Belles intégrales

We have $\begin{aligned} \int_1^\infty \frac{\{x\}}{x^3}\,dx & = \; \sum_{n=1}^\infty \int_0^1 \frac{x}{(n+x)^3}\,dx \; = \; \sum_{n=1}^\infty \int_0^1 \left( \frac{1}{(n+x)^2} - \frac{n}{(n+x)^3}\right)\,dx \\ & = \; \sum_{n=1}^\infty \Big[-\frac{1}{n+x} + \frac{n}{2(n+x)^2} \Big]_0^1 \; = \; \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1} + \frac{n}{2(n+1)^2} - \frac{1}{2n}\right) \\ & = \; \frac12\sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} - \frac{1}{(n+1)^2}\right) \; = \; \frac12\big(1 + (\zeta(2) - 1)\big) \; = \; 1 - \tfrac{1}{12}\pi^2 \end{aligned}$ while $\begin{aligned} \int_1^\infty \frac{\{x\} - \frac12}{x}\,dx & = \; \sum_{n=1}^\infty \int_0^1 \frac{x-\frac12}{n+x}\,dx \; = \; \sum_{n=1}^\infty \int_0^1 \left(1 - \frac{n+\frac12}{n+x}\right)\,dx \\ & = \; \sum_{n=1}^\infty \Big[x - (n+\tfrac12)\ln(n+x)\Big]_0^1 \; = \; \lim_{N \to \infty}\sum_{n=1}^N\left(1 - (n+\tfrac12)\ln(n+1) + (n+\tfrac12)\ln n\right) \\ & = \; \lim_{N\to\infty}\left\{ N - (N+\tfrac12)\ln(N+1) + \ln N! \right\} \; = \; \lim_{N \to \infty} \left\{ \ln\left(\frac{e^N N!}{N^{N+\frac12}}\right) - (N+\tfrac12)\ln\left(1 + \tfrac{1}{N}\right) \right\} \\ & = \; \ln\sqrt{2\pi} - 1 \end{aligned}$ using Stirling's Approximation and the result that $\big(1 + \tfrac{1}{n}\big)^n \to e$ as $n \to \infty$ . Thus the answer is $\sqrt{1+12+2+1} = \boxed{4}$ .