π/2 conjecture

The second conjecture in this blog is related to my recent work on Lévy processes.

Suppose that x and y are positive reals, and \psi is an increasing function. Then, whenever the integral
\displaystyle\int_0^\infty\frac{(x+y)t^2}{(x^2+t^2)(y^2+t^2)}\exp\Biggl(\frac{1}{\pi}\int_0^\infty\biggl(\frac{x}{x^2+s^2}+\frac{y}{y^2+s^2}\biggr)\log\frac{\psi'(t^2)(s^2-t^2)}{\psi(s^2)-\psi(t^2)}ds\Biggr)dt
makes sense, it is equal to \pi/2.

Although the statement is quite elementary, I fail to find any elementary proof. Even for simple \psi, except \psi(\xi) = \xi and perhaps \psi(\xi) = \sqrt{\xi}, the conjecture seems to be highly non-trivial.

I showed this conjecture to several people. Let me cite here two comments: ‘It is the worst formula for \pi/2 I have ever seen!’, and ‘Come on, this must be either elementary or false’.

The conjecture originated in my recent preprint, while I was studying spectral theory of the transition semigroup of a subordinate Brownian motion killed at the time of first exit from the half-line. Every subordinate Brownian motion corresponds to a Bernstein function \psi. For the theory developed in the article, I need \psi to be a complete Bernstein function and satisfy the above conjecture.

In my preprint I prove the conjecture for \psi(\xi) = \xi^\alpha, where 0 < \alpha \le 1, and a class of complete Bernstein functions, including for example \psi(\xi) = \sqrt{\xi + 1} and \psi(\xi) = \xi + \xi^\alpha for 0.441... < \alpha < 1. Unfortunately, the argument is rather involved. As far as I know, the problem is open for all other functions, even for \psi(\xi) = \xi^\alpha with \alpha > 1.

One can easily verify the conjecture numerically for various \psi. Try playing around with the following Mathematica code:

psi[t_] = t^2 + Exp[-t] + Sin[Sqrt[t]];
dpsi[t_] = D[psi[t], t];
x = 3;
y = 2;
2/Pi NIntegrate[
  Hold[
   t^2 (x + y)/(x^2 + t^2)/(y^2 + t^2)
    Exp[1/Pi
       NIntegrate[
       (x/(x^2 + s^2) + y/(y^2 + s^2))
        Log[dpsi[t^2] (t^2 - s^2)/(psi[t^2] - psi[s^2])],
       {s, 0, Infinity}
       ]
     ]
   ],
  {t, 0, Infinity}, WorkingPrecision -> 50]

I used this code to convince myself that the conjecture is true for all complete Bernstein functions. In fact I tried some other functions (that is, not complete Bernstein ones) only to check if the code works correctly. I was quite surprised to see that it works in the general case. Or perhaps I was just not smart enough to find a good counterexample?

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