Estimate for $\sin(x)/x$ and all its derivatives

Question

Consider the function $f(x) := \frac{\sin x}{x}, f(0) := 1$. It is easy to see that $$\lvert f(x) \rvert \leq \frac{2}{1+\lvert x \rvert}. $$ I am trying to prove that more generally, for each $k\in \mathbb N$ we have $$\lvert f^{(k)}(x) \rvert \leq \frac{C}{1+\lvert x \rvert}, $$ where $C$ may or may not depend on $k$, but not on $x$. From plotting the first few derivatives of $f$, this seems obvious. For large values of $\lvert x \rvert$ I was able to prove it, but the case where $x$ is near $0$ seems to be more subtle. For example, $$f'(x) = \frac{x\cos x - \sin (x)}{x^2},$$ and the enumerator behaves like $x^3$ near $0$, so cancels out the singularity. I appreciate any approach for proving the inequality for general $k$!

Answer 1

From the integral representation $$ f(x) = \int_0^1 \cos(x u) \, du $$ we get by differentiation $$ \tag{$*$} f^{(k)}(x) = \int_0^1 u^k \cos^{(k)}(xu) \, du $$ which will be used to obtain upper bounds for $|f^{(k)}(x)|$. This is done in two different ways, one which is better for “small” values of $x$ and one which is better for “large” values of $x$.

First, for all $x \in \Bbb R$ and all $k \ge 0$ is $$ |f^{(k)}(x)| \le \int_0^1 u^k \, du = \frac{1}{k+1} \le 1 \, . $$

Second, for $x \ne 0$ and $k \ge 1$ we can do integration by parts in $(*)$ $$ f^{(k)}(x) = \frac{\cos^{(k-1)}(x)}{x} - \int_0^1 k u^{k-1} \frac{\cos^{(k-1)}(x)}{x} \, du $$ and estimate $$ |f^{(k)}(x)| \le \frac{1}{|x|} \left( 1 + \int_0^1 k u^{k-1} \, du\right) = \frac{2}{|x|} \, . $$ This holds in the case $k=0$ as well.

Combining these results we get that $$ \boxed{|f^{(k)}(x)| \le \min \left(1, \frac{2}{|x|} \right) \le \frac{3}{1+|x|}} $$ for all $x \in \Bbb R$ and all $k \ge 0$.

These bounds are not sharp, but of the right order of magnitude. Here is a plot of the functions $\pm \min(1, 2/x)$, $f(x), f'(x), f''(x), f'''(x)$ in the range $0 \le x \le 20$ (created with wxMaxima):