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Find the integral $$\int_1^{1000}\frac{dx}{x+⌊\log_{10}(x)⌋}$$ The logarithm is creating some problems along with floor function.

Jack D'Aurizio
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    I would rather say that without the floor function the logarithm would create problems. The floor function is locally constant, so can you see how to best split this into subintervals, where that applies? – Jyrki Lahtonen Feb 16 '15 at 11:05
  • Is that $\log(x)$ to base $10$, $e$, or another base? The easiest answer would come with base $10$. – Rory Daulton Feb 16 '15 at 11:38
  • Yep, its base 10. –  Feb 16 '15 at 11:40

2 Answers2

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We have that $\lfloor \log_{10}(x)\rfloor $ equals $0$ for $x\in(0,10)$, $1$ for $x\in[10,100)$ and $2$ for $x\in[100,1000)$, hence the value of the integral is just: $$ \int_{1}^{10}\frac{dx}{x}+\int_{10}^{100}\frac{dx}{x+1}+\int_{100}^{1000}\frac{dx}{x+2}=\log(10)+\log\frac{101}{11}+\log\frac{1002}{102}$$ or: $$ \color{red}{\log\frac{168670}{187}}.$$

Jack D'Aurizio
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    @TheGreatDuck What indefinite integral? The question is about a definite integral. This answer is correct. – Jean-Claude Arbaut Nov 07 '16 at 16:54
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    @TheGreatDuck Not the only way. By far. Never heard about the residue theorem, for instance? Often it's impossible to compute explicitly the antiderivative. Even if it's possible, it's often tricky and very convoluted. Here is another example: http://math.stackexchange.com/questions/1942983/find-int-02-pi-frac15-4-cos-x-dx – Jean-Claude Arbaut Nov 07 '16 at 17:28
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    @TheGreatDuck I disagree. And compared to the crystal clear solution in Jack d'Aurizio's answer, yours is a mess. The question is easy, your are making it overly complicated. – Jean-Claude Arbaut Nov 07 '16 at 17:31
  • @TheGreatDuck Without sounding rude or anything, you should check your facts before arguing. – StubbornAtom Nov 07 '16 at 17:41
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HINT:

$$n\le\log_{10}(10x)<n+1\iff10^n\le(10x)<10^{n+1}$$

For example, $n=1,10\le(10x)<100\iff1\le x<10$

lab bhattacharjee
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