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My question below arises from the linked question $ \int\limits_{-\infty}^{+\infty}\frac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \leq n^{3/2}\pi$ and the comments by jack and David Speyer under that question.

Let $p(x)=\prod_{k=1}^n(x-x_k)$, where $n\ge 1$ and $x_1,\dots,x_n\in\mathbb R$. Then for $f(t)=\frac{t^2}{1+t^2}$, the integral appearing in the linked question can be rewritten as $$\int_{-\infty}^{+\infty}f\left(\frac{p'(x)}{p(x)}\right)dx= \int_{-\infty}^{+\infty}f\left(\sum_{k=1}^n\frac{1}{x-x_k}\right)dx. $$ Then as noted by jack and David Speyer, the first question is: for $f(t)=\frac{t^2}{1+t^2}$, how to prove $$\int_{-\infty}^{+\infty}f\left(\sum_{k=1}^n\frac{1}{x-x_k}\right)dx=\int_{-\infty}^{+\infty}f\left(\frac{n}{x}\right)dx \tag{$*$} $$ always hold for every $n\ge 2$?

Moreover, it seems that $(*)$ also holds for more general form of $f$. Then the second question comes: for what $f$, $(*)$ holds for every $n\ge 2$?

Any suggestion is appreciated. Thanks in advance.

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23rd
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    A quick observation: It looks like the more general pattern is $\int_{-\infty}^{\infty} f\left( \sum \frac{a_k}{x-x_k} \right) dx = \int_{-\infty}^{\infty} f\left( \frac{\sum a_k}{x} \right) dx$. Of course, that follows from your formula when the $a_k$ are positive integers, but it seems to always be true. – David E Speyer Apr 29 '15 at 14:17
  • I believe this question has something to do with the argument principle in complex analysis. – Samrat Mukhopadhyay Apr 29 '15 at 14:22
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    Actually, it appears to be true for $a_k>0$, but not when the $a_k$ are negative. Curiouser and curiouser... We should also bear in mind that Mathematica doesn't always get definite integrals right. – David E Speyer Apr 29 '15 at 14:23
  • @SamratMukhopadhyay: Thank you, but I have no idea how to apply argument principle for this problem. – 23rd Apr 29 '15 at 15:02
  • @DavidSpeyer: Thank you for your comments. I agree with your observation. I think in fact the statement for general positive real $a_k$ can be deduced from the statement for integers. First, by change of variables for $x$, the statement holds for rational $a_k$; second, use rationals to approximate reals. – 23rd Apr 29 '15 at 15:05
  • Yes @23rd, initially I thought that argument principle might be helpful, but I think I was wrong. Nevertheless, the $\deg(p)\pi$ in the RHS is constantly making me suspect some kind of relation with complex analysis. – Samrat Mukhopadhyay Apr 29 '15 at 17:23
  • One simple observation is that if one uses residue theorem, answering the first question boils down to proving that the sum of residues of $\frac{(p'(x))^2}{(p'(x))^2+(p(x))^2}$ is $-in/2$ where $n$ is the degree of $p$. It now remains to see how to prove this apparently remarkable fact. – Samrat Mukhopadhyay Apr 29 '15 at 19:34
  • As long as we're throwing out ideas that we don't know what to do with, I'll point out that a rational function $q(x)$ is of the form $\sum_{m=1}^m \frac{a_k}{x-x_k}$ if and only if $q: \mathbb{RP}^1 \to \mathbb{RP}^1$ is a degree $m$ covering. – David E Speyer Apr 29 '15 at 19:35

1 Answers1

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I know what is happening now. Here is the general result:

Let $f(z)$ be a rational function with real coefficients and no real poles which vanishes to order $2$ at $x=\infty$. Let $a_1 < b_1 < a_2 < b_2 < \cdots < a_{n-1} < b_{n-1} < a_n$ be real numbers and set $r(x) = c \frac{\prod(x-a_i)}{\prod(x-b_i)}$ for $c>0$. Then

$$\int_{x=-\infty}^{\infty} f(r(x)) dx = \int_{x=-\infty}^{\infty} f(cx).$$

In particular, we can take $r(x) = p(x)/p'(x)$, in which case the inequalities on the $a$'s and $b$'s are a consequence of Rolle's theorem, and take $f(z) = 1/(1+z^2)$. (In this case, $c$ will be $1/n$.)

Proof: We first note that the integral makes sense at all. Since $f(z)$ has no real poles (including at $z=\infty$), the integrand has no real poles. As $x \to \infty$, we have $r(x) \sim x$ so $f(r(x)) = O(x^{-2})$, and thus the integral is convergent.

As observed before, $\int_{\infty}^{\infty} f(r(x))$ is $2 \pi i$ times the sum of the residues of the poles of $f(r(x))$ in the upper half plane. We have $f(r(x)) = \infty$ if and only if $r(x)$ is a pole of $f$. Since $f$ has no real poles, the poles of $f$ come in conjugate pairs $(\alpha_i, \overline{\alpha_i})$, where we always take $\alpha_i$ to lie in the upper half plane. So the poles of $f(r(x))$ are the solutions to $r(x) = \alpha_i$ or $\overline{\alpha_i}$.

Lemma For all $x \in \mathbb{C}$, the imaginary part of $r(x)$ has the same sign as the imaginary part of $x$.

Proof: Since $c>0$, this is true for $x=iT$ with $T$ large. Therefore, if it ever fails, there must be some $x_0 \not \in \mathbb{R}$ for which $r(x)$ is real. Suppose $r(x_0) = s \in \mathbb{R}$ and, without loss of generality, take $s>0$.

The function $r$ increases from $0$ to $\infty$ on each of the intervals $(a_1, b_1)$, $(a_2, b_2)$, ..., $(a_{n-1}, b_{n-1})$ and $(a_n, \infty)$. So there are already $n$ real roots of the equation $r(x) = s$, as well as the additional root $n+1$. But, clearing denominators, $r(x)=s$ is a polynomial of degree $n$, so it can only have $n$ roots. $\square$

We see that the poles of $f(r(x))$ in the upper half plane are precisely the solutions to $r(x) = \alpha_i$.

Write $f(x)$ in partial fraction expansion as $\sum \frac{\beta_i}{x-\alpha_i} + \frac{\overline{\beta}_i}{x-\overline{\alpha}_i}$ and set $g(x) = \sum \frac{\beta_i}{x-\alpha_i}$. So the poles of $f(r(x)) dx$ in the upper half plane are the same as the poles of $g(r(x)) dx$ in the entire complex plane, and with the same residues. The sum of the residues of a meromorphic differential form over the whole Riemann sphere is $0$, so the sum of the residues of $g(r(x))$ at $x \in \mathbb{C}$ is negative the residue of $g(r(x)) dx$ at $\infty$.

Using the same argument to study $\int f(cx) dx$, we are reduced to showing that $g(r(x)) dx$ and $g(cx) dx$ have the same residue at $\infty$. Note that $g$ vanishes at $\infty$ and $dx$ has a double pole at $\infty$, so we are talking about a differential with a simple pole. So this follows from the fact that $r(x) = cx+O(1)$ near $\infty$. $\square$.

David E Speyer
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  • Dear David Speyer, thank you for your excellent answer! I still have one question. The displayed equality in your answer also seems to hold for some $f$ that fails to be holomorphic somewhere, say $f(z)=\frac{1}{1+|z|^\alpha}$ for real $\alpha>2$. Could such $f$ be covered as a simple corollary of your result? – 23rd May 02 '15 at 02:23
  • That's odd. Perhaps such an $f$ is a uniform limit of the $f$'s I consider? – David E Speyer May 02 '15 at 02:44
  • I have noticed that Theorem 3 in sos440's answer to the original post about this topic has solved the problem in my last comment. Thank you! – 23rd May 02 '15 at 09:47
  • Very cool. And Achille Hui and orangekid apparently also deserve credit. I did not know this technique before now. – David E Speyer May 02 '15 at 21:49
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    For the record, the general result you prove is American Mathematical Monthly Problem 11234 - 06 by J. Brennan and R. Ehrenborg. – Martin Nicholson Dec 21 '17 at 10:21