# Sign of a function being integrated

I'm dealing with the following problem and I would appreciate any suggestions.

Let $$F, G:[1,\infty)\to\mathbb{R}$$ piecewise constant on every interval of the form $$[N,N+1)$$, with $$N$$ positive integer and $$F(s)=F(N)$$ and $$G(s)=G(N)$$, for all $$s\in[N,N+1)$$.

Suppose that $$G(s)\leq 0$$ for all $$s\geq 1$$, $$\lim_{s\to\infty}G(s)=-1$$ and that $$\int_{1}^{x}\left(x^{1/2}F(s)s^{-3/2}-G(s)s^{-1/2}\right)\,ds=0,$$ for all $$x>1$$.

I want to show that $$F(s)\leq 0$$, for all sufficiently large $$s$$.

My idea: for each positive integer $$N$$ we have that $$\int_{1}^{N+1}\left((N+1)^{1/2}F(s)s^{-3/2}-G(s)s^{-1/2}\right)\,ds=0,$$ and $$\int_{1}^{N}\left(N^{1/2}F(s)s^{-3/2}-G(s)s^{-1/2}\right)\,ds=0.$$ Therefore, from this two previous identities we obtain $$\int_{1}^{N}((N+1)^{1/2}-N^{1/2})F(s)s^{-3/2}-G(s)s^{-1/2})\,ds+\int_{N}^{N+1}((N+1)^{1/2}F(s)s^{-3/2}-G(s)s^{-1/2})\,ds=0.$$ Since $$\lim_{N\to\infty}(N+1)^{1/2}-N^{1/2}=0$$ and since $$F$$ and $$G$$ are piecewise constant on each $$[N,N+1)$$ we have that

$$\lim_{N\to\infty}\,\left( -\int_{1}^{N}G(s)s^{-1/2}\,ds+(N+1)^{1/2}F(N)\int_{N}^{N+1}s^{-3/2}-G(N)\int_{N}^{N+1}s^{-1/2}\,ds\right)=0.$$ That is, $$\lim_{N\to\infty}\,\left( -\int_{1}^{N}G(s)s^{-1/2}\,ds+2(N+1)^{1/2}F(N)(N^{-1/2}-(N+1)^{-1/2}) -G(N)((N+1)^{1/2}-N^{1/2})\right)=0.$$ So, since $$(N+1)^{1/2}F(N)(N^{-1/2}-(N+1)^{-1/2})\sim \frac{1}{2}N^{-1} F(N),\,\,N\to\infty,$$ and $$\lim_{N\to\infty}G(N)((N+1)^{1/2}-N^{1/2})=0,$$ because $$\lim_{N\to\infty}G(N)=-1$$, we may conclude that $$\lim_{N\to\infty}\left(-\int_{1}^{N}G(s)s^{-1/2}\,ds+\frac{1}{2N}F(N) \right)=0.\,\,\,\,\,\,(1)$$ Again, since $$G(s)\leq 0$$ and $$\lim_{N\to\infty}G(N)=-1$$, we have that $$\int_{1}^{N}G(s)s^{-1/2}\,ds\leq 0$$ for all $$N>1$$ and therefore we see from (1) that $$F(N)\leq 0$$ for all $$N$$ sufficiently large.

• Do you mean "on every interval of the form $[N,N+1)$?" Otherwise they're just constant. – Dzoooks Jun 4 at 23:34
• Yes, I do. My original question was edited (by someone else) but I re-edited and now it is in its original form. – GoldSoundz Jun 5 at 0:43