I was reading an analysis paper and now I have a question about indefinite integrals. Let be $x_0\geq 0$ and $f,g:[x_0,+\infty)\to\Bbb{R}$ differentiable functions such that:

  1. There is $x_1\in(x_0,+\infty)$ with $f(x_1)=0$ such that $f<0$ in $[x_0,x_1)$ and $f>0$ in $[x_1,+\infty)$.
  2. $g(x_0)=0$ and $g>0$ in $(x_0,+\infty)$,

Question 1: Is $h(x)=\int_{x_0}^x\frac{f(s)}{g(s)}ds$ well defined in $(x_0,+\infty)$? Or it is necessary some conditions about $f$ and $g$? Is correct to say that $\lim_{x\to x_0} h(x)=0$?

Question 2: Is correct to say that $h$ is strictly decreasing in $(x_0,x_1)$ and stricly increasing $(x_1,+\infty)$?

My point about it is that $g$ is zero at $x_0$, so it is weird to consider $h$ like above.

Thanks any help!

  1. In general, that integral diverges. That will be the case if, say, $x_0=0$, $f(x)=x-1$ and $g(x)=x$. So, yes, some extra hypothesis is required here.
  2. Since (assuming that the definition of $h$ makes sense) $h'(x)=\dfrac{f(x)}{g(x)}$, and since this quotient is smaller than $0$ on $[x_0,x_1)$ and greater than $0$ on $(x_1,\infty)$, you are right.
  • $\begingroup$ Thanks for your answer! But in the first question, $h$ like above make sense? For the convergence part, if for example $L≤f/g$ for some constant L, we get that the $\lim h$ is finite, so can we infer that $h(x_0)=0$? Thanks again. $\endgroup$ – Irddo Jul 19 at 10:10
  • 1
    $\begingroup$ As my example shows, your function $h$ may well not make sense. $\endgroup$ – José Carlos Santos Jul 19 at 10:12

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