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I'm working with Exercise 4 of Chapter 4, Stein & Shakarchi "Complex Analysis".

The problem is

Suppose $Q$ is a polynomial of degree$\geq2$ with distinct roots, none lying on the real axis. Calculate $$F(\xi) = \int_{-\infty}^{\infty} \dfrac{e^{-2\pi i x \xi}}{Q(x)}\ dx,\space\xi\in\mathbb R$$ in terms of the roots of $Q$. What happens when several roots coincide?

For $\xi=0$, I approached by residue formula using upper half circle $C_R^+$ and $R\to\infty$ for the first summation and $C_R^-$ for the second summation and got the below: $$\dfrac{F(0)}{2\pi i} = \sum_{\Im(s_i)>0} \operatorname{Res}\dfrac{1}{Q(z)}=-\sum_{\Im(s_i)<0} \operatorname{Res}\dfrac{1}{Q(z)} $$ where $s_i$'s are distinct roots of $Q$. I think there would be a better representation for this using $Q(x)=A\prod(x-s_i)$ so that I can get a "Fourier transform formula" of $\frac{1}{Q(x)}$, but I can't go further.

For $\xi<0$ and $\xi>0$, I find that I should use $C_R^+$ and $C_R^-$ respectively but that's all. I don't know about the the role of coinciding several roots too.

How can I get some meaningful formulas for this problem?

Edit Thanks for hints and answer. Using all I got for $\xi\geq 0$, $$ F(\xi) = -2\pi i \sum_{\Im(s_i)<0} \operatorname{Res} \dfrac{e^{-2\pi i x \xi}}{Q(x)} = -2\pi i \sum_{\Im(s_i)<0} \dfrac{e^{-2\pi i s_i \xi}}{Q’(s_i)} $$

where the second equality holds when $Q(x)$ has all distinct roots and $Q’(s_i)=A\prod_{s_j\neq s_i}(s_j-s_i)$. Also for $\xi\leq 0$,

$$ F(\xi) = 2\pi i \sum_{\Im(s_i)>0}\operatorname{Res} \dfrac{e^{-2\pi i x \xi}}{Q(x)} = 2\pi i \sum_{\Im(s_i)>0} \dfrac{e^{-2\pi i s_i \xi}}{Q’(s_i)} $$

For multiple roots, the residue formula of $\frac{1}{Q(z)}$ where $Q(z)=A\prod(z-s_i)^{r_i}$ would be different, but again I can’t approach. The residue formula for $n$-th order poles requires $(n-1)$-times of differentiation, and I feel this direct route is too dirty. How can I approach this?

Jinmu You
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    $\frac{1}{Q(x)} = \sum_{j=1}^n \frac{c_j}{x-s_j}$ where $c_j = 1/Q'(s_j)$. And for $\Im(s_j) > 0$, $ \frac{1}{x-s_j}$ is the Fourier transform of $2i \pi e^{2 i\pi s_j \xi} 1_{x > 0}$. Do you see how this relates to the residue theorem ? – reuns May 28 '19 at 17:35
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    A worked example for a polynomial of degree 3 and a slightly different Fourier Transform convention: https://math.stackexchange.com/a/3086586/441161 – Andy Walls May 28 '19 at 22:13
  • @reuns I really needed that formula of $\frac{1}{Q(x)}$ of yours. How can I modify this for $Q(x)$ with coinciding roots? By the way in my textbook, the Fourier transform and its inverse is defined only for moderately decreasing function, so I can’t use the transform of Heaviside function. – Jinmu You May 29 '19 at 05:05
  • @AndyWalls Thanks for your answer. I’ll refer this to expand this problem for roots on the real axis. – Jinmu You May 29 '19 at 05:09
  • Your edit is likely wrong for $\xi = 0$. Your answer, for all roots distinct, should look something like $F(\xi) = 2\pi i \sum_{s_i} \dfrac{e^{-2\pi i s_i \xi}}{Q’(s_i)}\dfrac{1}{2}\left[\mathrm{sgn}(\xi) +\mathrm{sgn}\left(\Im\left[s_i\right]\right)\right]$ (I may have the $+$ sign wrong). That factor of $1/2$ makes a difference when $\xi =0$ as then $\mathrm{sgn}(\xi) =0 $. – Andy Walls May 29 '19 at 13:02
  • @AndyWalls Thanks, I missed that. By the way, when $\xi=0$ the summation on the upper half plane is equal to minus of summation on the lower half plane, so it would be equal to the half of summation on $\mathbb C$. – Jinmu You May 29 '19 at 13:19
  • Is that the case when the complex roots do not come in conjugate pairs (i.e. a polynomial with complex coefficients)? – Andy Walls May 29 '19 at 13:49
  • @AndyWalls I think yes because in the description of the problem, there is no such mention that $Q$ has only real coefficients. – Jinmu You May 29 '19 at 13:54
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    Actually $(n-1)$-times differentiation is not so dirty: as you accepted differentiating once for the distinct-root case, why are you so reluctant to accept multiple differentiations? Indeed, I cannot think of any better representation of the residue. – Lee May 30 '19 at 12:17
  • @Szeto Thanks. I’ll try it. – Jinmu You Jun 02 '19 at 07:35
  • @AndyWalls Why is it relevant that $Q$ has degree $\geq 2$? – CuriousKid7 Oct 22 '19 at 16:31
  • @CuriousKid7 I don't see where I said $Q$ has to have degree $\ge 2$. The degree 3 worked example I pointed to was just an example I knew I had lying around on MSE. – Andy Walls Oct 22 '19 at 21:19
  • @AndyWalls Right, but do you know why the problem in the OP's question gives it as an assumption? – CuriousKid7 Oct 22 '19 at 21:23
  • @CuriousKid7 Well, if one doesn't have at least 2 roots, there's not much to be said about what happens when the roots coincide. That's the only reason I see. – Andy Walls Oct 22 '19 at 23:42

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Your presumption that $C_R^+$ and $C_R^-$ can be used in combination with the residue formula when $\xi<0$ and $\xi>0$, respectively, is correct. In the case $\xi>0$, you can parameterize $C_R^+$ by $$t\rightarrow Re^{it}=R(\cos(t)+i\sin(t)),\ t\in(0,\pi)$$ and estimate \begin{align*} \bigg| \int_0^\pi\frac{e^{-2\pi i R (\cos(t)+i\sin(t))\xi}}{Q(R e^{it})} iRe^{it}dt \bigg| \leq \int_0^\pi\bigg|\frac{e^{2\pi R \sin(t)\xi}}{Q(R e^{it})}R \bigg|dt \rightarrow 0\ \text{as }R\rightarrow \infty\quad (\text{since }\xi<0). \end{align*} Consequently, you get via the residue formula \begin{align*} F(\xi) = 2\pi i \sum_{Im(s_i)>0}{{\mathrm{Res}}_{x=s_i}\frac{e^{-i2\pi x \xi}}{Q(x)}}. \end{align*} Using the hint from the comment above, you can compute these residuals.

StarBug
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  • Rereading your question, I think maybe this part was already clear to you. Were you asking specifically how to compute the residuals? – StarBug May 28 '19 at 18:34
  • Yes, I want to find a way to calculate that neatly and I’m working with the above comment. But still I don’t know how multiple roots work. – Jinmu You May 29 '19 at 04:17