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## Integral and power series

Calculus (Integrals, Series)
Tolaso J Kos
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### Integral and power series

Let $\displaystyle I=\int_{0}^{1}\left ( \left \lfloor \frac{2}{x} \right \rfloor-2\left \lfloor \frac{1}{x} \right \rfloor \right )\, {\rm d}x$.

If $\displaystyle e^{I+1}=\sum_{n=0}^{\infty }\left ( \frac{a}{b} \right )^n$ where $\displaystyle a, b$ are coprime numbers, then calculate the sum $\displaystyle S=a+b$.

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Grigorios Kostakos
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### Re: Integral and power series

We give a solution to some point:

\begin{align*}
I&=\int_{0}^{1}\left ( \left \lfloor \frac{2}{x} \right \rfloor-2\left \lfloor \frac{1}{x} \right \rfloor \right )dx\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\frac{1}{x}} \\
{dx\,=\,-\frac{1}{t^2}\,dt}
\end{subarray}}\,\int_{1}^{+\infty}\frac{1}{t^2}\bigl({\lfloor{2t}\rfloor-2\lfloor{t}\rfloor }\bigr)\,dt\\
&=\sum_{n=1}^{\infty }\int_{n+\frac{1}{2}}^{n+1}{\frac{1}{t^2}\,dt}\\
&=\sum_{n=1}^{\infty }\frac {1}{(n+1)(2n+1)}\\
&=2\log{2}-1\,.
\end{align*} So $e^{I+1}=e^{\log4-1+1}=4$ and the geometric series must be converges to $4$, i.e. $a<b$ and $\sum_{n=0}^{\infty }\left ( \frac{a}{b} \right )^n=\frac{b}{b-a}=4\,.$ So we have that $b=\frac{4}{3}\,a\,.$

Remains the evaluation of the number $S=a+\frac{4}{3}a=\frac{7}{3}a$ with the assumption that ${\rm{gcd}}\bigl(a,\tfrac{4}{3}a\bigr)=1\,.$

$\bullet$ In the graph below appears that the integral $\int_{1}^{+\infty}\frac{1}{t^2}\bigl({\lfloor{2t}\rfloor-2\lfloor{t}\rfloor }\bigr)\,dt$ is equal to the infinite sum of the integrals $\int_{n+\frac{1}{2}}^{n+1}{\frac{1}{t^2}\,dt}\,,\; n\in\mathbb{N}\,.$
floorsfunction.png (9.58 KiB) Viewed 946 times
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Grigorios Kostakos