For $$ f(x) = \frac{x}{(x+1)(x+2) - e^{-3x}} $$ I try to prove that
there exist $M > 0$ such that for all non-negative integers $n$ and all $x > 0$ we have $$ |f^{(n)}(x)| \le M n! \frac{1}{x^{n+1}}.\tag{1} $$
Numerical analysis shows that it should be true. However, none of the approaches I tried (guessing a formula for the $n$-th derivative, using Cauchy's integral formula or Mittag-Leffer's decomposition) works for me. I also know that (1) is a consequence of the following:
for all $a > 0$ there exists $K > 0$ such that for all $s > 0$ and all $n \in \{2,3,\dots\}$, $$ \biggl| \int_{-\infty}^{\infty} \frac{f(a+it)}{(1-ist)^n} dt \biggr| \le K. $$
Is there a way to estimate the integral?
This question is related to Estimate of the $n$-th derivative of $\frac{x}{(x+1)(x+2)-2e^{-3x}}$.