Who's Up To the Challenge?79

A couple of substitutions $y = x^{-1}$ and $z = y^{-1}$ give $\begin{aligned} \int_0^1 \left\{\tfrac{1}{x}\right\}^2\left\{\tfrac{1}{1-x}\right\}\,dx & = \; \int_1^\infty \left\{y\right\}^2 \left\{\tfrac{y}{y-1}\right\}\frac{dy}{y^2} \; = \; \int_1^\infty \left\{y\right\}^2 \left\{\tfrac{1}{y-1}\right\}\,\frac{dy}{y^2} \; = \; \int_0^\infty \left\{y+1\right\}^2\left\{\tfrac{1}{y}\right\}\,\frac{dy}{(y+1)^2} \\ & = \; \int_0^\infty \left\{y\right\}^2\left\{\tfrac{1}{y}\right\}\,\frac{dy}{(y+1)^2} \; = \; \sum_{k=0}^\infty \int_0^1 y^2 \left\{\tfrac{1}{y+k}\right\}\, \frac{dy}{(y+k+1)^2} \\ & = \; \int_0^1 \frac{y^2}{(y+1)^2}\left\{\tfrac{1}{y}\right\}\,dy + \sum_{k=1}^\infty \int_0^1 \frac{y^2}{(y+k)(y+k+1)^2}\,dy \\ & = \; \int_1^\infty \frac{\left\{z\right\}}{(z+1)^2}\,\frac{dz}{z^2} + \sum_{k=1}^\infty \int_0^1 \frac{y^2}{(y+k)(y+k+1)^2}\,dy \\ & = \; \sum_{k=1}^\infty \int_0^1 \frac{z}{(z+k)^2(z+k+1)^2}\,dz + \sum_{k=1}^\infty \int_0^1 \frac{y^2}{(y+k)(y+k+1)^2}\,dy \\ & = \; \sum_{k=1}^\infty \int_0^1 \frac{y(y^2+ky+1)}{(y+k)^2(y+k+1)^2}\,dy \end{aligned}$ By symmetry, the desired integral is $\int_0^1 \left\{\tfrac{1}{x}\right\}\left\{\tfrac{1}{1-x}\right\}\left(\left\{\tfrac{1}{x}\right\} + \left\{\tfrac{1}{1-x}\right\}\right)\,dx \; = \; 2\int_0^1 \left\{\tfrac{1}{x}\right\}^2\left\{\tfrac{1}{1-x}\right\}\,dx \; = \; 2\lim_{K \to \infty}Z_K$ where $\begin{aligned} Z_K & = \; \sum_{k=1}^K \int_0^1 \frac{y(y^2+ky+1)}{(y+k)^2(y+k+1)^2}\,dy \\ & = \; \sum_{k=1}^K \left(-k(k+2)\ln(k+2) + (2k^2+4k+1)\ln(k+1) - (k+1)^2\ln k - \frac{k+2}{k+1}\right) \\ & = \; -K(K+2)\ln(K+2) + (K^2+4K+2)\ln(K+1) - 2\ln K! - K - H_{K+1} + 1 \end{aligned}$ Now $\lim_{K \to \infty}S_K = \gamma$ , where $S_K = H_K - \ln K$ , and Stirling's approximation tells us that $X_K = \mathrm{O}(K^{-1})$ as $K \to \infty$ , where $X_K \; = \; 2\ln K! - (2K+1)\ln(K+1) + 2K+2 - \ln(2\pi)$ Now $Z_K \; = \; -K(K+2)\ln(K+2)+ K(K+2)\ln(K+1) + K + 3 - X_K - S_{K+1} - \ln(2\pi) \; = \; F(K) + 3 - X_K - S_{K+1} - \ln(2\pi)$ where $\begin{aligned} F(K) & = \; K - K(K+2)\ln\left(\tfrac{K+2}{K+1}\right) \; = \; \ln\left(\tfrac{K+2}{K+1}\right) + K - (K+1)^2\ln\left(1 + \tfrac{1}{K+1}\right) \\ & = \; \mathrm{O}(K^{-1}) + K - (K+1)^2\left(\tfrac{1}{K+1} - \tfrac{1}{2(K+1)^2} + \mathrm{O}(K^{-3})\right) \; =\; -\tfrac12 + \mathrm{O}(K^{-1}) \end{aligned}$ as $K \to \infty$ , and hence $\lim_{K \to \infty}F(K) = -\tfrac12$ . Thus $\lim_{K \to \infty}Z_K \; = \; \tfrac52 - \gamma - \ln(2\pi)$ and so the desired integral is $5 - 2\gamma - 2\ln(2\pi)$ , making the answer $5 + 2 + 2 +2 = \boxed{11}$ .