Who's up to the challenge? 48

Since $H_x \; = \; x\sum_{k=1}^\infty \frac{1}{k(k+x)} \qquad \qquad x > 0$ we have $xH_x \; = \; \lim_{K\to\infty}x \sum_{k=1}^K \frac{x}{k(k+x)} \; = \; \lim_{K \to \infty}x \sum_{k=1}^K \left(\frac{1}{k} - \frac{1}{k+x}\right)$ and so, since $\begin{array}{rcl} \displaystyle \int_0^1 x\sum_{k=1}^K\left(\frac{1}{k} - \frac{1}{x+k}\right)\,dx & = & \displaystyle \int_0^1 \left(xH_K - \sum_{k=1}^K \frac{x}{x+k}\right)\,dx \\ & = & \displaystyle\int_0^1 \left(xH_K - K + \sum_{k=1}^K \frac{k}{k+x}\right)\,dx \\ & = & \displaystyle\tfrac12H_K - K + \sum_{k=1}^K k\Big[\ln(k+x)\Big]_0^1 \\ & = & \displaystyle \tfrac12H_K - K + \sum_{k=1}^K k \ln(k+1) - \sum_{k=1}^K k \ln k \\ & = & \displaystyle \tfrac12H_K - K + \sum_{k=2}^{K+1}(k-1)\ln k - \sum_{k=1}^K k \ln k \\ & = & \displaystyle \tfrac12H_K - K + K\ln(K+1) - \sum_{k=2}^K \ln k \\ & = & \displaystyle \tfrac12\big[H_K - \ln K\big] - K + K\ln\big(1+\tfrac{1}{K}\big) + (K+\tfrac12)\ln K- \ln K! \\ & = & \displaystyle \tfrac12\big[H_K - \ln K\big] + \ln\big(1 + \tfrac{1}{K}\big)^K - \ln \left(\frac{K! \times e^K}{K^{K+\frac12}}\right) \end{array}$ we deduce (using the Monotone Convergence Theorem) that $\int_0^1 x H_x\,dx \; = \; \lim_{K \to \infty}\Big\{\tfrac12\big[H_K - \ln K\big] + \ln\big(1 + \tfrac{1}{K}\big)^K - \ln \left(\frac{K! \times e^K}{K^{K+\frac12}}\right)\Big\} \; = \; \tfrac12\gamma + 1 - \ln\sqrt{2\pi}$ making the answer $1 + 2 + 2 = \boxed{5}$ .

$\begin{aligned} &\int_0^1 x H_x dx &&= \int_0^1 x(\psi(x+1) + \gamma)dx && {\color{#D61F06} H_x = \psi(x+1) + \gamma} \\ &&&= \int_0^1 (x\psi(x+1) + \gamma \cdot x) dx\\ &&&= \int_0^1 ({\color{#3D99F6}x\psi(x)} + 1 + \gamma \cdot x) dx && {\color{#D61F06} \psi(x+1)=\psi(x) + \frac{1}{x} } \\ &&& {\color{#3D99F6}\begin{matrix} u=x & dv=\psi(x) \\ du= dx & v = \ln(\Gamma(x)) \end{matrix}} \\ & \color{#3D99F6} \int_0^1 x\psi(x) && \color{#3D99F6} = \left . x \ln(\Gamma(x)) \right |_0^1-\int_0^1 \ln(\Gamma(x)) dx && \color{#D61F06} \left . x \ln(\Gamma(x)) \right |_0^1 = 0\\ &&& \color{#3D99F6} = -\frac{1}{2}\int_0^1 \ln(\Gamma(x))+\ln(\Gamma(1-x)) dx && \color{#D61F06} \int_a^b f(x)dx=\int_a^b f(a+b-x)dx \\ &&& \color{#3D99F6} = -\frac{1}{2}\int_0^1 \ln(\Gamma(x) \Gamma(1-x)) dx \\ &&& \color{#3D99F6} = -\frac{1}{2}\int_0^1 \ln(\frac{\pi}{\sin(\pi x)}) dx && \color{#D61F06} \text{Euler's reflection formula}\\ &&& \color{#3D99F6} = -\frac{1}{2}\left (\int_0^1 \ln(\pi)dx - \int_0^1 \ln(\sin(\pi x)) dx \right ) \\ &&& \color{#3D99F6} = -\frac{1}{2}\left (\ln(\pi) - {\color{#D61F06} (-\ln(2))} \right )= -\ln(\sqrt{2 \pi}) && \color{#D61F06} let \ u=\pi x \text{ and use symmetry}\\ &&& \color{#20A900} \text{back to original integral} \\ &\int_0^1 x H_x dx && = {\color{#3D99F6} -\ln(\sqrt{2 \pi})} + 1 + \frac{\gamma}{2} \\ \end{aligned}$