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\[ I_2 = \displaystyle\int_0^\infty\dfrac{1}{\left(x+1+\left\lfloor2\sqrt{x}\right\rfloor\right)^2}dx \]
És pot veure fàcilment que
\[
\displaystyle\int_0^\infty
= \sum_{k=0}^{\infty}\displaystyle\int_{k^2/4}^{(k+1)^2/4}
=
\underbrace{\displaystyle\int_{0}^{1/4}}_{k=0}+
\underbrace{\displaystyle\int_{1/4}^{1}}_{k=1}+
\underbrace{\displaystyle\int_{1}^{9/4}}_{k=2}+
\underbrace{\displaystyle\int_{9/4}^{4}}_{k=3}+
\underbrace{\displaystyle\int_{4}^{25/4}}_{k=4}+
\underbrace{\displaystyle\int_{25/4}^{9}}_{k=5}+
\underbrace{\displaystyle\int_{9}^{49/4}}_{k=6}+...
\]
\noindent Així
\[
I_2 = \displaystyle\int_0^\infty\dfrac{1}{\left(x+1+\left\lfloor2\sqrt{x}\right\rfloor\right)^2}dx
= \sum_{k=0}^{\infty}\displaystyle\int_{k^2/4}^{(k+1)^2/4}\dfrac{1}{\left(x+1+\left\lfloor2\sqrt{x}\right\rfloor\right)^2}dx
\]
En cadascun d'aquest intervals \(\left(k^2/4,(k+1)^2/4\right)\) el valor de \( \left\lfloor x\right\rfloor=k^2/4\) i llavors \(\left\lfloor2\sqrt{x}\right\rfloor=\left\lfloor2\sqrt{k^2/4}\right\rfloor=\left\lfloor2k/2\right\rfloor=k\)
\[
I_2 = \sum_{k=0}^{\infty}\displaystyle\int_{k^2/4}^{(k+1)^2/4}\dfrac{1}{\left(x+1+k\right)^2}dx
\]
Ara cadascuna de les integrals dins del sumatori son trivials
\[
I_2 = \sum_{k=0}^{\infty}\left.-\dfrac{1}{x+1+k}\right|_{k^2/4}^{(k+1)^2/4}
\]
Utilitzant el signe de la primitiva per canviar l'ordre dels extrems d'integració
\[
\begin{array}{rl}
I_2 = \displaystyle\sum_{k=0}^{\infty}\left.\dfrac{1}{x+1+k}\right|^{k^2/4}_{(k+1)^2/4}
&= \displaystyle\sum_{k=0}^{\infty}\left.\dfrac{1}{x+1+k}\right|^{\frac{k^2}{4}}_{\frac{(k+1)^2}{4}}\\
&= \displaystyle\sum_{k=0}^{\infty}\left(\dfrac{1}{\dfrac{k^2}{4}+1+k}-\dfrac{1}{\dfrac{(k+1)^2}{4}+1+k}\right)\\
&= \displaystyle\sum_{k=0}^{\infty}\left(\dfrac{4}{k^2+4+4k}-\dfrac{4}{(k+1)^2+4+4k}\right)\\
&= 4\displaystyle\sum_{k=0}^{\infty}\left(\dfrac{1}{(k+2)^2}-\dfrac{1}{k^2+2k+1+4+4k}\right)\\
&= 4\displaystyle\sum_{k=0}^{\infty}\left(\dfrac{1}{(k+2)^2}-\dfrac{1}{k^2+6k+5}\right)
\end{array}
\]
El primer sumatori esta relacionat amb la famosa serie de Basel
\[
\displaystyle\sum_{k=0}^{\infty}\dfrac{1}{k^2}
=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...=\dfrac{\pi^2}{6}
\]
així
\[
\displaystyle\sum_{k=0}^{\infty}\dfrac{1}{(k+2)^2}
=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...=\dfrac{\pi^2}{6}-1
\]
El segon és pot calcular degut a les cancel.lacions que apareixen
\[
\begin{array}{rl}
\displaystyle\sum_{k=0}^{\infty}\dfrac{1}{k^2+6k+5}
&=\displaystyle\sum_{k=0}^{\infty}\dfrac{1}{(k+1)(k+5)}\\
&=\displaystyle\sum_{k=0}^{\infty}\dfrac{1}{4}\left(\dfrac{1}{k+1}-\dfrac{1}{k+5}\right)\\
&=\dfrac{1}{4}\left(\displaystyle\sum_{k=0}^{\infty}\dfrac{1}{k+1}-\displaystyle\sum_{k=0}^{\infty}\dfrac{1}{k+5}\right)\\
&=\dfrac{1}{4}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...
-\left\{\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...\right\}
\right)\\
&=\dfrac{1}{4}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)\\
&=\dfrac{1}{4}\left(\dfrac{12}{12}+\dfrac{6}{12}+\dfrac{4}{12}+\dfrac{3}{12}\right)\\
&=\dfrac{1}{4}\dfrac{25}{12}\\
&=\dfrac{25}{48}
\end{array}
\]
Per tant la nostra integral
\[
\begin{array}{rl}
I_2 &= 4\displaystyle\sum_{k=0}^{\infty}\left(\dfrac{1}{(k+2)^2}-\dfrac{1}{k^2+6k+5}\right)\\
&= 4\left(\dfrac{\pi^2}{6}-1-\dfrac{25}{48}\right)\\
&= 4\left(\dfrac{\pi^2}{6}-\dfrac{73}{48}\right)\\
&= \dfrac{2\pi^2}{3}-\dfrac{73}{12}
\end{array}
\]
\[
\displaystyle\int_0^\infty\dfrac{1}{\left(x+1+\left\lfloor2\sqrt{x}\right\rfloor\right)^2}dx
= \dfrac{2\pi^2}{3}-\dfrac{73}{12}
\]
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