Hello I have an very elementary calculus problem.

Let $\phi(\eta)$ be a real value function satisfying \begin{equation} \phi(-\infty)=1,\quad \phi(+\infty)=0, \end{equation} Let $g$ be a positive function satisfying \begin{equation} g(0)=g(1)=0\quad\text{and}\quad g(u)>0\,\,\forall\,\,u\in (0,1). \end{equation} Now I want to know, where could the limit of the following integral lies in? \begin{equation} \lim_{\eta\to -\infty}\int_{\eta}^x g(\phi(s))\,ds \end{equation}

I was given an answer that the limit exist, and yes I can see this because I think $\frac{d}{d\eta}\int_{\eta}^x g(\phi(s))\,ds=-g(\eta)$ is telling me the above integral is a decreasing function on the half line $(-\infty, x)$ and so a the integral is a monotone and such function has a limit.

However I am not sure where exactly could the limit lies in? And I was told the answer, the limit can either be non-negative real value or $+\infty$. I am lost... what does $+\infty$ correspond to? and non-negative real value correspond to?

PS. Now if I let $x\to\infty$, can I claim the limit of this integral is strictly greater than 0? If not, what do I need to assume for the function $\phi$? $0<\varphi<1$?

  • $\begingroup$ Are we to assume $g(x)=0,\forall x \notin (0,1)$? $\endgroup$ – user76844 Jan 23 '15 at 1:56
  • $\begingroup$ no, there is nothing more than what is stated there for $g$. $\endgroup$ – math101 Jan 23 '15 at 2:07
  • $\begingroup$ Nevermind...I just realized that the argument to the integrand never goes beyond $(0,1)$ anyway...:=P $\endgroup$ – user76844 Jan 23 '15 at 2:25

As an example, let $g(x):= \frac{1}{x^2}-1$, in which case the integral could diverge to $+\infty$ (depending on how fast $\phi(x)$ moves away from zero and the value of $x$ in the limits.

In general,since $g(x)>0$ on $(0,1)$, and $\phi(x)\in (0,1),\;\forall x$ the integral must take only positive values. Thus, it must be either positive finite, or unbounded from above.

  • $\begingroup$ I am looking for a way that does not require me to guess what function could $g$ take, except the assumption given. Also, could you please edit your answer from "In general..." not quite clear what you mean... $\endgroup$ – math101 Jan 23 '15 at 3:21
  • $\begingroup$ @math101 my example was just that...I wanted to show how you can get $+\infty$. My second paragraph shows why in general, the integral will be $>0$ $\endgroup$ – user76844 Jan 23 '15 at 3:45
  • $\begingroup$ OK. "In general,..." is this a well known fact, to be honest , I've never heard of it til now from first year, what reference could help? $\endgroup$ – math101 Jan 23 '15 at 3:54
  • $\begingroup$ @math101 no references. How would the integral of a strictly positive integrand yield anything other than a value $>0$? $\endgroup$ – user76844 Jan 23 '15 at 4:02
  • $\begingroup$ back to the idea of integration: calculating the area under the curve. Well if the curve never lies below 0, then the area should alway be positive...thanks $\endgroup$ – math101 Jan 23 '15 at 4:11

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