1
$\begingroup$

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 $\phi$? $0<\varphi<1$?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.