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My teacher used integration by parts to solve the problem like so: $$\int_0^2 xd(\{x\}) =[x\{x\}]_0^2-\int_0^2 \{x\}dx\\ =0-\int_0^1 xdx-\int_1^2 (x-1)dx$$

which comes out to -1. But when I was solving I simplified the $d(\{x\})$ to just $(1)dx$ since the derivative of the $\{x\}$ is $1$ almost everywhere. $$\int_0^2 xd(\{x\})=\int_0^2 xdx$$ which comes out to 2, obviously not right. I have a feeling that the mistake is related to $\{x\}$ not being continuous everywhere, but my reasoning is that it shouldn't matter since we are integrating and hence finite number of discontinuities (in this case, only 1) shouldn't change the answer.

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    $\begingroup$ a) In the inegration by parts there is a sign error, it should be $- \int \{x\}\,dx$. b) Since $\{x\}$ has jumps, in $d\{x\}$ there appear Dirac distributions. $\endgroup$ Commented 14 hours ago
  • $\begingroup$ @DermotCraddock a) sorry, fixed it b) could you elaborate? I'm not familiar with that $\endgroup$ Commented 14 hours ago
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    $\begingroup$ Your teacher is correct (assuming the result was $-1$). Your method yields $2$ because you ignored the "jumps" in the function $\{x\}$. When integrating $d(\{x\})$, you must account for both the continuous slope and the instantaneous drops at the integers. $\endgroup$ Commented 14 hours ago
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    $\begingroup$ You wrote "it shouldn't matter since we are integrating and hence finite number of discontinuities (in this case, only 1) shouldn't change the answer." That's true for Riemann integrals, but here you have a Riemann-Stieltjes integral. $\endgroup$ Commented 14 hours ago
  • $\begingroup$ It boils down to $$\int_0^2 f(x)\,d\{x\} = \int_0^2 f(x)\,dx - f(1) - f(2)$$ for all appropriate $f$. $\endgroup$ Commented 14 hours ago

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