# Integral $\int\frac{(1+x) \lfloor(1-x+x^2)(1+x+x^2)+x^2\rfloor}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$

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.

• Does the integrand also contain the greatest integer function - also represented by []? – TheSimpliFire Jul 18 '18 at 13:17
• Yes the integrand also contains the box function. – MathsLearner Jul 18 '18 at 17:50

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,

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

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

$$\implies I \approx 3.2$$

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

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