For $r \in (0, 2)$, I would like to estimate the decay of $f(k) = \int_0^1 (1 - x^r) \cos(kx) dx$ as $k \to \infty$. Indeed, according to Riemann–Lebesgue lemma, $\lim_{k \to \infty} f(k) = 0$.
For $r = 1$, one can compute explicitely $f$ and obtain that $|f(k)| \leq k^{-2}$ for $k \geq 1$. For $r \in (0, 1)$, performing an integration by part, I have proved that $|f(k)| \leq 1/k^r $, for $k \geq 1$. Using also integration by part, for $r \in (1, 2)$, I was only able to prove that $|f(k)| \leq r/k$, for $k \geq 1$.
However, regarding the case $r = 1$, I think that the bounds for $r \neq 1$ are not tight at all.
My question is: can we prove that $|f(k)| \leq C/k^{1+r}$ for $k \geq1$ and where $C > 0$ is a constant ?
Edit : I have an argument that shows that the bounds that I obtain are not tight. By integration by part: $f(k) = \frac{r}k \int_0^1 x^{r-1} \sin(kx) dx$. According to Riemann–Lebesgue lemma the integral goes to $0$ as $k \to \infty$. Thus $f(k) = o(1/k)$.
So the problem is equivalent to $I(k) = \int_0^1 x^{r-1} \sin(kx) dx = O(1/k^r)$. By change of variable $I(k) = k^{-r} \int_0^k y^{r-1} \sin(y) dy$. I do know how to handle this improper integral.
The integral can be simplified as following: $$\begin{align*} f(k) &= \int_0^{1} \cos{(kx)} dx -\int_{0}^{1} x^r \cos{(kx)}dx \\ &= \frac{\sin{k}}{k} - \left[ \left. x^r \frac{\sin{kx}}{k}\right|_{0}^{1}-\frac{r}{k}\int_{0}^{1}x^{r-1}\sin{(kx)}dx\right]\\ &= \frac{r}{k}\int_{0}^{1}x^{r-1}\sin{(kx)}dx \end{align*}$$ we stop here if $r\in (0,1]$. Now if $r \in (1,2)$, then we can do integration by parts again to have $$\begin{align*} f(k)&= \frac{r}{k}\int_{0}^{1}x^{r-1}\sin{(kx)}dx = \frac{r}{k^2}\left[-\cos{(k)}+(r-1)\int_{0}^{1}x^{r-2}\cos{(kx)}dx \right],\\ |f(k)|&\le \frac{r}{k^2}\left[1+(r-1)\int_{0}^{1}x^{r-2}dx\right] = \frac{2r}{k^2}. \end{align*}$$ Sadly that's the best we can do in this case. Going back to $r\in (0,1]$ case, we have $$f(k)=\frac{r}{k}Im\left\{\int_{0}^{1}x^{r-1}e^{ikx}dx\right\} $$ and since $|Im(z)| \le |z|$ for a complex number $z$, we obtain the bound $|f(k)|\le \frac{r}{k}\left|\int_{0}^{1}x^{r-1}e^{ikx}dx\right|$. Now consider the following contour,
Defining the integrand $g(z) = z^{r-1}e^{ikz}$ , show that $\int_{C_{\epsilon}} g(z) dz \to 0$ as $\epsilon \to 0$ and $\int_{C_{2}} g(z) dz = O\left( \frac{1}{k}\right)$. Thus, you will get the result that $$\int_{0}^{1}x^{r-1}e^{ikx}dx = ie^{i\frac{\pi}{2}(r-1)}\int_{0}^{1} t^{r-1}e^{-kt}dt+O\left( \frac{1}{k}\right) = O\left( \frac{1}{k^r}\right) $$
and $|f(k)| \le \frac{r!}{k^{r+1}}$.