Who's up to the challenge? 57

Consider the integral $\displaystyle J(a) = \int_{0}^{\infty} e^{-2x}x^{a}dx$ , which implies $\displaystyle J(a) = \frac{\Gamma(a+1)}{2^{a+1}}$

We want to evaluate $\displaystyle J''(0)$ , $\displaystyle J''(a) = \frac{1}{2^{a+1}}[\Gamma''(a+1)-2\Gamma'(a+1)ln2+\Gamma(a+1)(ln2)^2]$

Using $\displaystyle \Gamma'(1)=-\gamma,\Gamma''(1)=\gamma^2+\frac{\pi^2}{6}$ we obtain ,

$\displaystyle J''(0) = \frac{\pi^2}{12}+\frac{\gamma^2}{2}+\gamma ln2 + \frac{(ln2)^2}{2}$ , Thus $\boxed{A+B+C+D+E+F+G+H+I=28}$

The integral is equal to $\tfrac12\int_0^\infty (\ln u - \ln 2)^2e^{-u}\,du \; = \; \tfrac12\big[\Gamma''[1] - 2\ln 2 \Gamma'[1] + (\ln 2)^2\big]$ which is equal to $\tfrac1{12}\pi^2 + \tfrac12\gamma^2 + \gamma \ln 2 + \tfrac12(\ln2)^2$ This makes the answer $\boxed{28}$ .