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I am teaching myself how convolution works, there's a question which looks like this - find the convolution of the following two functions $f$ and $g$.

graphics of f and g

I understand the problem intuitively that the resulting function should be essentially the product of $f$ sweeping over $g$, and since the functions are quite simple, I can find key points like when $x = 0, 1, 2, 3$ and interpolate the graph of the resulting function easily.

While reading through the "solution" of this problem in my textbook, for the intersection of $0 \le x \lt 1$, the author wrote this: for $0\leq x<1$, $$\int_{-\infty}^{\infty} g(t)\cdot f(x-t)dt=\int_{0}^x 2t\cdot dt=\frac{x^2}{2}\cdot 2.$$ which I'm having trouble to understand. How exactly did he replace the $\infty$ and $-\infty$ with $0$ and $x$, and how exactly did he turn the whole $g(t) \cdot f(x-t)$ into $2t$?

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Hint. Note that $f(x-t)=2$ when $0\leq x-t\leq 1$, i.e. $x-1\leq t\leq x$, otherwise it is zero. Hence $$\int_{-\infty}^{\infty} g(t)f(x-t)dt=\int_{x-1}^x g(t)2 dt.$$ Now if $x\in[0,1)$ then what is $g(t)$ for $t\in [x-1,0]$? And for $t\in [0,x]$?

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  • $\begingroup$ So let me think out loud - if $x \in [0,1)$ then $g(t) = t$ for the positive part $t \in [0,x)$, but $0$ for the negative part $t \in [x-1, 0)$, and we can write that in two parts because $0$ must be inside $[x-1, x]$. So it makes sense to break $\int_{x-1}^{x} g(t) 2dt$ into two parts - $\int_{x-1}^{0} g(t) 2dt$ and $\int_{0}^{x} g(t) 2dt$, where the first part is simply $0$, no matter what $x$ we choose. Hence our original integral is now $\int_{0}^{x} g(t) 2dt$ which is actually $\int_{0}^{x} t\cdot 2dt$ due to $g(t) = t$ for the positive part. Is my reasoning correct? :D $\endgroup$ – maranic Sep 8 at 13:13
  • $\begingroup$ @maranic It's perfect! $\endgroup$ – Robert Z Sep 8 at 13:15
  • $\begingroup$ It was quite tough to wrap my head around this, because we are dealing with 2 variables here. :D $\endgroup$ – maranic Sep 8 at 13:22
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Observe that integrand $g(t)f(x-t)$ (where $x$ is fixed and $t$ is ranging over $\mathbb R$) takes value $0$ for every $t\notin[0,x]$.

This justifies to replace $\int_{-\infty}^{\infty}\cdots$ by $\int_0^x\cdots$.

Further for any fixed $x\in[0,1)$ it is true that $g(t)f(x-t)=t2$ on interval $[0,x]$.

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I'm assuming

$$ f(x)=\begin{cases} 2 & x\in [0,1] \\ 0 & else \end{cases} $$

And

$$ g(x)=\begin{cases} x & x\in [0,1] \\ 2-x & x\in [1,2)\\ 1 & x\in [2,3] \\0 & else \end{cases} $$

Adding this together, we see $$ \int_{-\infty}^{\infty} g(t)f(x-t)\textrm{d}t=2\int_{1-x}^x g(t)\textrm{d}t, $$ since for these values of $t,$ $f(x-t)=2$ and for all other values of $t$, $f(x-t)$ is $0$.

Now, $g(t)=0$ for $t\leq 0$ so $$ 2\int_{1-x}^x g(t)\textrm{d}t=2\int_0^x g(t)\textrm{d}t=2\int_0^x t\textrm{d}t, $$ simply by plugging into the definition of $g$.

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  • $\begingroup$ You're correct. $\endgroup$ – WoolierThanThou Sep 8 at 12:34

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