Oscillatory integrals with a decaying factor in the integrand Lemma 4.5 of Titchmarsh's book The Theory of the Riemann Zeta function says (slightly rephrased):
Let $F$ be a twice differentiable real function such that $ F''(x) \geq r > 0$ for all $x$ in $[a,b]$ or $ F''(x) \leq -r < 0$ for all $x$ in $[a,b]$. Let $G$ be a real function such that $G(x)/F'(x)$ is monotonic and $|G(x)| \leq M$ for all $x$ in $[a,b]$. Then we have
$$ \left| \int_a^b G(x) e^{iF(x)} dx \right| \leq \frac{8M}{\sqrt{r}} .$$
I am considering the case
$$ I_\alpha(T) = \left| \int_1^T x^\alpha e^{iF(x)} dx \right| $$
where $ F''(x) \geq 1/T$ and $x^\alpha/F'(x)$ is monotonic for all $x$ in $[1, T]$ and for all $\alpha > -1$.
Titchmarsh's Lemma (for $\alpha > -1/2$) and the trivial bound (for $-1/2 \geq \alpha > -1$) give
$$ I_\alpha(T) \leq \begin{cases} 8T^{\alpha + 1/2} \ &\text{if} \ \alpha \geq 0 \\ 8T^{1/2} \ &\text{if} \ 0 >\alpha > -\frac{1}{2} \\ \frac{T^{\alpha + 1}}{\alpha+1} &\text{if} -\frac{1}{2} \geq \alpha > -1 \end{cases} $$
I am wondering if one can do better (asymptotically in $T$) in the cases when $0 > \alpha > -1$ by combining the oscillatory cancellation from $e^{iF(x)}$ with the decay from $x^{\alpha}$.
 A: $\newcommand\R{\mathbb R}\newcommand{\al}{\alpha}$If a function $h\colon\R\to\R$ is monotonic and does not change sign on an interval $[a,b]$, then it is easy to see (make a picture) that for all real $s$
\begin{equation*}
    \Big|\int_{[a,b]}h(u)\sin(u+s)\,du\Big|\le\sup_{c\in\R}\int_{[a,b]\cap[c,c+\pi]}|h(u)|\,du,
\end{equation*}
whence
\begin{equation*}
    \Big|\int_{[a,b]}h(u)e^{iu}\,du\Big|\le\sqrt2\,\sup_{c\in\R}\int_{[a,b]\cap[c,c+\pi]}|h(u)|\,du. 
\end{equation*}
Since $x^\al/F'(x)$ is monotonic in $x\in[1,T]$, $F'$ cannot change its (nonzero) sign on $[1,T]$. Making now the substitution $u=F(x)$, letting here
\begin{equation*}
    h(u):=\frac{g(u)^\al}{F'(g(u))} \text{ with } g:=F^{-1},  
\end{equation*}
and then making the inverse substitution $x=g(u)$, we get
\begin{equation*}
\begin{aligned}
    I_\al(T)=
    &\Big|\int_{F([1,T])}h(u)e^{iu}\,du\Big| \\
    &\le\sqrt2\,\sup_{c\in\R}\int_{F([1,T])\cap[c,c+\pi]}|h(u)|\,du \\ 
    &=\sqrt2\,\sup_{c\in\R}\int_{[1,T]\cap g([c,c+\pi])}x^\al\,dx. 
\end{aligned}
\tag{1}
\end{equation*}
Recall that $F'$ cannot change its (nonzero) sign on $[1,T]$.
If $F'>0$ on $[1,T]$, then for any $x,y$ in $[1,T]$ such that $x\le y$
\begin{equation*}
    |F(y)-F(x)|=F(y)-F(x)\ge F'(x)(y-x)+\frac{(y-x)^2}{2T}\ge\frac{(y-x)^2}{2T}, 
\end{equation*}
since $F'>0$ and $F''\ge1/T$.
Similarly, if $F'<0$ on $[1,T]$, then for any $x,y$ in $[1,T]$ such that $x\le y$
\begin{equation*}
    |F(y)-F(x)|=F(x)-F(y)\ge F'(y)(x-y)+\frac{(x-y)^2}{2T}\ge\frac{(y-x)^2}{2T}, 
\end{equation*}
since $F'<0$ and $F''\ge1/T$.
So, in either case, $|y-x|\le\sqrt{2T|F(y)-F(x)|}$ for any $x,y$ in $[1,T]$. So, denoting by $l$ the length of the interval $[1,T]\cap g([c,c+\pi])$, we have $l\le\sqrt{2\pi T}$.
Hence, by (1),
\begin{equation*}
    I_\al(T)
    \le\sqrt2\,\int_0^{\sqrt{2\pi T}}x^\al\,dx
\le\frac C{1+\al}\, T^{(1+\al)/2},
\end{equation*}
whence $C$ is a universal positive real constant.
The latter bound is indeed an improvement of the corresponding bound in your post.
Morever, the latter bound is optimal, as it is asymptotically attained (as $T\to\infty$, up to a positive real constant factor depending only on $\al$) if $F(x)=\dfrac{x^2}{2T}$:

