- More generally, let \(R ( x ) = P ( x ) / Q ( x )\) be a rational function with (degree \(Q ) \geq\) (degreeP )+2 and \(Q ( x ) \neq 0\) on the real axis.
- Prove that if \(\alpha _ { 1 } , \ldots , \alpha _ { k }\) are the roots of R in the upper half-plane, then there exists polynomials \(P _ { j } ( \boldsymbol { \xi } )\) of degree less than the multiplicity of \(\alpha _ { j }\) so that
\[ \int_ {- \infty} ^ {\infty} R (x) e ^ {- 2 \pi i x \xi} d x = \sum_ {j = 1} ^ {k} P _ {j} (\xi) e ^ {- 2 \pi i \alpha_ {j} \xi}, \quad \text {when} \xi < 0. \]
In particular, if \(Q ( z )\) has no zeros in the upper half-plane, then \(\begin{array} { r } { \int _ { - \infty } ^ { \infty } R ( x ) e ^ { - 2 \pi i x \xi } d x = 0 } \end{array}\) for \(\xi < 0 .\)
Show that similar results hold in the case \(\xi > 0\)
Show that
\[ \int_ {- \infty} ^ {\infty} R (x) e ^ {- 2 \pi i x \xi} d x = O (e ^ {- a | \xi |}), \quad \xi \in \mathbb {R} \]
as \(| \xi | \to \infty\) for some \(a > 0\) . Determine the best possible \(a \mathrm { { s } }\) in terms of the roots of R.
[Hint: For part \(\mathrm { ( a ) }\) , use residues. The powers of \(\xi\) appear when one diferentiates the function \(f ( z ) = R ( z ) e ^ { - 2 \pi i z \xi }\) (as in the formula of Theorem 1.4 in the previous chapter). For part (c) argue in the lower half-plane.]
Solution
Write the poles of \(R=P/Q\) in the upper half-plane as \(\alpha_j\), with multiplicities \(m_j\). For \(\xi<0\), close the contour in the upper half-plane. The assumption \[ \deg Q\ge \deg P+2 \] ensures that the large semicircle contributes \(0\). Hence \[ \int_{\mathbb R}R(x)e^{-2\pi i x\xi}\,dx =2\pi i\sum_j\operatorname{Res}_{z=\alpha_j} \left(R(z)e^{-2\pi i z\xi}\right). \tag{1} \] At a pole \(\alpha_j\) of order \(m_j\), the higher-order residue formula differentiates at most \(m_j-1\) times. Differentiating the exponential produces powers of \(\xi\), so the corresponding residue has the form \[ P_j(\xi)e^{-2\pi i\alpha_j\xi}, \qquad \deg P_j<m_j. \] This proves (a). If there are no poles in the upper half-plane, the sum in (1) is empty, proving (b).
For \(\xi>0\), close instead in the lower half-plane. Since the contour is clockwise, \[ \int_{\mathbb R}R(x)e^{-2\pi i x\xi}\,dx =-2\pi i\sum_{\Im\beta_j<0} Q_j(\xi)e^{-2\pi i\beta_j\xi}, \tag{2} \] where \(\deg Q_j\) is less than the multiplicity of the pole \(\beta_j\). This proves (c).
For (d), let \[ \delta_+=\min_{\Im\alpha_j>0}\Im\alpha_j, \qquad \delta_- =\min_{\Im\beta_j<0}|\Im\beta_j|, \] omitting a quantity when the corresponding half-plane contains no pole. From (1), as \(\xi\to-\infty\), each term has magnitude bounded by a polynomial times \[ e^{-2\pi(\Im\alpha_j)|\xi|}, \] so for every \(a<2\pi\delta_+\) the transform is \(O(e^{-a|\xi|})\) on that side. Similarly, for every \(a<2\pi\delta_-\) it is \(O(e^{-a|\xi|})\) as \(\xi\to+\infty\). If one half-plane contains no poles, the transform is identically zero on the corresponding side.
Consequently a two-sided estimate \[ \int_{\mathbb R}R(x)e^{-2\pi i x\xi}\,dx=O(e^{-a|\xi|}) \] holds for every \[ a<2\pi\,\operatorname{dist}(\{\text{poles of }R\},\mathbb R). \] The supremal possible exponential rate is therefore \[ 2\pi\,\operatorname{dist}(\{\text{poles of }R\},\mathbb R). \] At the endpoint itself the estimate holds when every pole at minimal distance is simple; higher multiplicity may introduce a nonconstant polynomial factor, so in general the endpoint rate is only a supremum.