<s>[INST] Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$. Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$.
Let $[x]$ denote $\lceil{x-\frac{1}{2}}\rceil$.
Using Desmos, I got $2.46736022133$ and WolframAlpha does not give me a solution. My intuition tells me that it might be possible to find an exact value using Trapezoidal Reimann Sums but I am not really sure how to go about doing it. After my attempt, I got stuck but I was at a point where I could plug it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. Why did it come out so nicely?
My attempt:
Where $A_n$ denotes the area of the $nth$ trapezoid from the right:
$$A=\frac{h}{2}(a+b)$$
$$A_n=\frac{\frac{2}{2n-1}-\frac{2}{2n+1}}{2}(\frac{2n}{2n-1}+\frac{2n}{2n+1})$$
$$A_n=\frac{\frac{4n+2}{4n^{2}-1}-\frac{4n-2}{4n^{2}-1}}{2}\left(\frac{4n^{2}+2n}{4n^{2}-1}+\frac{4n^{2}-2n}{4n^{2}-1}\right)$$
$$A_n=\frac{2}{4n^{2}-1}\left(\frac{8n^{2}}{4n^{2}-1}\right)$$
$$A_n=\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
Then:
$$\int_{0}^{2}x[\frac{1}{x}]dx=\sum_{n=1}^{\infty}A_n=\sum_{n=1}^{\infty}\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
I do not know how to solve this infinite summation so I plugged it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. How did it get to this conclusion? Is there a more efficient way to solve this?
 [/INST] Another way to get the sum of the series
$\displaystyle\sum_{n=1}^{\infty}\frac{16n^2}{\left(4n^2-1\right)^2}$
without double summations and without switching the order of the sums.
$\displaystyle\sum_{n=1}^N\frac{16n^2}{\left(4n^2-1\right)^2}=$
$=\displaystyle\sum_{n=1}^N\left[\frac{2n}{\left(2n-1\right)^2}-\frac{2n}{\left(2n+1\right)^2}\right]=$
$=\displaystyle2\sum_{n=1}^N\frac1{\left(2n-1\right)^2}+\sum_{n=1}^N\left[\frac{2(n-1)}{\left(2n-1\right)^2}-\frac{2n}{\left(2n+1\right)^2}\right]=$
$=\displaystyle2\left(\sum_{n=1}^{2N-1}\frac1{n^2}-\sum_{n=1}^{N-1}\frac1{(2n)^2}\right)+\left[0-\frac2{3^2}+\frac2{3^2}-\frac4{5^2}+\ldots+\\\quad+\frac{2(N-1)}{\left(2N-1\right)^2}-\frac{2N}{\left(2N+1\right)^2}\right]=$
$=\displaystyle2\left(\sum_{n=1}^{2N-1}\frac1{n^2}-\frac14\sum_{n=1}^{N-1}\frac1{n^2}\right)-\frac{2N}{\left(2N+1\right)^2}\,.$
Hence ,
$\displaystyle\sum_{n=1}^{\infty}\frac{16n^2}{\left(4n^2-1\right)^2}=$
$=\displaystyle\lim_{N\to\infty}\left[\sum_{n=1}^N\frac{16n^2}{\left(4n^2-1\right)^2}\right]=$
$=\displaystyle\lim_{N\to\infty}\left[2\left(\sum_{n=1}^{2N-1}\frac1{n^2}-\frac14\sum_{n=1}^{N-1}\frac1{n^2}\right)-\frac{2N}{\left(2N+1\right)^2}\right]=$
$=\displaystyle2\left(\sum_{n=1}^{\infty}\frac1{n^2}-\frac14\sum_{n=1}^{\infty}\frac1{n^2}\right)=$
$=\displaystyle2\left(\frac{\pi^2}6-\frac14\!\cdot\!\frac{\pi^2}6\right)=\displaystyle2\left(1-\frac14\right)\frac{\pi^2}6=$
$\displaystyle=2\!\cdot\!\frac34\!\cdot\!\frac{\pi^2}6=\frac{\pi^2}4\,.$
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