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If $$f\left(x\right)=\begin{cases} f_{1}\left(x\right), & x\in[0,1]\\ f_{2}\left(x\right), & x\in[1,\sqrt{5}]\\ 0, & \mbox{elsewhere} \end{cases}$$ what does the piecewise-defined function $f\star f$ look like or how to express the pieces of $f\star f$ by the integrals (or the sum of integrals) of the products of $f_{1}$ and $f_{2}$? Thanks in advance.

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2 Answers 2

Why don't you use this one instead:

If $$f\left(x\right)=\begin{cases} f_{1}\left(x\right), & x\in[0,a_1)\\ f_{2}\left(x\right), & x\in[a_1,a_2)\\ f_3(x), & x\geq a_2 \end{cases}$$ then by using the unit step function, we can write $f(x)$ as the following summation: $$f(x)=f_1(x)+\left(f_2(x)-f_1(x)\right)u_{a_1}(x)+\left(f_3(x)-f_2(x)\right)u_{a_2}(x)$$ wherein $$u_c(x)=\begin{cases} 0\left(x\right), & x\in[0,c)\\ 1\left(x\right), & x\geq c\\ \end{cases}$$ is step unit function.

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+1 nice suggestion! –  amWhy Feb 22 '13 at 2:04

In order to solve such a problem, try to work straight forward from the definition of the convolution: The integrand is a product of which one factor only depends on the integration variable. Segment the integral so that you can express that factor by its piecewise definitions. Then use commutativity of the convolution and repeat that process for the other factor that now only depends on the new integration variable for each of the integral summands. Then simplify the sum of integrals.

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