Suppose there is a function, $f(t)$, not explicitly defined. Is the convolution of $f(t)$ with Heaviside Eq.1 or Eq.2? $$ \int_{0}^{t}f(\tau)H(\tau-t)d\tau = -\int_{0}^{t}f(t)dt \ \Leftarrow Eq.1 $$
$$ \int_{0}^{t}f(\tau)H(\tau-t)d\tau= \int_{0}^{t}f(t)dt \ \Leftarrow Eq.2 $$
I have doubts because doesn't it depend on whether $f(t)$ is odd or even? and My knowledge of convolution is sadly very limited. My knowledge of convolution mainly comes from this document.
If the answer is neither, then I wish someone can explain to me what convolution with Heaviside can be simplified into, especially if the integral's limit is from 0 to some finite t.
Your problem is that you don't write convolution in the convenient way ; you should write:
$$\int_{0}^{t}f(\boxed{\tau})H(\boxed{t-\tau})d\tau = \int_{0}^{t}f(\tau)d\tau \tag{1}$$
Mnemonic : the sum of the two boxed expressions must simplify into the value of the "final" variable, here $t$:
$$\boxed{\tau}+\boxed{t-\tau}=\boxed{t}$$
Remark 1: please note that, on the RHS, I have modified the "dumb" variable into $\tau$ in order it is not the same as the "final" variable $t$.
Remark 2: if we write (1) under the form
$$f \star H = F$$ where $F$ is the primitive function that takes value $0$ in $0$, we can apply to it the formula for the differentiation of a convolution product:
$$(f \star g)'= f \star (g')$$
giving:
$$(f \star H)' = f \star \delta = f$$
which is a proof of (1).