Fractional Parts in 2019

First note that $\int_0^1 \left\{\tfrac{1}{x}\right\} \,\left\{2x\right\}\,\frac{dx}{x} \; = \; \int_0^{\frac12} \left\{\tfrac{1}{x}\right\} \,2x\,\frac{dx}{x} + \int_{\frac12}^1 \left\{\tfrac{1}{x}\right\} \,(2x-1)\,\frac{dx}{x} \; = \; 2\int_0^1 \left\{\tfrac{1}{x}\right\}\,dx - \int_{\frac12}^1 \left\{\tfrac{1}{x}\right\} \,\frac{dx}{x}$ and $\int_{\frac{1}{N+1}}^1 \left\{\tfrac{1}{x}\right\}\,dx \; = \; \sum_{n=1}^N \int_{\frac{1}{n+1}}^{\frac{1}{n}}\left\{\tfrac{1}{x}\right\}\,dx \; = \; \sum_{n=1}^N \int_{\frac{1}{n+1}}^{\frac{1}{n}}\left(\tfrac{1}{x} - n\right)\,dx \; = \; \int_{\frac{1}{N+1}}^1 \frac{dx}{x} - \sum_{n=1}^N n\left(\frac{1}{n}- \frac{1}{n+1}\right) \; = \; \ln(N+1) - H_{N+1} + 1$ for any $N \in \mathbb{N}$ . Letting $N \to \infty$ gives $\int_0^1 \left\{\tfrac{1}{x}\right\}\,\frac{dx}{x} \; = \; 1 - \gamma$ Since $\int_{\frac12}^1 \left\{\tfrac{1}{x}\right\}\,\frac{dx}{x} \; = \; \int_{\frac12}^1 \left(\tfrac{1}{x}-1\right)\,\frac{dx}{x} \; = \; 1 - \ln2$ we deduce that $\int_0^1 \left\{\tfrac{1}{x}\right\} \,\left\{2x\right\}\,\frac{dx}{x} \; = \; 2(1-\gamma) - (1 - \ln2) \; = \; 1 + \ln2 - 2\gamma$ making the answer $1 + 2 + 2 = \boxed{5}$ .