Formalism in integration Let's say we have some $y(t)$. The derivative of $y$ along time axis will be $y'(t)=\frac{dy(t)}{dt}=\frac{dy}{dt}$. So I will integrate like this over time:
$\require{cancel}$
$\int_{t=0}^{+\infty}\frac{dy}{\cancel{d\tau}}\cancel{d\tau} = \int_{t=0}^{+\infty}dy=y|_{t=0}^{+\infty}=y(\infty)-y(0)$ , (since $y = y(t)$)
Even though the result being correct, is this procedure correct according to formalism?
In other words:


*

*can I cancel the differentials? Thinking of interval that makes sense to me.

*can the integral variable be different from the differential (second step)? I've almost never see the integration variable explicitly shown like I put there.

*I solved the integral in the third step as my variable was $y$ even though it was actually $\tau$. Can that be done?


We chose the fraction-looking notation for derivatives because many "apparent laws", like this cancellation, really do hold. But the notation alone is not a proof of this.
This question is really a question about what many call "u-substition" in disguise. u-substitution is an integral statement of the chain rule for differentiation, and this is why you can "cancel the $\mathrm{d}\tau$ terms." For a proof that this works, you might look up proofs of u-substition. I happen to have written a proof in another answer, and in that answer the equality that you want is colored in red (suggesting that this is a common source of confusion).
The bounds of integration are treated exactly as you have treated them. 
When you ask about integrating with respect to $y$ "even though it was actually $\tau$," you are in fact integrating against $y$. This is the same question as your first question, and the proof is also contained in the other answer. 
In general, we have that
$$ \int_a^b f(g(x))g'(x)\mathrm{d}x = \int_{g(a)}^{g(b)} f(y)\mathrm{d}y,$$
for sufficiently nice $f,g$.
