Is $\int_t^{t+s}f(y)\mathrm{d}y=sf(t)(1+O(1))$?

Question

I am now reading a book about the introduction of stochastic processes, and there is an equation like this

\begin{eqnarray*} \int_t^{t+s}f(y)\mathrm{d}y=sf(t)(1+O(1))\rightarrow 0 \end{eqnarray*} when $s\rightarrow 0$ ($f$ is continuous). My question is, how could the integral become $sf(t)(1+O(1))$? I thought that it was a result of Taylor Expansion of $f$ around $y=t$. However, if it was, then instead of $sf(t)O(1)$, it should be $f'(t)O(s)$.

Thank you so much in advance for your comments.

Answer 1

$$\begin{align}\int_t^{t+s} f(y) dy&=\int_t^{t+s}\left[f(t)+(y-t)f'(t)+O((y-t)^2)\right]dy\\&=\left[yf(t)+\frac{(y-t)^2}2f'(t)+O((y-t)^3)\right]_{t}^{t+s}\\&=sf(t)+\frac {s^2}2f'(t)+O(s^3)\end{align}$$ So we get the $sf(t)$ term as desired. I am not sure why it is written as $(1+O(1))$, since $1$ is $O(1)$. I suspect that is a typo and they meant $(1+O(s))$.

Answer 2

Proof without assuming differntiability: $\frac 1 s \int_t^{t+s} f(y)dy \to f(t)$ by continuity of $f$ at t. [Write $\frac 1 s \int_t^{t+s} f(y)dy - f(t)=\frac 1 s \int_t^{t+s} \{f(y)-f(t)\}dy $ to see why this is true]. Now just divide by $f(t)$.