# Commutativity of convolutions implies $\int\limits_{-\infty}^\infty f(t)dt = \int\limits_{-\infty}^\infty f(t-a)dt$?

As I understand it, to convolve $f$ and $g$ means to find $\displaystyle \int_{\mathbb R} f(a)g(t-a)da$, which is also apparently commutative, and therefore $\displaystyle \int_{\mathbb R}f(a)g(t-a)da = \displaystyle \int_{\mathbb R}f(t-a)g(a)da$

That means, if $f(a) = 1$, then $\displaystyle \int_{\mathbb R}g(t-a)da = \displaystyle \int_{\mathbb R}g(a)da$. For $g(t) = t$, for example, is not true. So what am I misunderstanding?

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Neither integral converges for g(t)=t. –  AppliedSide Mar 1 '11 at 3:36
For your purposes you should only consider functions with finite integral.. preferably their absolute value has finite integral. Otherwise you're comparing infinities which gets icky and sometimes not well-defined. –  Zarrax Mar 1 '11 at 3:37

\begin{align*} \\&\int_0^tf(t-u)\text du \\\text{let } v=t-u\\ \text dv=-\text du\\ \\ u=0\rightarrow v=t\\u=t\rightarrow v=0\\ &=\int_t^0f(v)\cdot -\text dv\\ &=\int_0^tf(v)\text dv\\ \\&\\ \int_0^tf(t-u)\text du&=\int_0^tf(u)\text du\\ \end{align*} These integrals are equal because they are summing over the same region, the only difference is one is starting at the other end and working backwards

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I think the answers in the comments are slightly besides the point. It's true that the integrals aren't well-defined in the case of $g(t)=t$, but the more fundamental point here seems to me that

$$\int_{-\infty}^{+\infty} g(t-a)\mathrm{d}a=\int_{-\infty}^{+\infty} g(a)\mathrm{d}a$$

is in fact true, and whenever one of the sides is well-defined the other is, too. It follows without any mention of convolutions from a simple change of variables $u=t-a$, which leaves the limits of integration at infinity unchanged. So it seems that to answer the question needs to involve addressing why Joey thought that these two were not equal in the first place, which probably had nothing to do with lack of convergence. Joey, I'm guessing that you didn't see that making the substitution from $a$ to $t-a$ leaves the limits at infinity unchanged?

Another way of seeing this is that the area under $g(t-a)$ is just the shifted mirror image of the area under $g(a)$.

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