Ratio with exponential integral function

Question

Let $$F(r)=1+\int_{0}^{r}{\frac{se^{-1/s}}{V(s)}}ds$$ where $V(s)$ is a positive function. I want to show that $$F(r)\approx 1+\left(\int_{1}^{r}{\frac{s}{V(s)}}ds\right)_{+},$$ where $(.)_{+}=max\{0, (.)\}$ and $\approx$ means that the ratio is bounded by above an below by a constant respectively. Can someone help?

Answer 1

Let $G(r):=1+\left(\int_1^r \frac{s}{V(s)}\, \mathrm{d}s\right)_+$. First, suppose $r\geq 1.$ Then since the exponential is bounded by $1,$ $F(r)\leq G(r)+\int_0^1 \frac{se^{-\frac{1}{s}}}{V(s)}\,\mathrm{d}s$. This means $\frac{F(r)}{G(r)}\leq 1+\frac{\int_0^1 \frac{se^{-\frac{1}{s}}}{V(s)}\,\mathrm{d}s}{G(r)}$. Since $G(r)\geq 1,$ we have that $$ \frac{F(r)}{G(r)}\leq 1+\int_0^1 \frac{se^{-\frac{1}{s}}}{V(s)}\,\mathrm{d}s=F(1). $$ Since everything is positive, this also means $$ \frac{G(r)}{F(r)}\geq \frac{1}{1+\int_0^1 \frac{se^{-\frac{1}{s}}}{V(s)}\,\mathrm{d}s}=\frac{1}{F(1)}. $$

Now, suppose $r<1.$ Then $G(r)\equiv 1$ since $$ \int_1^r \frac{s}{V(s)}\, \mathrm{d}s=-\int_r^1 \frac{s}{V(s)}\, \mathrm{d}s<0. $$ It is then trivial to see that $\frac{F(r)}{G(r)}=F(r),$ which is increasing so $\frac{F(r)}{G(r)}\leq F(1)$ and obviously, $$ \frac{G(r)}{F(r)}\geq \frac{1}{F(1)}. $$

EDIT: An easy, one line proof of the claim that you proposed is that $G(r)\geq 1$ so $\frac{F(r)}{G(r)}\leq F(r)< F(\infty)$ and $\frac{G(r)}{F(r)}\geq \frac{1}{F(r)}> \frac{1}{F(\infty)}$. The advantage of my solution above is that $F(1)$ is a better bound.