$ \int_{0}^3 f(x)\,d ([x]+x) $ where $ f (x) = [x] $ if $ 0\le x < 3/2 $ and $ f (x) = e^x$ if $ 3/2\le x \le2 $

I could break this into $ \int_{0}^3 f(x)\,d ([x]) $ and $ \int_{0}^3 f(x)\,d x $ and solve the 2nd part as $ \int_{0}^{3/2} x\,d x $ and $ \int_{3/2}^2 e^x\,d x $ and I am not sure on how ti compute $ \int_{0}^3 f(x)\,d ([x]) $ as both $f(x)$ and $[x]$ is discontinous on the interval and intergration by parts for Riemann Stieltjes integral cannot be used.

Any help on how to do that would be appreciated

where $[x]$ denotes the greatest integer function

  • $\begingroup$ Does $[x]$ refer to the "greatest integer function"? $\endgroup$ Dec 16, 2013 at 6:21
  • $\begingroup$ yes it does! edited it now $\endgroup$
    – user83369
    Dec 16, 2013 at 6:27

1 Answer 1


Assuming I understand your notation, you should have $$ \int_0^3 f(x)\,d[x] = f(1) + f(2) + f(3) $$ Which can be deduced following the definition of the Riemann Stieltjes integral (hint: consider the partition given by $\{0,1-\delta,1+\delta,2-\delta,2+\delta,3-\delta,3\}$ for a suitably small $\delta$).

With this, you should be able to answer the question fully. In general, the R-S integral $\int_a^b f(x)\, d[x]$ will exist so long as $f$ is continuous (from the left) at each integer on $[a,b]$.

  • $\begingroup$ Than you! I have only problem as to if it wasn't given that f(x) is integrable with respect to $\alpha$ can we use that partition and claim it is integrable if not how could that be proved? $\endgroup$
    – user83369
    Dec 16, 2013 at 11:45
  • $\begingroup$ Using this partition, you can show that $f$ is integrable with respect to $[x]$. In particular, for any $\epsilon>0,$ you can choose a $\delta$ so that $U(f,P,\alpha)-L(f,P,\alpha)<\epsilon$. $\endgroup$ Dec 16, 2013 at 13:58
  • $\begingroup$ $f$ is not integrable wrt $g$ because $f$ and $g$ are discontinuous at $x = 1$. The Riemann-Stieltjes sum can contain the term $f(1 - \epsilon/2) (g(1 + \epsilon) - g(1 - \epsilon)) \to 0$ or the term $f(1 + \epsilon/2) (g(1 + \epsilon) - g(1 - \epsilon)) \to 1$. The limit of the sum doesn't exist. $\endgroup$
    – Maxim
    May 29, 2018 at 12:04

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