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I am reading a textbook which claims that we can obtain by partial integration, for CDF $F(x)$:$$\int_{t}^{\infty} 1-F(x) \frac{dx}{x}=\int_{t}^{\infty} (\log u -\log t) dF(u) $$

I am aware that the latter integral is a Riemann-Stieltjes integral, but I am not sure how to go from the first to the latter via the partial integration formula I am familiar with, I obtain:

$$\int_{t}^{\infty} 1-F(x) \frac{dx}{x}=\log(x)(1-F(x)|_{t}^{\infty} -\int_{t}^{\infty}\log(x)(1-F(x) dx.$$

Joogs
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1 Answers1

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Hope this helps: \begin{align} \int_t^\infty(1-F(x))\frac{dx}{x} & = (1-F(x))\log x|_t^\infty-\int_t^\infty (-dF(x)) \log x\\ & = -(1-F(t))\log t+\int_t^\infty\log u dF(u)\\ & = -\int_t^\infty dF(u)\log t+\int_t^\infty\log u dF(u)\\ & = \int_t^\infty(\log u-\log t)dF(u). \end{align}

AddSup
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  • I am only wondering why we can change the variable of integration from x (dx) to the Riemann-Stieltjes kind with dF(x). – Joogs Oct 15 '18 at 10:58
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    @Joogs We have integration by parts formulas for Lebesgue-Stieltjes integration as well. See, e.g., https://math.stackexchange.com/questions/221521/integration-by-parts-and-lebesgue-stieltjes-integrals and a link therein. To be clear, both in my answer and in your question, $\int_t^\infty$ means $\int_{(t,\infty)}$ (as opposed to $\int_{[t,\infty)}$). – AddSup Oct 15 '18 at 11:50
  • @Joogs I just noticed that you were interpreting the integrals as Riemann-Stieltjes integrals. (Once you learn Lebesgue integration, you grow out of Riemann integration, I guess that's why I didn't notice Riemann's name.) Since Lebesgue integration generalizes Riemann integration, Riemann-Stieltjes, too, naturally has integration by parts formulas, which are simpler. See https://math.stackexchange.com/questions/211552/riemann-stieltjes-integral-integration-by-parts-rudin. – AddSup Oct 15 '18 at 12:01