I was reading the book interesting integrals and this came up: $$\int_{j}^{j+1} \frac{n-j}{x} \mathrm{d}x$$ it then goes on to say that $j$ is equal to floor $x$ because the integration interval $j \leq x < j+1$. I get this but aren't the limits of integration on the interval $j\leq x\leq j+1$?


It's irrelevant. Adding or removing a single point will not change the integral; if you think about it, a single point will belong to a single very small division in the partition of the interval, so its contribution to the value of the integral is zero.

  • $\begingroup$ Great thanks, would that also be because the point has 0 area in that line? $\endgroup$ – jake walsh May 7 '16 at 15:39
  • $\begingroup$ Yes, that's one way to see it. But if you want to justify it formally, you should probably go for the definition of integral with the partitions. $\endgroup$ – Martin Argerami May 7 '16 at 15:41
  • $\begingroup$ Glad I could help. $\endgroup$ – Martin Argerami May 7 '16 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.