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I'm trying to evaluate the integral $$\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y$$ where $\chi_{[0,1]}(x)=1$ is the characteristic function, i.e. equals $1$ for $x \in [0,1]$ and $0$ otherwise. How do I evaluate this integral? So far I've done $$\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y = \int_{0}^1 \chi_{[0,1]} (x-y) \, \mathrm{d}y.$$ I've tried substitution but I don't find it yielding anything useful. The answer should be the function $$f(x)= \left\{ \begin{array}{l l} x & \quad \text{if $x \in [0,1]$}\\ 2-x & \quad \text{if $x \in [1,2]$}\\ 0 & \quad \text{otherwise.} \end{array} \right.$$

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    $\begingroup$ There's no trick to be done. The value of this integral depends completely on $x$. Try plotting the function $\chi_{[0,1]}(x-y)$, $y$ varying in $[0,1]$, for different fixed values of $x$. You'll find that it is $0$ outside of some interval where it may be constantly $1$. i.e. it will look like a rectangular bump. And of course you know how to integrate piecewise constant functions. $\endgroup$ – Gyu Eun Lee May 21 '13 at 20:51
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Change variables $x-y=s$, thus the new domain is $(x,x-1)$. You have $$ \int_0^1\Xi(x-y)dy=\int_{x-1}^x\Xi(s)ds. $$ Now, if $x-1>1$ and if $x<0$ the integral vanishes. Moreover, you have $$ \int_{x-1}^x\Xi(s)ds=\text{length}\left((x-1,x)\cap(0,1)\right). $$ If $0<x<1$ $$ \text{length}\left((x-1,x)\cap(0,1)\right)=x. $$ Otherwise, if $1<x<2$ $$ \text{length}\left((x-1,x)\cap(0,1)\right)=1-(x-1)=2-x $$

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  • $\begingroup$ Thanks for your great answer! $\endgroup$ – Numbersandsoon Jun 14 '13 at 22:57
  • $\begingroup$ you are welcome :-) $\endgroup$ – guacho Jun 15 '13 at 8:10
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You're off to a good start. Next note that $$ \chi_{[0,1]}(x-y) = \begin{cases} 1, & \text{for $x-1 \le y \le x$} \\ 0, & \text{otherwise}\end{cases}. $$ Draw some graphs!

Hence, if $0< x < 1$, then $\chi_{[0,1]}(x-y)$ on the whole of $[0,1]$. Thus the integral is $0$.

If $0 < x < 1$, then $\chi_{[0,1]}(x-y) = 1$ on $[0,x]$ and $=0$ on the rest of $[0,1]$. In this case, the integral will be $$ \int_0^x 1\,dy = x. $$

Third case, if $1 < x < 2$, then $\chi_{[0,1]}(x-y) = 1$ on $[x-1,1]$ and $=0$ on the rest of $[0,1]$, so the integral will be $$ \int_{x-1}^1 1\,dy = 2-x.$$

Finally, if $x > 0$, then $\chi_{[0,1]}(x-y) = 0$ on $[0,1]$ again.

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  • $\begingroup$ Thanks for your great answer! $\endgroup$ – Numbersandsoon Jun 14 '13 at 22:58

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