Hot Integral - 16

If we define $F_n(u) \; = \; \int_0^\infty x^u J_n(x)\,dx$ then standard integration tables tell us that $F_n(u) \; =\; 2^u \frac{\Gamma\big(\tfrac12(1+n+u)\big)}{\Gamma\big(\tfrac12(1+n-u)\big)}\;.$ If we define $G_\alpha(u) \; = \; \frac{\Gamma(\alpha+u)}{\Gamma(\alpha-u)} \;,$ then $\begin{array}{rcl} G_\alpha'(u) & = & G_\alpha(u)\left[ \psi(\alpha+u) + \psi(\alpha-u)\right] \\ G_\alpha''(u) & = & G_\alpha(u)\left[\psi(\alpha+u) + \psi(\alpha-u)\right]^2 + G_\alpha(u)\left[\psi'(\alpha+u) - \psi'(\alpha-u)\right] \\ G_\alpha'''(u) & = & G_\alpha(u)\left[\psi(\alpha+u) + \psi(\alpha-u)\right]^3 + 3G_\alpha(u)\left[\psi(\alpha+u)+\psi(\alpha-u)\right]\left[\psi'(\alpha+u) - \psi'(\alpha-u)\right] \\ & & {}+ G_\alpha(u)\left[\psi''(\alpha+u) + \psi''(\alpha-u)\right] \end{array}$ and so $\begin{array}{rclcrcl} G_\alpha(0) & = & 1 & \qquad \qquad & G_\alpha'(0) & = & 2\psi(\alpha) \\ G_\alpha''(0) & = & 4\psi(\alpha)^2 & \qquad \qquad & G_\alpha'''(0) & = & 8\psi(\alpha)^3 + 2\psi''(\alpha) \end{array}$ Thus, with $\alpha = \tfrac12(1+n)$ , we have $F_n(u) \,=\, 2^uG_\alpha(\tfrac12u)$ and hence $\begin{array}{rcl} \displaystyle\int_0^\infty (\ln x)^3 J_n(x)\,dx & = & F_n'''(0) \; = \; (\ln2)^3 G_\alpha(0) + \tfrac32(\ln2)^2G_\alpha'(0) + \tfrac34(\ln2)^2G_\alpha''(0) + \tfrac18G_\alpha'''(0) \\ & = & (\ln 2)^3 + (\ln 2)^2 \psi(\alpha) + 3(\ln 2)\psi(\alpha)^2 + \psi(\alpha)^3 + \tfrac14\psi''(\alpha) \\ & = & \left(\psi^{(0)}(\tfrac12(n+1)\big) + \ln 2\right)^3 + \tfrac14\psi^{(2)}(\tfrac12(n+1)\big) \;, \end{array}$ and so the answer is $0+1+2+2+3+1+4+2+1+2 \,=\, 18$ .