# Indefinite integral of $f/g$ with $g(x_0)=0$

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.
• 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. – Irddo Jul 19 at 10:10
• As my example shows, your function $h$ may well not make sense. – José Carlos Santos Jul 19 at 10:12