# 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.

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

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)

• Thank you so much! That is awesome. Just to confirm the $G(0)\text{log}(1)=0$ because $\text{log}(1)=0$, right? Nov 25 at 11:54
• Precisely. Since G(0) = 1, and thus finite, G(0)log(1) is zero. Nov 25 at 12:05
• Thanks once more. Nov 25 at 12:06
• 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. Nov 26 at 9:38
• 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) $$[ \log_2(1+x)A(x) ]_0^ \infty$$ Nov 26 at 9:41