How to perform integration by parts when the upper integral limit is infinity?

Define $ G(x) $ with derivative $g(x)$ such that (assume $\zeta\phi\rho=b$)

$$ G(x)= \begin{array}{cc} \left\{ \begin{array}{cc} 0 & x< 0 \\ 1-F(x) &x\ge0 \end{array} \right. \end{array} $$

$$ \implies g(x)= -f(x) $$ Now $\log_2(1+x)=\log_2(e)\cdot\log_e(1+x)$

Taking away $ \frac{-\log_2(e)}{2} $ the integral reduces to $$ I = \int_0^\infty \log (1+x) g(x) \,dx $$ We evaluate the improper integral, by taking the limit of the definite integral $$ I = \lim \limits_{y\to\infty} \left[ \int_0^y \log (1+x) g(x) \,dx \right] $$

Simplifying by parts

[EDIT: This is where the utility of introducing an auxiliary function is most useful. Due to the definition of $G(x)$, the indefinite integral of its derivative $g(x)$ is identical to it. Hence the integral evaluation by parts is elegant.]

$$ I = \lim \limits_{y\to\infty} \left[ G(y)\log(1+y)-G(0)log(1) - \int_0^y \frac{G(x)}{1+x} \,dx \right] $$ $$ I = \lim \limits_{y\to\infty} G(y)\log(1+y) - \int_0^\infty \frac{G(x)}{1+x} \,dx $$ By application of L'hospital rule, we can evaluate the limit to be zero.

Hence, $I$ reduces to $$ I =-\int_0^\infty \frac{G(x)}{1+x} \,dx = -\int_0^\infty \frac{e^{-\frac{x}{b}}}{(1+x)(1+ax)} \,dx $$ $$ = \frac{1}{a-1} \int_0^\infty \frac{(1-a)e^{-\frac{x}{b}}}{(1+x)(1+ax)} \,dx = \frac{1}{a-1} \int_0^\infty \left(\frac{1}{1+x}-\frac{a}{1+ax}\right) e^{-\frac{x}{b}} \,dx $$ Multiplying with $ \frac{-\log_2(e)}{2} $, we get result (2)