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Question

Find the exact value of $$\int_{1}^{2016}\frac {\lfloor \ln x \rfloor} {x}\ \mathrm{d}x\ .$$

My working

My intuition is to treat the integral without the floor function and integrate, then "put back" the floor function after, so $$\int_{1}^{2016}\frac {\lfloor \ln x \rfloor} {x}\ \mathrm{d}x = \frac 1 2 [(\lfloor \ln x \rfloor)^2]^{x = 2016}_{x = 1}\ .$$

May I know if my intuition is correct? If not, what would be the right way to do this? This is my first time encountering an integral with the floor function. Any help/suggestions would be greatly appreciated :)

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  • $\begingroup$ Well, one thing which strikes me at first for solving this would be as follows. ln1=0, ln2016 = 7.6 (approx). there would be points along the x axis where the floor function is discontinuous. Break your integral about those points $\endgroup$
    – Aman Kumar
    Nov 15, 2020 at 7:47

1 Answer 1

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enter image description here

This is a plot of floor(ln(x)) along x axis.Find these points of discontinuity and break your integral at these points. The the numerator in all of these sub parts will be a constant and you will have just $\frac{some-constant}{x}$ as the integrand.

You can find these points of discontinuity using desmos.com or a scientific calculator (which is easily available on smartphones or computers these days).

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  • $\begingroup$ The key takeaway here is that there is no direct method for integrating curves which are discontinuous (like floor, ceil, piece-wise defined). Integration is done simply by applying the concept that integration in 2d curves is just the area under the defined curves. $\endgroup$
    – Aman Kumar
    Nov 15, 2020 at 7:56
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    $\begingroup$ I had the same idea but he wants the exact value. How do you compute the exact values of the discontinuities ? $\endgroup$
    – nicomezi
    Nov 15, 2020 at 8:33
  • $\begingroup$ Thank you! I believe I have solved it :) $\endgroup$
    – Ethan Mark
    Nov 15, 2020 at 9:54
  • $\begingroup$ How did you find the exact values of the discontinuities ? @EthanMark $\endgroup$
    – nicomezi
    Nov 15, 2020 at 10:08
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    $\begingroup$ @nicomezi $\int^{2016}_1 \frac {\lfloor \ln x \rfloor} {x}\ \mathrm{d}x = \int^{e^1}_{e^0} \frac 0 x\ \mathrm{d}x + \int^{e^2}_{e^1} \frac 1 x\ \mathrm{d}x + \ldots + \int^{e^7}_{e^6} \frac 6 x\ \mathrm{d}x + \int^{2016}_{e^7} \frac 7 x\ \mathrm{d}x$. The "pattern" can actually be generalised to any integral involving a floor function. I hope this satisfies your curiosity! $\endgroup$
    – Ethan Mark
    Jan 20, 2021 at 18:32

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