# Factoring Bessel functions into an amplitude and a phase

Take some $\nu>0$. Let $J_\nu(x)$ be the Bessel function of the first kind. Let's restrict its domain to $\mathbb R^+$. Is it possible to find a pair of functions $A_\nu(x), \phi_\nu(x):\mathbb R^+\to\mathbb R$ that are real-analytic and completely monotone (i.e. the function itself and all its derivatives are monotone) such that $$J_\nu(x)=A_\nu(x)\sin(\phi_\nu(x)),\quad A_\nu(x)>0?$$

Is such a pair of functions uniquely determined by $\nu$ (modulo constant term $2\pi n$ in $\phi_\nu$)?

The same question applies to other decaying oscillating functions such as the cosine integral $\operatorname{Ci}(x)$ or the Airy function $\operatorname{Ai}(-x)$.

• I presume this is not the amplitude/phase factorization you are looking for? – Carlo Beenakker May 10 '16 at 9:43
• @CarloBeenakker Thanks. Yes, I recall seeing this representation in some paper. This might be the representation I am looking for. But the question remains: are those amplitude and phase functions completely monotonic (the paper says just they are simply monotonic), and whether such a factorization unique. – Vladimir Reshetnikov May 10 '16 at 18:18
• It cannot be done for the Airy function, I suppose. If you write $y=Ae^{i\phi}$ for a solution of $\frac {d^2y}{dx^2} \pm xy=0$, you get two equations for $A$ and $\phi$, by equating real and imaginary parts. One of them is $2A'\phi '+A\phi ''=0$. But this cannot hold if the functions are completely monotone, because the left side would be positive and thus not $0$. – FusRoDah Nov 17 '19 at 14:06

Not an answer but too large for a comment.

In my paper with van de Lune On the exact location of the non-trivial zeros f Riemann's zeta function, it is proved.

Theorem. If $$f\colon\mathbf{R}\to\mathbf{C}$$ is real analytic, then there are two real analytic functions $$U\colon\mathbf{R}\to\mathbf{R}$$ and $$\varphi\colon\mathbf{R}\to\mathbf{R}$$ such that $$f(t)=U(t)e^{i\varphi(t)}$$. Given two such representations, $$f=U_1e^{i\varphi_1}$$ and $$f=U_2e^{i\varphi_2}$$ we have either $$U_1=U_2$$ and $$\varphi_1-\varphi_2=2k\pi$$ or $$U_1=-U_2$$ and $$\varphi_1-\varphi_2=(2k+1)\pi i$$ for some integer $$k$$.

Your question is slightly different. If a real analytic function can be represented as $$f(x)=A(x)\sin(\phi(x))$$ with $$A$$ and $$\phi$$ real analytic, then $$f(x)=\Im (A(x)e^{i\phi(x)})$$. The function $$F(x) =A(x)e^{i\phi(x)}$$ will be a complex real analytic function with $$f(x)=\Im(F(x))$$. But this $$F$$ is not unique. Any real and real analytic function $$h(x)$$ gives us $$F(x)+h(x)$$ with the same property. By the above theorem we will have $$F(x)+h(x)=B(x)e^{i\phi_h(x)}$$ with $$B$$ and $$h$$ real real analytic functions. Therefore we obtain $$f(x)=B(x) \sin(\phi_h(x))$$. All representation to your functions are of this type, starting from the one given in the comment by Carlo Beenakker.

The problem whether we may get $$\phi_h(x)$$ completely monotone appear to be complicated. In simple cases as $$\Gamma(s/2)=|\Gamma(s/2)|e^{i\vartheta(t)+i\frac{t}{2}\log\pi}$$, we get the representation $$\Im(\Gamma(\tfrac14+i\tfrac{t}{2}))=|\Gamma(\tfrac14+i\tfrac{t}{2})|\sin(\vartheta(t)+\tfrac{t}{2}\log\pi),\qquad s=\tfrac12+it$$ the phase is not monotonous for $$t>0$$. Certainly $$\vartheta(t)+\frac{t}{2}\log\pi$$ appear to be almost completely monotonous. I will be very surprised if in this case there is a completely monotonous phase.

Also too long for a comment. Let us discuss an equivalent representation $$J_\nu(x) = A_\nu(x) \cos \phi_\nu(x)$$ (with a cosine rather than sine). Theorem 5 in:

K.S. Miller, S.G. Samko, Completely monotonic functions, Integral Transforms and Special Functions 12(4) (2001): 389–402, DOI: 10.1080/10652460108819360

asserts that the function $$(A_\nu(x))^2 = (J_\nu(x))^2 + (Y_\nu(x))^2$$ is completely monotone. Furthermore, the phase function $$\phi_\nu$$ satisfies $$\phi_\nu'(x) = \tfrac{2}{\pi x} (A_\nu(x))^{-2}$$, as it is observed in the reference pointed by Carlo Beenakker in his comment:

M. Goldstein, R.M. Thaler, Bessel Functions for Large Arguments, Mathematical Tables and Other Aids to Computation 12(61) (1958): 18–26, DOI: 10.2307/2002123

Some random remarks:

• Complete monotonicity of $$(A_\nu(x))^2$$ does not automatically imply complete monotonicity of $$A_\nu(x)$$. This would be the case if $$(A_\nu(x))^2$$ were a Stieltjes function, but unfortunately it is not.

• The function $$\phi_\nu(x)$$ is a Bernstein function if and only if $$\tfrac{1}{x} (A_\nu(x))^{-2}$$ is completely monotone. This is only possible when $$\nu \leqslant \tfrac{1}{2}$$. For $$\nu > \tfrac{1}{2}$$, $$\tfrac{1}{x} (A_\nu(x))^{-2}$$ is increasing (possibly a Bernstein function), and therefore $$\phi_\nu(x)$$ is neither completely monotone nor Bernstein.

• A closely related function $$\tfrac{1}{x} (A_\nu(\sqrt{x}))^{-2}$$ appears in the representation of $$\sqrt{x} K_\nu(\sqrt{x}) / K_{\nu-1}(\sqrt{x})$$ as a Stieltjes transform, see entry 116 in Section 16.8 in:

R. Schilling, R. Song, Z. Vondraček, Bernstein functions: theory and applications, De Gruyter, 2012.

(In the above I use the term "completely monotone" for $$(-1)^n f \geqslant 0$$ for $$n = 0, 1, \ldots$$, and "Bernstein" for $$f \geqslant 0$$ and $$f'$$ completely monotone.)