<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] An alternate solution:
$$
\begin{align}
\int_{0}^{2}x\operatorname{round}\left(\frac{1}{x}\right)dx &= \int_0^2 x\Biggl\lfloor{\frac{1}{x}+\frac{1}{2}\Biggr\rfloor}dx \tag{1}\\
&= 8\int_{1}^{\infty}\frac{\lfloor{x\rfloor}}{\left(2x-1\right)^{3}}dx \tag{2}\\
&= 8\sum_{n=1}^{\infty}n\int_{n}^{n+1}\frac{dx}{\left(2x-1\right)^{3}} \\
&= 16\sum_{n=1}^{\infty}\frac{n^{2}}{\left(1-4n^{2}\right)^{2}} \\
\end{align}
$$
where in $(1)$ we used the Desmos interpretation $\displaystyle \operatorname{round}\left(\frac{1}{x}\right) := \Biggl\lfloor{\frac{1}{x}+\frac{1}{2}\Biggr\rfloor}$ and in $(2)$ we used the mapping $\displaystyle \frac{1}{x}+\frac{1}{2} \mapsto x$.
Next, let $f$ be a piecewise smooth function on $[0,L]$. Then the Fourier sine expansion is given by
$$f\left(x\right)\ =\ \sum_{n=1}^{\infty}b_{n}\sin\left(\frac{\pi nx}{L}\right)$$
where we have the sequence
$$b_{n}=\frac{2}{L}\int_{0}^{L}f\left(x\right)\sin\left(\frac{\pi nx}{L}\right)dx.$$
Let $f(x) = \cos(x)$ and $L = \dfrac{\pi}{2}$. Then using some basic integration, we can prove that
$$b_{n}=\frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\cos\left(x\right)\sin\left(2nx\right)dx\ =\ \frac{4}{\pi}\cdot\frac{2n}{4n^{2}-1}.$$
This means
$$
\begin{align}
\cos\left(x\right) &= \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{2n}{4n^{2}-1}\sin\left(2nx\right) \\
\implies \cos^2(x) &=  \frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n\cos\left(x\right)\sin\left(2nx\right)}{4n^{2}-1} \\
\implies \int_{0}^{\frac{\pi}{2}}\cos^{2}\left(x\right)dx &= \frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n}{4n^{2}-1}\int_{0}^{\frac{\pi}{2}}\cos\left(x\right)\sin\left(2nx\right)dx \\
&= \frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n}{4n^{2}-1}\cdot\frac{2n}{4n^{2}-1}. \\
&= \frac{16}{\pi}\sum_{n=1}^{\infty}\frac{n^{2}}{\left(4n^{2}-1\right)^{2}} \\
\end{align}
$$
But $\displaystyle \int_{0}^{\frac{\pi}{2}}\cos^{2}\left(x\right)dx=\frac{\pi}{4}$. By transitivity, we get
$$\frac{\pi}{4}=\frac{16}{\pi}\sum_{n=1}^{\infty}\frac{n^{2}}{\left(4n^{2}-1\right)^{2}}.$$
Therefore,
$$\int_{0}^{2}x\operatorname{round}\left(\frac{1}{x}\right)dx = \frac{\pi^{2}}{4}.$$
</s>