric field and $p$ the susceptibility. But note that \begin{equation*} \int_{0}^{t}dsq(s)p(t-s)=\int_{0}^{t}dsq(t-s)p(s) \end{equation*} In general we have \begin{equation*} \int_{-\infty }^{+\infty }dsp(t,s)q(s) \end{equation*} But if we have time-translation invariance this becomes \begin{equation*} \int_{-\infty }^{+\infty }dsp(t-s)q(s) \end{equation*} In addition we require the response not to preceed the cause so \begin{equation*} \int_{-\infty }^{t}dsp(t-s)q(s) \end{equation*} and if we switch on $q$ at time $t=0$ \begin{equation*} \int_{0}^{t}dsp(t-s)q(s) \end{equation*} which is your expression. Your final expression \begin{equation*} \int_{0}^{t}dsp(s)q(s) \end{equation*} does not seem to have an interpretation along these lines.