## π/2 conjecture solved

A few posts ago I stated the pi-over-two c﻿﻿oncecture. Recently, my friend and collegue Jacek Małecki proved it in full generality. (He did it while I was on my vacation in November — sometimes it is good to take a few weeks off!) This is one of the results of our joint preprint with Michał Ryznar, available here. The solution is quite elementary, but is somehow hidden in the rather technical paper. For this reason, I would like to describe it briefly in this note.

Let us recall the statement of the conjecture, which is now a theorem:

Theorem (Jacek Małecki, 2010)
Suppose that $x$ and $y$ are positive reals, and $\psi$ is a nonnegative, increasing, continuously differentiable function on $(0, \infty)$. Suppose furthermore that $(1 / s^2) \log \psi(s^2)$ is integrable at infinity. Then

$\qquad \qquad \begin{array}{l} \displaystyle\int_0^\infty\frac{(x+y) t^4 \psi'(t^2)}{(x^2+t^2)(y^2+t^2) \psi(t^2)} \times \\ \displaystyle \qquad \times \exp\left(-\frac{1}{\pi}\int_0^\infty\biggl(\frac{x}{x^2+s^2}+\frac{y}{y^2+s^2}\biggr) \log\frac{1 - \psi(s^2)/\psi(t^2)}{1 - s^2/t^2} \, ds\right)dt = \frac{\pi}{2} \, . \end{array} \hspace{\stretch{1}} (1)$

(Compared to the original formulation, we rearranged the integral using the fact that $\int_0^\infty (x/(x^2 + s^2) + y/(y^2 + s^2)) ds = \pi$.)

Proof. The proof basically identifies the integrand with a jump of a holomorphic function along the branch cut on the negative half-axis, and then uses contour integration and Cauchy theorem to find the integral. With no loss of generality we may assume that $\psi(0) = 0$. For simplicity, we assume that $\psi$ is unbounded, the argument for bounded $\psi$ is very similar. Below we prove (1) in three steps.

1. We begin with a brief study of the following auxiliary function:

$\displaystyle f(x, z) = \exp\left(-\frac{1}{\pi} \int_0^\infty \frac{x \log (1 + \psi(s^2) / z)}{x^2 + s^2} \, ds\right) .$

Here $\log$ means the principal branch of the complex logarithm, so that the definition of $f(x, z)$ is well-formed for any $x > 0$ and $z \in \mathbf{C} \setminus (-\infty, 0]$. Of course we need to assume that the integral converges, that is, that $\log(\psi(s^2)) / s^2$ is integrable at infinity.

When $x > 0$ and $z > 0$, then $0 < f(x, z) < 1$, and the limits of $f(x, z)$ as $z \searrow 0$ and $z \nearrow \infty$ are $0$ and $1$ respectively. Furthermore, if $x > 0$ and $\mathrm{Im} \, z > 0$, we have

$\displaystyle \arg f(x, z) = -\frac{1}{\pi} \int_0^\infty \frac{x \arg (1 + \psi(s^2) / z)}{x^2 + s^2} \, ds \in (0, \pi/2) .$

Here we use the relations $\arg (1 + \psi(s^2) / z) \in (-\pi, 0)$ and $\int_0^\infty x/(x^2 + s^2) ds = \pi / 2$. Since $f(x,\bar{z}) = \overline{f(x, z)}$, we have $\arg f(x, z) \in (-\pi/2, 0)$ when $\mathrm{Im} \, z < 0$. It follows that for any $x, y > 0$,

\begin{aligned} & \mathrm{Im} (f(x, z) f(y, z)) > 0 && \qquad \text{when } \mathrm{Im} \, z > 0 , \\ & \mathrm{Im} (f(x, z) f(y, z)) < 0 && \qquad \text{when } \mathrm{Im} \, z < 0 , \\ & 0 < f(x, z) f(y, z) < 1 && \qquad \text{when } z > 0 . \end{aligned}

(A function $g$ satisfying the above conditions (except perhaps $g(z) < 1$) is said to be a complete Bernstein function, see Appendix below.)

2. Let us find the boundary values of $f(x, z)$ at $z \in (-\infty, 0)$. Thanks to the identity $f(x,\bar{z}) = \overline{f(x, z)}$ it is enough to consider the limit approached from the upper half-plane, denoted $f^+(x, z)$. For $x, z > 0$ we have (with $\log^-$ being the boundary limit of the principal branch of the logarithm approached from the lower half-plane)

$\displaystyle f^+(x, -z) = \exp\left(-\frac{1}{\pi} \int_0^\infty \frac{x \log^- (1 - \psi(s^2) / z)}{x^2 + s^2} \, ds\right) .$

Note that $\log^-(1 - u) = \log |1 - u| + i \pi \mathbf{1}_{(1, \infty)}(u)$ when $u > 0$. Therefore, with $z = \psi(t^2)$,

$\displaystyle f^+(x, -\psi(t^2)) = \exp\left(-\frac{1}{\pi} \int_0^\infty \frac{x \log |1 - \psi(s^2) / \psi(t^2)|}{x^2 + s^2} \, ds + i \int_t^\infty \frac{x}{x^2 + s^2} \, ds\right) . \qquad \hspace{\stretch{1}} (2)$

The second integral on the right hand side of (2) is simply equal to $\pi/2 - \arctan(t/x) = \arctan(x/t)$, and therefore

$\displaystyle \exp\left(i \int_t^\infty \frac{x}{x^2 + s^2} \, ds \right) = \cos\left(\arctan \frac{x}{t}\right) + i \sin\left(\arctan \frac{x}{t}\right) = \frac{t + i x}{\sqrt{t^2 + x^2}} \, . \qquad \hspace{\stretch{1}} (3)$

Furthermore, we have the following result.

Proposition
For $x, t > 0$ we have

$\displaystyle \frac{1}{\pi} \int_0^\infty \frac{x \log |1 - s^2 / t^2|}{x^2 + s^2} \, ds = \frac{1}{2} \, \log \left(1 + \frac{x^2}{t^2}\right) .$

Proof. The function $h(z) = \log |1 - z^2 / t^2| = \mathrm{Re} \, \log (1 - z^2 / t^2)$ is positive and harmonic in the upper complex half-plane, and $h(z) / z \to 0$ as $|z| \to \infty$. Hence, by Poisson’s integral formula (see also Appendix below),

$\displaystyle h(i x) = \frac{1}{\pi} \int_{-\infty}^\infty \frac{x \, h(s)}{x^2 + s^2} \, ds = \frac{2}{\pi} \int_0^\infty \frac{x \, h(s)}{x^2 + s^2} \, ds . \qquad \square$

By combining (2), (3) and the above proposition, we obtain

$\displaystyle f^+(x, -\psi(t^2)) = \frac{t (t + i x)}{t^2 + x^2} \exp\left(-\frac{1}{\pi} \int_0^\infty \frac{x}{x^2 + s^2} \, \log \frac{1 - \psi(s^2) / \psi(t^2)}{1 - s^2 / t^2} \, ds\right) .$

Note that $\mathrm{Im}((t + i x)(t + i y)) = t (x + y)$. Hence, the conjecture (1) states that

$\displaystyle \int_0^\infty \mathrm{Im} \bigl(f^+(x, -\psi(t^2)) f^+(y, -\psi(t^2))\bigr) \, \frac{t \psi'(t^2)}{\psi(t^2)} \, dt = \frac{\pi}{2} \, .$

Substituting $u = \psi(t^2)$, we obtain the following equivalent form of (1):

$\displaystyle \frac{1}{\pi} \int_0^\infty \mathrm{Im} \bigl(f^+(x, -u) f^+(y, -u)\bigr) \, \frac{du}{u} = 1 . \hspace{\stretch{1}} (4)$

3. This part is rather informal. For technical details, see the discussion below and Appendix. From the identity $f(x,\bar{z}) = \overline{f(x, z)}$ it follows that the integrand in (4) is, up to the factor $-i/2$, the jump of $f(x, z) f(y, z) / z$ along the branch cut $z \in (-\infty, 0]$. By considering an appropriate family of contours, shown in the figure on the right, one can prove that:

$\begin{array}{l} \displaystyle \int_0^\infty \bigl(f^+(x, -u) f^+(y, -u) - f^+(x, -u) f^+(y, -u)\bigr) \frac{du}{-i u} \\ \displaystyle \qquad = \lim_{\genfrac{}{}{0pt}{}{R \nearrow \infty}{r \searrow 0}} \int_{-\pi}^\pi \bigl(f(x, R e^{i \alpha}) f(y, R e^{i \alpha}) - f(x, r e^{i \alpha}) f(y, r e^{i \alpha}) \bigr) d\alpha . \end{array} \hspace{\stretch{1}} (5)$

The left hand side of (5) is equal to the left hand side of (4) multiplied by $(2 \pi)$. By inspecting the definition of $f$, one can show that $f(x, z) \to 0$ as $|z| \to 0$ and $f(x, z) \to 1$ as $|z| \to \infty$, and hence the right hand side of (5) is equal to $2 \pi$. The proof of the $\pi/2$ conjecture is complete. $\qquad \square$

The last part of the proof is not rigorous. It can be made formal by a careful limiting procedure, accompanied by estimates of the integrand near the branch cut, and this is what Jacek primarily did. However, I prefer the following `soft’ argument: $g(z) = f(x, z) f(y, z)$ is a complete Bernstein function, and (4) is simply a property of complete Bernstein functions. Instead of giving references, let me explain this in detail right here.

## Appendix: complete Bernstein functions

Definition
A holomorphic function $g(z)$, $z \in \mathbf{C} \setminus (-\infty, 0)$, is a complete Bernstein function (CBF) if:
(a) $g(z) \ge 0$ for $z > 0$;
(b) $\mathrm{Im} \, g(z) > 0$ when $\mathrm{Im} \, z > 0$,
(c) $\mathrm{Im} \, g(z) < 0$ when $\mathrm{Im} \, z < 0$.

There are several alternative names for the class of complete Bernstein functions, one of them is operator monotone functions. This is because $g(z)$ is a CBF if and only if we have $g(A) \le g(B)$ whenever $A, B$ are linear operators satisfying $0 \le A \le B$. A Stieltjes function is a closely related concept. Despite its beauty and plenty of applications, the theory of complete Bernstein functions is not very widely known. There is an outstanding book by Rene Schilling, Renming Song and Zoran Vondraček covering this area (book’s website, publisher site, Google Books preview). I also like the exposition in the first volume of the Niels Jacob’s book on pseudo-differential operators and Markov processes (publisher site, Google Books preview).

Examples of CBFs include $g(z) = z^a$ for $a \in [0, 1]$, $g(z) = \log(1 + z)$ and $g(z) = z / (1 + z)$. Furthermore, if $g_1$ and $g_2$ are CBFs, then $c_1 g_1 + c_2 g_2$ ($c_1, c_2 \ge 0$), $\sqrt{g_1} \sqrt{g_2}$ (or, more generally, $g_1^p g_2^{1-p}$ for $p \in [0, 1]$) and $g_1 \circ g_2$ are again CBFs. Below we prove a fundamental representation theorem for CBFs. In fact this is usually taken as the definition of a complete Bernstein function, and a similar result has been proved by several authors in the early 20th century.

Theorem
Every complete Bernstein function has the following form:

$\displaystyle \qquad g(z) = a + b z + \int_0^\infty \frac{z}{t + z} \, \frac{\mu(dt)}{t} \, , \hspace{\stretch{1}} (6)$

where $a, b \ge 0$ and $\mu(dt)$ is a nonnegative measure on $(0, \infty)$, for which the function $\min(1/t, 1/t^2)$ is integrable. Furthermore, the measure $\mu(dt)$ can be recovered as the (distributional) jump of the imaginary part of $g(z)$ along $(-\infty, 0)$, that is,

$\displaystyle \qquad \mu(dt) = \frac{1}{\pi} \, \lim_{\varepsilon \searrow 0} \bigl(\mathrm{Im} \, g(-t + i \varepsilon) dt\bigr) . \hspace{\stretch{1}} (7)$

Proof. The imaginary part of a CBF $g(z)$ is a nonnegative harmonic function in the upper half-plane. This is a very classical object (see, for example, the book by Sheldon Axler, Paul Bourdon and Wade Ramey, available online here). In particular, we have the representation theorem (Th. 7.26 in the book),

$\displaystyle \mathrm{Im} \, g(u + i v) = b v + \frac{1}{\pi} \int_{\mathbf{R}} \frac{v}{(t - u)^2 + v^2} \, \sigma(dt) , \hspace{\stretch{1}} (8)$

where $b \ge 0$ and the nonnegative measure $\sigma(dt)$ is the weak-* limit of absolutely contiunous measures $\mathrm{Im} \, g(t + i \varepsilon) dt$ as $\varepsilon \searrow 0$. A priori, $\sigma(dt)$ can be an arbitrary measure for which $\min(1, 1/t^2)$ is integrable. (Compare this result with the proof of Proposition above.)

Since $g(z)$ is continuous on $(0, \infty)$ and $\mathrm{Im} \, g(t) = 0$ for $t > 0$, the measure $\sigma$ is concentrated on $(-\infty, 0]$. We write $(1/\pi) \sigma(-dt) = \mu(dt) + c \delta_0(dt)$ where $\mu$ is concentrated on $(0, \infty)$ and $c \ge 0$. Formula (8) can be rewritten as

$\displaystyle \mathrm{Im} \, g(z) = b \, \mathrm{Im} \, z + \frac{1}{\pi} \int_{(-\infty, 0]} \mathrm{Im} \frac{1}{t - z} \, \sigma(dt) = b \, \mathrm{Im} \, z - c \, \mathrm{Im} \frac{1}{z} - \int_{(0, \infty)} \mathrm{Im} \frac{1}{t + z} \, \mu(dt) .$

Since $1 / (t + z) = (1/t) (1 - z / (t + z))$, we obtain

$\displaystyle \mathrm{Im} \, g(z) = b \, \mathrm{Im} \, z - c \, \mathrm{Im} \frac{1}{z} + \int_{(0, \infty)} \mathrm{Im} \frac{z}{t + z} \, \frac{\mu(dt)}{t} \, .$

Two holomorphic functions have equal imaginary part if and only if their difference is a real constant. Hence, for some real $a$,

$\displaystyle g(z) = a + b z - \frac{c}{z} + \int_{(0, \infty)} \left(\frac{z}{t + z} - \frac{1}{t + 1} \right) \frac{\mu(dt)}{t} \, .$

Note that the term $1 /(t + 1)$ is necessary, because we do not know a priori that $\min(1, 1/t)$ is integrable with respect to $\mu(dt)$. It remains to prove that this is indeed a case, that $c = 0$, and that $a - \int_0^\infty 1 / (t + 1) \mu(dt) \ge 0$.

When $z \in (0, 1)$, clearly $0 \le g(z) \le a + b z - c/z$, which proves that $c = 0$. Furthermore, by monotone convergence,

$\displaystyle 0 \le \lim_{z \searrow 0} g(z) = a - \int_{(0, \infty)} \frac{1}{t + 1} \frac{\mu(dt)}{t} \, . \qquad \square$

The $\pi/2$ conjecture follows easily from the following simple consequence of (6) and (7): when $b = 0$, we have

$\displaystyle \int_0^\infty \frac{\mu(dt)}{t} = \lim_{z \nearrow \infty} g(z) - \lim_{z \searrow 0} g(z) . \hspace{\stretch{1}} (9)$

We apply the above identity for $g(z) = f(x, z) f(y, z)$. Since $g$ is bounded on $(0, \infty)$, clearly $b = 0$. By (7), $\mu(dt) = (1/\pi) \mathrm{Im} \, g^+(-t) dt$. Furthermore, the right-hand side of (9) is equal to $1$. This proves (4).