Boundedness of $\int_a^b e^{iF(x)} dx$ for $F''(x) \ge r > 0$

Question

My question is regarding the last sentence of Lemma 4.4. of Titchmarsh's book The Theory of the Riemann Zeta-Function:

First I approached by taking three intervals as the text has done but for new $\delta$: $(a,2c-\delta)$, $(2c-\delta,\delta)$ and $(\delta, b)$ but after replacing $\delta = 2 r^{-\frac12}$ I could not get the bound $8 r^{-\frac12}$. Then I took two intervals $(a,c)$ and $(c,b)$ and by use of WolframAlpha I found out that in order to get the mentioned upper bound, $c>b$ must hold which is impossible. So how to prove Eq. 4.4.1 for $c<a+\delta$ or $c>b-\delta$?

Answer 1

Let $a\leq c<a+\delta$. You take $I_2=\int_a^{c+\delta}$ and $I_3=\int_{c+\delta}^b$. Now we still have $\vert I_2\vert\leq2\delta$ and on $I_3$ we still have $(x-c)\geq\delta$. (No $I_1$ here, just named them so they match with the other case.)

If $F'$ doesn't vanish and is positive (negative is handled similarly) take $I_2=\int_a^{a+\delta}$ and $I_3=\int_{a+\delta}^b$ so $\vert I_2\vert\leq\delta$ and on $I_3$ we have $$F'(x)-F'(a)=\int_a^xF''(t)dt\geq r(x-a)\geq r\delta$$ so $$F'(x)\geq r\delta+F'(a)$$ where $F'(a)>0$ so $$\vert I_3\vert\leq 4/(r\delta+F'(a))\leq4/(r\delta).$$