<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] $$\frac{16 n^2}{\left(4 n^2-1\right)^2}=-\frac{1}{2 n+1}+\frac{1}{(2 n+1)^2}+\frac{1}{2 n-1}+\frac{1}{(2 n-1)^2}$$
$$\sum_{n=1}^p\frac{16 n^2}{\left(4 n^2-1\right)^2}=\frac{1}{2} \left(\psi ^{(0)}\left(\frac{3}{2}\right)-\psi
   ^{(0)}\left(p+\frac{3}{2}\right)\right)+$$
$$\frac{1}{8} \left(-2 \psi ^{(1)}\left(p+\frac{3}{2}\right)+\pi
   ^2-8\right)+$$ $$\frac{1}{2} \left(\psi ^{(0)}\left(p+\frac{1}{2}\right)-\psi
   ^{(0)}\left(\frac{1}{2}\right)\right)+
\frac{1}{8} \left(\pi ^2-2 \psi
   ^{(1)}\left(p+\frac{1}{2}\right)\right)$$
Using the asymptotics
$$\sum_{n=1}^p\frac{16 n^2}{\left(4 n^2-1\right)^2}=\frac{\pi ^2}{4}-\frac{1}{p}+\frac{1}{2
   p^2}+O\left(\frac{1}{p^3}\right)$$
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