How to perform integration by parts when the upper integral limit is infinity? I came across this definite integral in a journal article:
\begin{align*}
&{\mathcal{ E}} = \int _{0}^{+\infty } \frac {1}{2} \log _{2} (1 +x) f(x) dx, \tag{1}
\end{align*}
where, $f(x)=\frac{d}{dx}F(x)$ is the PDF associated with $F(x)=1-\frac{1}{1+ax} e^{-\frac{x}{\zeta \phi \rho}}$, $a=\frac{\phi_2\rho_2}{\zeta \phi \rho}$.
The authors show that integral can be reduced to
\begin{align*}
&{\mathcal{ E}} = \frac {\log _{2}(e)}{2(1-a) } \int _{0}^{+\infty }\left ({\frac {1}{1+x} - \frac {a}{1+ a x} }\right) e^{-\frac {x}{\zeta \phi \rho}} dx \tag{2}
\end{align*}
I understand that $f(x)=\frac{d}{dx}\left[1-\frac{1}{1+ax} e^{-\frac{x}{\zeta \phi \rho}}\right] = \dfrac{\mathrm{e}^{-\frac{x}{\phi{\rho}{\zeta}}}}{\phi{\rho}{\zeta}}+a$.
In my previous post, which I have now deleted to avoid duplication, it was pointed out to be an integration by parts problem.
So, here is what I have done.
For simplicity, I will keep the $\frac{1}{2}$ out of the integral.
\begin{align*}
\int _{0}^{+\infty }&  \log _{2} (1 +x) f(x) dx \\&= \log _{2} (1 +x) \int _{0}^{+\infty } f(x) dx -\int _{0}^{+\infty } \frac {d} {dx} \log _{2} (1 +x) \left( \int _{0}^{+\infty } f(x) dx \right)dx \tag{3}
\end{align*}
Subsituting $\frac {d} {dx} \log _{2} (1 +x) = \frac{1}{\text{ln}(2)(1+x)}$ in (3), we obtain
\begin{align*}
= \log _{2} (1 +x) \int _{0}^{+\infty } f(x) dx -\int _{0}^{+\infty } \frac{1}{\text{ln}(2)(1+x)} \left( \int _{0}^{+\infty } f(x) dx \right)dx \tag{4}
\end{align*}
Now, I am stuck here as I think the integral terms would become $\infty$ after substituting $f(x)= \dfrac{\mathrm{e}^{-\frac{x}{\phi{\rho}{\zeta}}}}{\phi{\rho}{\zeta}}+a$!
How do I circumvent this problem? I mean how did the authors arrive at (2) under this condition? Please let me know if I am mistaken.
Any help or tips would be appreciated.
 A: 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)
