How to show that $f_a(x) = \frac{1}{2\pi} \frac{2^{a-1}}{\Gamma(a)} \bigl|\Gamma\bigl(\frac{a+i x}{2}\bigr)\bigr|^2 e^{x(\tau - \pi/2)}$ is stable?

Question

Define $f_a(x) = \frac{1}{2\pi} \frac{2^{a-1}}{\Gamma(a)} \Gamma\bigl(\frac{a+i x}{2}\bigr) \Gamma\bigl(\frac{a-i x}{2}\bigr) e^{x(\tau - \pi/2)}$. Then it seems that $f_a$ is stable under convolution in the sense that $f_a\ast f_b = f_{a+b}$, where I define the convolution as $$(f\ast g)(x) = \int_\mathbb{R} f(y)g(x-y)\,dy.$$ Numerically, this even holds for complex $a,b$ so long as $\Re (a),\Re(b)>0$.

I ``found'' this identity more or less by accent during a long struggle to solve a more difficult problem, but it is essentially just an educated guess.

Do any of the integration wizards on this site have an idea how to show this analytically? I presume it would be sufficient to compute the Fourier transform of $f_a$, as then the identity should be somehow obvious...

Alternatively, is this perhaps already known in literature (in a different form?) that I just didn't recognize?

Here is some MWE for numerically checking the claim:

from scipy import special
from scipy.integrate import quad
g = special.gamma
cquad = lambda *args: quad(*args, complex_func=True)
tau=0.735; a=0.351-.16j; b=0.661+.1j; k=1.555; L=200
f_a = lambda x,a: 1/(2*np.pi) * 2**(a-1)/g(a) * g((a+1j*x)/2) * g((a-1j*x)/2) * np.exp(x*(tau-np.pi/2))
print(cquad(lambda xp: f_a(xp,a) * f_a(x-xp,b), -L,L)[0], f_a(x,a+b))
#out: 
(0.02381444903180519-0.0013008602568538431j) (0.02381444903180523-0.0013008602568538425j)

Answer 1

Without Fourier transform $$\int_{\mathbb{R}}f_a(y)f_b(x-y)\,\mathrm dy\\ =\frac{2^{a+b-2}e^{x(\tau-\pi/2)}}{4\pi^2\Gamma(a)\Gamma(b)}\int_{\mathbb{R}}\Gamma\left(\frac{a+iy}{2}\right)\Gamma\left(\frac{a-iy}{2}\right)\Gamma\left(\frac{b+ix-iy}{2}\right)\Gamma\left(\frac{b-ix+iy}{2}\right)\,\mathrm dy\\ \overset{yi/2\to s}{=}\frac{2^{a+b-1}e^{x(\tau-\pi/2)}}{4\pi^2i\Gamma(a)\Gamma(b)}\int_{-i\infty}^{i\infty}\Gamma\left(\frac{a}{2}+s\right)\Gamma\left(\frac{a}{2}-s\right)\Gamma\left(\frac{b+ix}{2}-s\right)\Gamma\left(\frac{b-ix}{2}+s\right)\,\mathrm ds$$ By the first Barnes lemma we have: $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\Gamma\left(\frac{a}{2}+s\right)\Gamma\left(\frac{a}{2}-s\right)\Gamma\left(\frac{b+ix}{2}-s\right)\Gamma\left(\frac{b-ix}{2}+s\right)\,\mathrm ds\\ =\frac{\Gamma\left(\frac{a+b+ix}{2}\right)\Gamma(a)\Gamma\left(\frac{a+b-ix}{2}\right)\Gamma(b)}{\Gamma(a+b)}$$ Which is valid since $\Re(a),\Re(b)>0$ Therefore $$\int_{\mathbb{R}}f_a(y)f_b(x-y)\,\mathrm dy=\frac{2^{a+b-1}e^{x(\tau-\pi/2)}\Gamma\left(\frac{a+b+ix}{2}\right)\Gamma\left(\frac{a+b-ix}{2}\right)}{2\pi\Gamma(a+b)}=f_{a+b}(x)$$

Answer 2

Fix $\tau$ with $0<\tau<\pi$, and assume $\Re(a),\Re(b)>0$. For each such parameter, define the horizontal contour $H:=\left\{u=t+i\left(\frac{\pi}{2}-\tau\right):t\in\mathbb{R}\right\}$ and set $$F_a(u):=\frac{1}{\cosh^a(u)}$$ with $u\in H$. Since $\cosh(z)$ has no zeros or singularities for $|\Im(z)|<\frac{\pi}{2}$, $F_a(u)$ is analytic on and near the strip containing $H$. Moreover, as $|t|\to\infty$, $$\left|\cosh(t+i\theta)\right|=\frac{1}{2}\left|e^{t+i\theta}+e^{-t-i\theta}\right|\sim\frac{1}{2}e^{|t|},$$ so $\left|F_a\left(t+i\left(\frac{\pi}{2}-\tau\right)\right)\right|=O\left(e^{-a|t|}\right)$, and in particular, $F_a\in L^1(H)$. By the standard Fourier inversion theorem for $L^1$-kernels on a horizontal line, $$f_a(x)=\frac{1}{2\pi}\int_{u\in H}F_a(u)e^{ixu}\,\mathrm{d}u=\frac{1}{2\pi}\int_{-\infty+i\left(\frac{\pi}{2}-\tau\right)}^{\infty+i\left(\frac{\pi}{2}-\tau\right)}\frac{e^{ixu}}{\cosh^a(u)}\,\mathrm{d}u,$$ i.e. we generalise the identity $$\frac{2^{a-1}}{\Gamma(a)}\Gamma\left(\frac{a+ix}{2}\right)\Gamma\left(\frac{a-ix}{2}\right)=\int_{-\infty}^{\infty}\frac{e^{ixy}}{\cosh^a(y)}\,\mathrm{d}y$$ valid for $\Re(a)>0$, obtainable through the beta function with the substitution $e^{2y}=\frac{t}{1-t}$. Now, \begin{align*}\left(f_a * f_b\right)(x)&=\int_{y=-\infty}^{\infty}f_a(y)f_b(x-y)\,\mathrm{d}y\\ &=\int_{-\infty}^{\infty}\left[\frac{1}{2\pi}\int_{u\in H}F_a(u)e^{iyu}\,\mathrm{d}u\right]\left[\frac{1}{2\pi}\int_{v\in H}F_b(v)e^{i(x-y)v}\,\mathrm{d}v\right]\mathrm{d}y\\ &=\frac{1}{\left(2\pi\right)^2}\int_{y\in\mathbb{R}}\int_{u\in H}\int_{v\in H}F_a(u)F_b(v)e^{iyu}e^{i(x-y)v}\,\mathrm{d}v\,\mathrm{d}u\,\mathrm{d}y.\end{align*} We now check for absolute convergence of the triple integral. Write $u=t+i\alpha$, $v=s+i\alpha$ with $\alpha=\frac{\pi}{2}-\tau$. Let $\varepsilon>0$, then, $$\left|e^{iyu}\right|=e^{-y\Im(u)}=e^{-y\alpha},\quad\left|e^{i\left(x-y\right)v}\right|=e^{-\left(x-y\right)\alpha},$$ and $|F_a(u)|=O\left(e^{-a|t|}\right)$, $|F_b(v)|=O\left(e^{-b|s|}\right)$. Hence, $$\int_{\mathbb{R}^3}\left|F_a(u)F_b(v)e^{iyu}e^{i(x-y)v}e^{-\varepsilon y^2}\right|\mathrm{d}t\,\mathrm{d}s\,\mathrm{d}y\leq Ce^{-x\alpha}\int_{\mathbb{R}^3}e^{-a|t|-b|s|}e^{-\varepsilon y^2}\,\mathrm{d}t\,\mathrm{d}s\,\mathrm{d}y$$ for some $C>0$, which factorises as product of three convergent one-dimensional integrals, so by Fubini-Tonelli, we may freely reorder the $u,v$, and $y$ integrals. Evaluating the $y$ integral first gives $$\int_{y=-\infty}^{\infty}e^{iyu}e^{-iyv}e^{-\varepsilon y^2}\,\mathrm{d}y=\int_{-\infty}^{\infty}e^{iy(u-v)}e^{-\varepsilon y^2}\,\mathrm{d}y=\sqrt{\frac{\pi}{\varepsilon}}\exp\left(-\frac{\left(u-v\right)^2}{4\varepsilon}\right).$$ Now one may show that by the dominated convergence theorem, we may pass the limit $\varepsilon\to 0^+$ through the double integral to recover our original integral of interest. In the sense of distributions, recall that $$\lim_{\varepsilon\to 0^+}\sqrt{\frac{\pi}{\varepsilon}}\exp\left(-\frac{\left(u-v\right)^2}{4\varepsilon}\right)=2\pi\,\delta(u-v).$$ Hence, in the limit $\varepsilon\to 0^+$, the triple integral collapses to \begin{align*}\left(f_a*f_b\right)(x)&=\frac{1}{\left(2\pi\right)^2}\int_{u\in H}\int_{v\in H}F_a(u)F_b(v)e^{ixv}\cdot 2\pi\,\delta(u-v)\,\mathrm{d}v\,\mathrm{d}u\\ &=\frac{1}{2\pi}\int_{u\in H}F_a(u)F_b(u)e^{ixu}\,\mathrm{d}u.\end{align*} Now since $$F_a(u)F_b(u)=\left(\cosh^{-a}(u)\right)\left(\cosh^{-b}(u)\right)=\cosh^{-(a+b)}(u)=F_{a+b}(u),$$ we conclude that the last line is exactly $$\frac{1}{2\pi}\int_{u\in H}\frac{e^{ixu}}{\cosh^{a+b}(u)}\,\mathrm{d}u=f_{a+b}(x).$$ Thus indeed $f_a*f_b=f_{a+b}$, as claimed. $\square$

Answer 3

Here is another solution taken from my class on exponential families in statistics.

Theorem:

Let $t>0.$ The natural exponential family $F_t$ with domain of the means $\mathbb{R}$ and variance function $t(1+\frac{m^2}{t^2})$ is generated by the probability $$\mu_t(dx)=\frac{2^{t-2}}{\pi}\left|\Gamma(\frac{t+ix}{2})\right|^2\frac{1}{\Gamma(t)}dx.$$ Its Laplace transform is defined on $\Theta(\mu_t)=(-\frac{\pi}{2},\frac{\pi}{2})$ and is $(\cos \theta)^{-t}.$ In particular, for $t=1$ and $t=2$ we have $$\mu_1(dx)=\frac{dx}{2\cosh \frac{\pi x}{2}},\ \ \ \ \mu_2(dx)=\frac{xdx}{2\sinh \frac{\pi x}{2}}$$

Proof:

We adopt the notations of the Letac-Mora paper (1991): if $\mu$ is a positive non Dirac Radon measure on $\mathbb{R}$ consider its Laplace transform $$L_{\mu}(\theta)=\int_{-\infty}^{\infty}e^{\theta x}\mu(dx)\leq \infty.$$ Assume that the interior $\Theta(\mu)$ of the interval $D(\mu)=\{\theta\in \mathbb{R}; L_{\mu}(\theta)<\infty\}$ is not empty. Write $k_{\mu}=\log L_{\mu}.$ Then the family of probabilities $$F=F(\mu)=\{P(\theta,\mu)\ ;\ \theta\in \Theta(\mu)\}$$ where $$P(\theta,\mu)(dx)=e^{\theta x-k_{\mu}(\theta)}\mu(dx)$$ is called the natural exponential family (NEF) generated by $\mu.$ Note that $F(\mu)=F(\nu)$ if and only if there exists $a$ and $b$ such that $\nu(dx)=e^{ax+b}\mu(dx).$ This implies that $F(\mu)$ can be generated by one of its members as well as sometimes an unbounded measure.

Two basic results are $k'_{\mu}(\theta)=\int_{-\infty}^{\infty}xP(\theta,\mu)(dx)$ and the fact that $k'_{\mu}$ is increasing (or that $k_{\mu}$ is convex. The set $k'_{\mu}(\Theta(\mu))=M_F$ is called the domain of the means. We denote by $\psi_{\mu}:M_F\rightarrow \Theta(\mu)$ the reciprocal function of $k'_{\mu}$. Thus $F(\mu)$ can be parametrized by $M_F$ by the map from $M_F$ to $F$ which is $$m\mapsto P(\psi_{\mu}(m),\mu)=P(m,F).$$ One can prove that the variance $V_F(m)$ of $P(m,F$ is $$V_F(m)=k''_{\mu}(\psi_{\mu}(m))=\frac{1}{\psi'_{\mu}(m)}.\ \ (*)$$ The map $m\mapsto V_F(m)$ from $M_F$ to $(0,\infty)$ is called the variance function and characterizes $F.$

Let us apply these concepts to the finding of a generating measure $\mu$ for the variance function $t(1+\frac{m^2}{t^2}).$ We use (*) for writing with $m=tu$ $$d\theta=\psi'_{\mu}(m)dm=\frac{dm}{t(1+\frac{m^2}{t^2})}=\frac{du}{1+u^2}=d \arctan u.$$ Thus $m=t\tan \theta=k'_{\mu}(\theta)$ and $k_{\mu}(\theta)=-t\log \cos \theta$ on $\Theta(\mu)=(-\frac{\pi}{2},\frac{\pi}{2}).$ The remark about $F(\mu)=F(\nu)$ leads us to ignore the two integration constants in the process. Note that here $k_{\mu}(0)=0$ implies that the chosen $\mu$ is a probability if it exists.

The interesting point is now to prove that $\mu$ indeed exists and to compute it. For this the analytic character of the Laplace transform enables us to declare that the Fourier transform of $\mu$ is $$s\mapsto \varphi(s)=\frac{1}{(\cosh s)^t}.$$ Luckily, this is an integrable function and the Fourier inversion formula applies: the density of $\mu$ should exists and should be $$f_t(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-isx}\frac{1}{(\cosh s)^t}ds.$$ The trick to compute this integral in a elementary way is to write $v=e^{2s}.$ We get

$$f_t(x)=\frac{2^{t-2}}{\pi}\int_{0}^{\infty} \frac{v^{\frac{t-ix}{2}-1}dv}{(1+v)^t}$$ We rely now on two formulas about the Gamma function

$$\int_{0}^{\infty} \frac{v^{a-1}dv}{(1+v)^{a+b}}=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},\ \ \Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin \pi p},$$ where the real or complex numbers $a,b,p,1-p$ have a positive real part. Applying the first formula to $a=\frac{t-ix}{2}$ and to $b=\bar{a}=\frac{t+ix}{2}$ and using the fact that $\Gamma(a)=\overline{\Gamma(\bar{a})}$ we get the announced value of $f_t(x)dx=\mu_t(dx)$ which is indeed positive. To get the last formulas for $t=1$ use $\Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin \pi p}$ with $p=\frac{1-ix}{2}$ and the sinus of a complex argument. For $t=2$ use $p=ix/2$ and $\Gamma(1+\frac{ix}{2})=\frac{ix}{2}\Gamma(\frac{ix}{2}).$

Answer 4

I just wanted to mention the link with the Beta function, supplementing KStar's answer. For simplicity, we consider the case $\tau = \pi/2$. \begin{align} \int_{-\infty}^\infty\frac{e^{ixy}}{\cosh^a(y)}\text dy &= 2^a\int_{-\infty}^\infty \frac{e^{ixy}}{e^{-ay}(1+e^{2y})^a}\text dy \\ &= 2^a\int_{-\infty}^\infty(e^y)^{a+ix} \left(1-\frac{e^{2y}}{1+e^{2y}}\right)^a\text dy \end{align} Setting $u = \frac{e^{2y}}{1+e^{2y}}$ (or equivalently $e^{2y} = \frac{u}{1-u}$), we have $\text du = 2\frac{e^{2y}}{(1+e^{2y})^2} \text dy = 2u(1-u)\text dy$, therefore \begin{align} \int_{-\infty}^\infty\frac{e^{ixy}}{\cosh^a(y)}\text dy &= 2^a\int_0^1 \left(\frac{u}{1-u}\right)^{\frac{a+ix}{2}} (1-u)^a\frac{\text du}{2u(1-u)} \\ &= 2^{a-1} \int_0^1 u^{\frac{a+ix}{2}-1} (1-u)^{\frac{a-ix}{2}-1} \text du \\ &= 2^{a-1} \mathop {\text B}\left(\frac{a+ix}{2},\frac{a-ix}{2}\right) \\ &= 2^{a-1} \frac{\Gamma\left(\frac{a+ix}{2}\right) \Gamma\left(\frac{a-ix}{2}\right)}{\Gamma(a)}. \end{align} And by the simple computation in the Fourier domain that KStar did, i.e., $\hat{f_a}\cdot\hat{f_b} = \hat{f}_{a+b}$, where $\hat{f_a}(x) = 1/(2\pi \cosh(x)^a)$, $f_a * f_b = f_{a+b}$ follows.