Fractional part fun

$\displaystyle \begin{array}{c}\\ \int_1^{\infty} \frac{ \{ x \} }{x^3} \text{ d}x && = \sum_{n=1}^{\infty} \int_n^{n+1} \frac{ \{ x \} }{x^3} \text{ d}x \\ && = \sum_{n=1}^{\infty} \int_n^{n+1} \frac{ x- \lfloor x \rfloor }{x^3} \text{ d}x \\ && = \sum_{n=1}^{\infty} \int_n^{n+1} \frac{1}{x^2} - \frac{n}{x^3} \text{ d}x \\ && = \sum_{n=1}^{\infty} \left[ \frac{-1}{x} + \frac{n}{2x^2} \right]_n^{n+1} \\ && = \sum_{n=1}^{\infty} \frac{1}{2n} - \frac{1}{n+1} + \frac{n}{2(n+1)^2} \\ && = \sum_{n=1}^{\infty} \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+1} - \frac{1}{(n+1)^2} \right) \\ && = \frac{1}{2} \left( 1 - \left( \zeta (2) - 1 \right) \right) \\ \end{array}$

( Using the fact that $\displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 1$ )

$\displaystyle \therefore \int_1^{\infty} \frac{ \{ x \} }{x^3} \text{ d}x = 1 - \frac{1}{2} \zeta (2) = 1 - \frac{\pi^2}{12}$ .

On comparing the coefficients ans summing them up, we get the answer as $\boxed{15}$ .

$\begin{array}{c}\text{I} &=\int_1^{\infty}\frac{\{x\}}{x^3} \text{d}x \\ &= \sum_{n=1}^{\infty} \left( \int_{n}^{n+1}\frac{x-\left \lfloor x \right \rfloor}{x^3} \text{d}x \right) \\ &=\sum_{n=1}^{\infty} \left( \int_{n}^{n+1}\frac{x-n}{x^3} \text{d}x \right) \\ &=\sum_{n=1}^{\infty} \left( -\frac{1}{x}+\frac{n}{2x^2} \right)_{n}^{n+1} \\ &=\sum_{n=1}^{\infty} \left( \frac{1}{2n}- \frac{1}{2(n+1)}-\frac{1}{2(n+1)^2} \right) \\ &=\frac{1}{2}-\frac{1}{2} \left(\frac{\pi^2}{6}-1 \right)\\ &=1-\frac{\pi^2}{12} \end{array}$