Can convolution be written as $(f \ast g)(t)=\int_{\tau = -\infty}^{\infty}g(t-\tau)f(\tau)\,\mathrm{d}\tau$? In every textbook that I read the definition of convolution is almost always given by 
$$(f \ast g)(t)=\int_{\tau = -\infty}^{\infty} f(\tau)\,g(t-\tau)\,\mathrm{d}\tau\tag{1}$$ 
apart from a significant number of definitions having a factor of $1/\sqrt{2 \pi}$ out front.
In the course that I am taking I am required to use the definition given by $(1)$.
So the question that I have is; can I re-write the $(1)$ as 
$$(f \ast g)(t)=\int_{\tau = -\infty}^{\infty}g(t-\tau)f(\tau)\,\mathrm{d}\tau\tag{2}?$$
I have simply switched the order of the factors in the integrand and since muliplication is commutative by my logic this equation $(2)$ is plausible.
The reason I asked this question is because it just makes more sense for several reasons: $f(\tau)$ is usually a signal and $g(t- \tau)$ is the resolution function that 'sweeps' across the signal $f(\tau)$; so the statement '$f$ convolves with $g$' or 'convolution of $f$ with $g$' now makes sense intuitively. Also, $f$ comes before $g$ alphabetically so it's easier to remember the formula $(2)$ (provided it is indeed correct). 
I thought carefully before asking this, as the answer may be a trivial 'yes of course! STUPID question'. But, I have only just started learning convolution so it is not entirely obvious to me.

So to summarise,
Could someone please confirm that equation $(2)$ is correct (and if not why)? 
Thanks.
 A: Yes, this follows from change of variables.  Consider the case when $f(t) = g(t) = 0$ if $t < 0$.  Then, for each $t \geq 0$, using the substitution $s = t - \tau$, we obtain
\begin{equation*}
(f*g)(t) = \int_{0}^{t} f(\tau)g(t - \tau) \, d\tau = \int_{0}^{t} f(t - s) g(s) \, ds.
\end{equation*}
The same idea works in the general case when the integral is an improper (Riemann) integral (or a Lebesgue integral).
A: This is a question about style. For example, we tend to write a quadratic equation, to solve for $x$, as $ax^2+bx+c=0$ rather than $x^2a+xb+c=0$. The latter looks odd, and would look even more odd in a concrete case such as $x^23-x4+1=0$. The rationale is that the primary variable or unknown elements are "weighted" by fixed factors, or by factors that considered at least temporarily constant or whose variability has at least some known restraints relative to the primary variable. Perhaps this ordering derives ultimately from the conventional ordering of nouns and adjectives in English and other natural languages that have influenced mathematical writing: for example, "three ducks" rather than "ducks three".
If we think of convolution as an abstract binary operation on a function space, it would look perverse to reverse the order of $f$ and $g$ in the defining integral. However, if $f$ is the more variable element—the focus of interest—while $g$ tends to remain in the background as a less variable element, then your reversal is the natural way of writing.
