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I was trying to solve the following simple integration involving indicator function $I_{(a,b]}$ in a journal article. Here are the equations (in LaTeX notation): $$ f(u) = \int_{0}^{1} (I_{(0,s]}(u) - s)\; ds\tag{1} $$ $$ g(u,v) = \int_{0}^{1} (I_{(0,s]}(u) - s)(I_{(0,s]}(v) - s)\; ds\tag{2} $$ where $0 < u, v < 1$. I was thinking that the integration will be simply just $$ f(u) = \int_{0}^{1} (1 - s)\; ds\tag{1} $$ $$ g(u,v) = \int_{0}^{1} (1 - s)(1 - s)\; ds \tag{2} $$ But, I'm not so sure about this. The constraint on both $u$ and $v$ confused me. Any pointer to this solution?

Thanks

Wayan

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  • $\begingroup$ The integral gives $f$ as a function of $u$, so your formula which just gives $f$ as a number would seem unlikely. $\endgroup$ – Gerry Myerson Apr 2 '12 at 12:30
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The integrals are functions of parameters, so I advise you to interchange the set and the variable in the indicator function: for a fixed $u$ we have $$ I_{(0,s]}(u) = \begin{cases} 1,&\text{ if }0<u\leq s \\ 0,&\text{ otherwise}. \end{cases}\Rightarrow I_{(0,s]}(u) = I_{[u,\infty)}(s) $$ where we also used that $u>0$. Hence $$ f(u) = \int\limits_{0}^1(I_{(0,s]}(u) - s)ds = \int\limits_0^1(I_{[u,\infty)}(s) - s)ds = (1-u)-\int\limits_0^1 sds = \frac12-u $$ and similar for the function $g$: when you open the brackets you get 4 terms - of which 3 you should know how to compute (they involve at most 1 indicator function) and for the last term you have: $$ \int\limits_0^1I_{(0,s]}(u)\cdot I_{(0,s]}(v)ds = \int\limits_0^1I_{[u,\infty)}(s)I_{[v,\infty)}(s)ds = \int\limits_0^1I_{[\max(u,v),\infty)}(s)ds = 1-\max(u,v). $$

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  • $\begingroup$ Where did $t$ come from? $\endgroup$ – Gerry Myerson Apr 2 '12 at 12:29
  • $\begingroup$ Thanks Ilya. But, I am still a bit confused how do you get $\int_0^1 I_{[u,\infty)}(s) = (1 - u)$? Is it $\int_0^1 1 ds = s|_{s = u}^{s = 1}$ since $0 < s \leq 1$? $\endgroup$ – Wayan Apr 2 '12 at 12:30
  • $\begingroup$ @GerryMyerson: I don't know how to say it in English. Apparently, from nowhere - it was a typo. $\endgroup$ – Ilya Apr 2 '12 at 12:32
  • $\begingroup$ @Wayan: sure, $\int\limits_0^1 I_{[u,\infty)}(s)ds = \int\limits_u^1 ds = 1-u$ - and you had a typo in your comment, in the lower limit of the latter integral $\endgroup$ – Ilya Apr 2 '12 at 12:33
  • $\begingroup$ @Ilya: Thank you for pointing the typo. What about in the second equation where $$\int_{0}^{1} I_{(0,s]}(u)I_{(0,s]}(v)?$$ If I do it similar like $f$ then $$\int_{u}^{1} I_{u,\infty]}(s)I_{(v,\infty]}(s)?$$ But what about the condition $0< u, v< 1$. Is it safe to assume $u < v$? $\endgroup$ – Wayan Apr 2 '12 at 13:20

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