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If $$F(x) = \int\frac{(1+x) \lfloor(1-x+x^2)(1+x+x^2)+x^2\rfloor}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$$then find the value of $\lfloor F(99)- F(3)\rfloor$

My attempt : I could only reduce the portion of the numerator within $\lfloor\rfloor$ the floor function. $$F(x) = \int\frac{(1+x)\lfloor1+2x^2 + x^4\rfloor}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$$ I could not understand how to reduce further. Any help will be gratefully acknowledged.

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  • $\begingroup$ Does the integrand also contain the greatest integer function - also represented by []? $\endgroup$
    – TheSimpliFire
    Commented Jul 18, 2018 at 13:17
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    $\begingroup$ Yes the integrand also contains the box function. $\endgroup$ Commented Jul 18, 2018 at 17:50

1 Answer 1

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After some factorisation, what we're left to figure out is:

$$ I= \left \lfloor \int_3^{99} \frac{\lfloor (1+x^2)^2 \rfloor}{(1+x)(1+x^2)^2} dx \right \rfloor $$

But first, for $g(x) = (1+x^2)^2$, let's study $ \lfloor g(x) \rfloor$ and $ \frac{\lfloor g(x) \rfloor}{g(x)}$.

Notice that for some positive integer $n$, $ \frac{\lfloor g(x) \rfloor}{g(x)}$ has a jump discontinuity at every $ x=\sqrt{\sqrt{n+1} -1}$ and that $ g(x)-1 \leq \lfloor g(x) \rfloor \leq g(x) $.

Consequently, for the interval $ \sqrt{\sqrt{n} -1} \leq x < \sqrt{\sqrt{n+1} -1}$, we have

$$ \frac{n}{n+1} < \frac{\lfloor (1+x^2)^2 \rfloor}{(1+x^2)^2} \leq 1 $$

So, since $ \frac{n}{n+1}$ increases as $n$ increases, $ \frac{99}{100} \leq \frac{\lfloor (1+x^2)^2 \rfloor}{(1+x^2)^2} \leq 1$ for all $ x \geq 3$.

Therefore,

$$ \int_3^{99} \frac{99}{100(1+x)} dx < I < \int_3^{99} \frac{1}{1+x} dx $$

$$ \frac{99}{100} \ln(25) < I < \ln(25) $$

$$ \implies I \approx 3.2 $$

$$ \implies \lfloor I \rfloor = 3 $$

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    $\begingroup$ I think you mean $\frac{n}{n+1}$ increases as $n$ increases. $\endgroup$
    – David K
    Commented Jul 18, 2018 at 18:55
  • $\begingroup$ Oof yes thank you $\endgroup$ Commented Jul 18, 2018 at 19:02

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