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Let $f$ be a nonnegative valued-measurable function on $[0,\infty)$ and $0<r<\infty$

By Fubini's theorem, we have

$$ \begin{align} \int_0^\infty\left(\int_0^t f(s)ds\right)t^{-r-1}dt &=\int_0^\infty\int_s^\infty t^{-r-1}f(s) dtds\\&=\frac1r\int_0^\infty\left[s f(s)\right]s^{-r-1}ds \end{align} $$

I don't understand how they apply the theorem of fubini.

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    $\begingroup$ You have at least a couple errors/typos. Presumably $0 < r$? Also maybe the first integral should be $\int_0^\infty\left(\int_0^t f(s)ds\right)t^{-r-1}dt$? If so, you are really applying Tonelli's theorem (similar to Fubini, but it requires the integrand be non-negative rather than integrable) $\endgroup$ Nov 30 '19 at 20:45
  • $\begingroup$ @BrianMoehring i made the corrections, but can you explain me more $\endgroup$
    – Vrouvrou
    Nov 30 '19 at 20:53
  • $\begingroup$ suppose that it is integrable and the forget it in the book how to obtain this? $\endgroup$
    – Vrouvrou
    Nov 30 '19 at 21:01
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We will use $\int$ as a shorthand for $\int_{\Bbb R }$ and for simplicity I will assume that $t\geqslant 0$. You have that $$ \begin{align*} \int\left(\int_{[0,t]} f(s)ds\right)t^{-r-1}dt &=\iint t^{-r-1}f(s)\chi_{\{(s,t)\in \Bbb R ^2:0\leqslant s\leqslant t\}}(s,t) \,\mathrm d s\,\mathrm d t\\ &\overset{(*)}{=}\iint t^{-r-1}f(s)\chi_{\{(s,t)\in \Bbb R ^2:0\leqslant s\leqslant t\}}(s,t) \,\mathrm d t\,\mathrm d s\\ &=\int_{[0,\infty )}\left(\int _{[s,\infty )}t^{-r-1}\,\mathrm d t\right)f(s)\,\mathrm d s\\ \end{align*} $$ where in $(*)$ we assumed that we can use Fubini's theorem or Tonelli's theorem. By last if $r>0$ then it follow that $$ \int_{[0,\infty )}\left(\int _{[s,\infty )}t^{-r-1}\,\mathrm d t\right)f(s)\,\mathrm d s=\frac1r\int_{[0,\infty )} s^{-r} f(s)\,\mathrm d s=\frac1r\int_{[0,\infty )} [sf(s)]s^{-r-1}\,\mathrm d s $$

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  • $\begingroup$ how you find $\int_s^\infty$ $\endgroup$
    – Vrouvrou
    Dec 1 '19 at 7:42
  • $\begingroup$ just integrate first respect to $t$ and use the definition of $\chi _{\{(s,t)\in \Bbb R ^2:0\leqslant s\leqslant t\}}$. That is: for fixed $s$ then the values where $\chi _{\{(s,t)\in \Bbb R ^2:0\leqslant s\leqslant t\}}=1$ are when $t\in[s,\infty)$ $\endgroup$
    – Masacroso
    Dec 1 '19 at 15:36

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