The definition of convolution is known as the integral of the product of two functions $$(f*g)(t)\int_{-\infty}^{\infty} f(t -\tau)g(\tau)\,\mathrm d\tau$$
But what does the product of the functions give? Why are is it being integrated on negative infinity to infinity?
What is the physical significance of the convolution? (For eg. integration of a curve gives the area of the curve within the limit)
If $$f(t) = \sin(t) ;\ g(t) = \cos t$$ then their convolution is $$\frac12 t\sin(t).$$ What does this function have to do with the initial functions?
I have been newly introduced to this concept and had many questions
Thanks for the help.
This is best answered by examples. If $g(x)=\begin{cases}\frac1a&\text{if $0\le x\le a$}\\0&\text{otherwise}\end{cases}$. then $$(f*g)(t)=\int_{-\infty}^\infty f(t-\tau)g(\tau)\,\mathrm d\tau=\frac1a\int_0^af(t-\tau)\,\mathrm d\tau$$ that is, folding any (integrable) $f$ with this $g$ replaces $f$ with its average over the preceeding interval of length $a$ at each point. Most applications are with "such" functions $g$, i.e., they have compact support (which allows you to replace $\int_{-\infty}^\infty$ with an integral with finite bounds); and the integral of $g$ is $1$ (so that calling the result averaging is justified; if $f$ is constant, this guarantees $f*g=f$). However, usually in such applications $g$ is chosen smooth, which results in $f*g$ being smooth even if $f$ is not (so $f*g$ is a much friendlier approximation of $f$).
Also very importantly, if you learn Fourier analysis, you will learn that the pointwise product of two functions corresponds to folding theri Fourier transforms and vice versa. There is a similar effect in the theory of polynomials: If $f(X)=\sum_{k\ge 0} a_k X^k$ and $g(X)=\sum_{k\ge 0} b_k X^k$ are polynomials, then their product is a polynomial $h(X)=\sum c_k X^k$, and howo do you compute its coefficients? By folding: $$c_k=\sum_{i+j=k}a_ib_j=\sum_{j\in\mathbb Z}a_{k-j}b_j.$$ I assume you notice the correspondence with function folding, and here too the sum over $j\in\mathbb Z$ is in fact finite because we actually need only consider $0\le j\le\min\{\deg f,\deg g\}$.