How to show this big O equality. Let $R(x) = P (x)/Q(x)$ be a rational function with $(\text{degree}\: Q)≥ (\text{degree}\: P )+2$ and $Q(x) \not= 0$ on the real axis. Then I want to prove that 
$$\int_{-\infty}^{\infty}R(x) e^{-2 \pi i x ξ}dx=O(e^{-a|ξ|})$$
and determine the best possibles $a's$  in terms of the roots of $Q$.
I know that if $α_1 , . . . , α_k$ are the roots of Q in the upper half-plane, then there exists polynomials $P_j(ξ)$ of degree less than the multiplicity of $α_j$ so that
$$\int_{-\infty}^{\infty}R(x) e^{-2 \pi i x ξ}dx= \sum_{j=1}^k P_j(ξ)e^{-2 \pi i \alpha_j ξ}$$
with $ξ<0$ and in the case $ξ>0$ we have:
$$\int_{-\infty}^{\infty}R(x) e^{-2 \pi i x ξ}dx= \sum_{j=1}^{l} H_j(ξ)e^{2 \pi i \beta_j ξ}$$
now the $\beta_j$ are the roots but in the lower half plane. But I don't know how to use this to prove an inequality in the integral.The first thing I've got is that I have to prove that
$$\sum_{j=1}^k |P_j(ξ)e^{-2 \pi i \alpha_j ξ}|< M_1e^{-a|ξ|},\;\;\; \xi < 0$$
$$\sum_{j=1}^{l} |H_{j}(ξ)e^{2 \pi i \beta_j ξ}|< M_2e^{-a|ξ|},\;\;\; \xi > 0$$
So I have to bound the function $ξ^n e^{-c|ξ|}$ or prove that $ξ^n e^{-c|ξ|}=O(e^{-c|ξ|})$ but I don't know how to do that and determine the best possibles $a's$
Can someone help me with this issue?
Thanks a lot in advance. 
 A: You're dealing exclusively with expressions of a real variable $\xi$. Every continuous function on a closed, bounded interval has a maximum value. For a fixe $r > 0$, the expression $f(x)=|x|^{n}e^{-r|x|}$ tends to $0$ as $r\rightarrow\pm\infty$:
\begin{align}
       \lim_{x\rightarrow\infty}e^{-rx}x^{n} & =\lim_{x\rightarrow\infty}\frac{x^n}{e^{rx}} = \lim_{x\rightarrow\infty}\frac{nx^{n-1}}{(-r)e^{rx}} \\
   & = \cdots = \lim_{x\rightarrow\infty}\frac{n!}{(-r)^{n}e^{rx}} = 0.
\end{align}
Therefore, you can choosen $X > 0$ such that $|e^{-rx}x^{n}| < 1$ whenever $x > X$. If you then find the maximum of $|e^{-rx}x^{n}|$ on $[0,2X]$--let's say that maximum is $M$--then $|e^{-rx}x^n| \le \max\{M,1\}$. So there exists a positive constant $C_{r,n}$ such that
$$
                          |e^{-r|x|}x^{n}| \le C_{r,n},\;\;\; 
$$
Therefore, if $p(x)$ is a polynomial in $x$, then there is a constant $C_{r,p}$ such that
$$
                      |p(x)e^{-r|x|}| \le C_{r,n}
$$
If $a > 0$ and $\epsilon > 0$ with $\epsilon < a$, then
$$
             |p(x)e^{-a|x|}| = |p(x)e^{-\epsilon|x|}|e^{-(r-\epsilon)|x|}
      \le C_{\epsilon,p}e^{-(r-\epsilon)|x|}
$$
You can't do any better than that. You an get a bound in terms of an exponent that is as close as you want to $a$, but you cannot get a bound using $a$ on the right instead of $a-\epsilon$. If you could, you would reach the contradiction that $|p(x)| \le C$ for some $C$. What happens is that as you take smaller and smaller values of $\epsilon$, the constant $C_{\epsilon,p}$ must be taken to be larger and larger. So you're not going to find a perfect bound--it's a little bit of a juggling act between $e^{-(r-\epsilon)|x|}$ and the constant $C_{\epsilon,p}$.
In your case, you have several different exponents to consider. In the case where $\xi < 0$, you must consider the roots $\alpha_j$ of $Q$ that are in the upper half plane, where the imaginary part satisfies $\Im\alpha_j > 0$.
\begin{align}
 |\exp\{-2\pi i\alpha_j \xi\}| & = |\exp\{ -2\pi i(\Re\alpha_j+i\Im\alpha_j)\xi\}| \\
  & = |\exp\{2\pi\Im\alpha_j\xi\}| = e^{-2\pi\Im\alpha_j|\xi|}
\end{align}
If you let $a = 2\pi\min\{\Im\alpha_1,\Im\alpha_2,\cdots,\Im\alpha_k\}$, then
$|e^{-2\pi i\alpha_j\xi}| \le e^{-a|\xi|}$ for $1 \le j \le k$ and for all $\xi$. Then you can combine these bounds with the bounds of the previous paragraph to get what you want for any $a$ strictly less than the minimum of $\Im\alpha_j$ for $1\le j \le k$ and less than the minimum of $-\Im\beta_j$ for $1 \le j \le l$.
