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    The Fourier transform of $1/Q(x)$ in terms of the roots of $Q$

    E-SS4.EX-4

    Area
    Complex Analysis
    Topics
    Fourier Transform, Residues
    Status
    draft
    1. Suppose \(Q\) is a polynomial of degree \(\geq 2\) with distinct roots, none lying on the real axis. Calculate

    \[ \int_ {- \infty} ^ {\infty} \frac {e ^ {- 2 \pi i x \xi}}{Q (x)} d x, \quad \xi \in \mathbb {R} \]

    in terms of the roots of \(Q\) . What happens when several roots coincide? [Hint: Consider separately the cases \(\xi < 0 , \xi = 0\) , and \(\xi > 0\) . Use residues.]

    Solution

    Let the distinct roots of \(Q\) be \(\alpha_1,\ldots,\alpha_m\), none real, and set \[ I(\xi)=\int_{-\infty}^{\infty}\frac{e^{-2\pi i x\xi}}{Q(x)}\,dx. \] Because \(\deg Q\ge2\), the semicircular arc contribution tends to \(0\) in every case below.

    If \(\xi<0\), close in the upper half-plane, where the exponential decays. The residue theorem gives \[ I(\xi)=2\pi i\sum_{\Im\alpha_j>0} \frac{e^{-2\pi i\alpha_j\xi}}{Q'(\alpha_j)}. \tag{1} \] If \(\xi>0\), close in the lower half-plane. The contour is clockwise, hence \[ I(\xi)=-2\pi i\sum_{\Im\alpha_j<0} \frac{e^{-2\pi i\alpha_j\xi}}{Q'(\alpha_j)}. \tag{2} \] For \(\xi=0\) one may close in either half-plane; for example \[ I(0)=2\pi i\sum_{\Im\alpha_j>0}\frac1{Q'(\alpha_j)}. \tag{3} \]

    If roots coincide, the same contour argument holds but the poles are of higher order. A root \(\alpha\) of multiplicity \(r\) contributes its higher-order residue \[ \operatorname{Res}_{z=\alpha} \frac{e^{-2\pi i z\xi}}{Q(z)}, \] which is \(e^{-2\pi i\alpha\xi}\) times a polynomial in \(\xi\) of degree at most \(r-1\). Thus (1)–(2) are replaced by finite sums of terms \[ P_\alpha(\xi)e^{-2\pi i\alpha\xi}, \qquad \deg P_\alpha<r. \]