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Integral with floor and ceiling function

Let \left \lfloor \cdot \right \rfloor denote the floor function and \left \{ \cdot \right \} denote the ceiling function. Evaluate the integral

    \[\mathcal{J}= \int_{0}^{1}x \left \lfloor \frac{1}{x} \right \rfloor\left \{ \frac{1}{x} \right \} \, {\rm d}x\]

Solution

We have successively

    \begin{align*} \mathcal{J} &= \int_{0}^{1}x \left \lfloor \frac{1}{x} \right \rfloor\left \{ \frac{1}{x} \right \} \, {\rm d}x \\ &\!\!\!\! \overset{x \mapsto 1/x}{=\! =\! =\! =\!} \int_{1}^{\infty} \frac{\left \lfloor x \right \rfloor \left \{ x \right \}}{x^3} \, {\rm d}x\\ &= \sum_{n=1}^{\infty} \int_{n}^{n+1} \frac{n(x-n)}{x^3} \, {\rm d}x\\ &= \sum_{n=1}^{\infty} \left [ \int_{n}^{n+1} \frac{n}{x^2} \, {\rm d}x - \int_{n}^{n+1} \frac{n^2}{x^3} \, {\rm d}x \right ] \\ &= \sum_{n=1}^{\infty} \left ( \frac{1}{n+1} - \frac{2n+1}{2(n+1)^2} \right ) \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(n+1)^2}\\ &= \frac{1}{2} \left ( \zeta(2) - 1 \right ) \\ &= \frac{\pi^2}{12} - \frac{1}{2} \end{align*}

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