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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.

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  • $\begingroup$ $\int_c^{\infty} f(x)dx = \lim_{m\rightarrow \infty} f(x) dx$. If applying this limit gives you $\infty$, then you just substitute that in and accept that your integral diverges. $\endgroup$
    – Prometheus
    Nov 23 at 7:51
  • $\begingroup$ @Prometheus But if that is the case, then how did the authors arrive at (2)? I have edited the concerned text for more clarity. $\endgroup$
    – nashynash
    Nov 23 at 7:57
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    $\begingroup$ For legibilty, avoid composite constants such as $\zeta \phi \rho$. $\endgroup$
    – user995027
    Nov 25 at 10:40
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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)

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    $\begingroup$ Thank you so much! That is awesome. Just to confirm the $G(0)\text{log}(1)=0$ because $\text{log}(1)=0$, right? $\endgroup$
    – nashynash
    Nov 25 at 11:54
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    $\begingroup$ Precisely. Since G(0) = 1, and thus finite, G(0)log(1) is zero. $\endgroup$ Nov 25 at 12:05
  • $\begingroup$ Thanks once more. $\endgroup$
    – nashynash
    Nov 25 at 12:06
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    $\begingroup$ Thanks for the comment. I have edited my answer to clarify. But the truth is, there is no need of $G(x)$. You could substitute the expression directly and evaluate. Abstracting it away using the right function makes it neat, that's all. $\endgroup$ Nov 26 at 9:38
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    $\begingroup$ In fact, that's the mistake you made in evaluating the integral. In step (3), the first integral is (assuming A(x) is the antiderivative) $$[ \log_2(1+x)A(x) ]_0^ \infty $$ $\endgroup$ Nov 26 at 9:41

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