Positivity of $ int_{-infty}^{infty} left{{2^{1/beta-1/2} over v}right}^{it} { Gamma{(it+1)/beta}over Gamma{(it+1)/2} }dt$

$newcommandGaGamma newcommand{R}{mathbb{R}} newcommand{de}{delta} newcommand{ga}{gamma} newcommand{Si}{Sigma}$ We have to show that for $a:=-ln(2^{1/b-1/2}/v)inR$ and $b:=betain(1,2)$, begin{equation*} I(a):=int_{-infty}^{infty} e^{-iat}R(t),dt>0, tag{1} end{equation*} where begin{equation*} R(t):=frac{Gabig((1+it)/bbig)}{Gabig((1+it)/2big)}. tag{2} end{equation*}

The key is Euler's product formula
begin{equation*} Ga(z)=frac1z,prod_{j=1}^inftyfrac{(1+1/j)^z}{1+z/j} end{equation*} for $zinmathbb Csetminus{0,-1,-2,dots}$, which yields begin{equation*} frac{Ga(s+it)}{Ga(s)}=prod_{j=1}^infty(1+1/j)^{it} Big/prod_{j=0}^inftyBig(1+frac{it}{j+s}Big); tag{3} end{equation*} here and in what follows, $s$ is any positive real number and $t$ is any real number.

Based on (3), it is easy to obtain

Lemma 1: $ln|Ga(s+it)|sim-pi|t|/2$ as $|t|toinfty$.

The proof of Lemma 1 will be given at the end of this answer.

It also follows from (3) that begin{equation*} R(t)=cprod_{j=1}^infty(1+1/j)^{iht}f_j(t), tag{4} end{equation*} where $c:=Ga(1/b)/Ga(1/2)>0$, begin{equation} h:=frac1b-frac12=frac{2-b}{2b}, end{equation} and begin{equation} f_j(t):=frac{1+it/(1+2j)}{1+it/(1+bj)}, end{equation} so that $f_j$ is the characteristic function (c.f.) of a random variable (r.v.) $X_jsim p_jde_0+(1-p_j)Exp(-1/(1+bj))$, where in turn $p_j:=(1+bj)/(1+2j)in(0,1)$, $de_0$ is the Dirac distribution supported on the set ${0}$, and $Exp(-1/(1+bj))$ is the exponential distribution with mean $-1/(1+bj)$, supported on the interval $(-infty,0]$. Here and in what follows, $j$ is any natural number. Note that $EX_j=-frac{hj}{(j+1/b)(j+1/2)}$ and $Var,X_jle1/(bj)^2le1/j^2$. So, the series begin{equation} sum_{j=1}^infty(X_j-EX_j)=:S end{equation} converges almost surely. Hence, by (4) begin{equation*} R(t)=ce^{ihc_1t}f_S(t), end{equation*} where $f_S$ is the c.f. of the r.v. $S$ and begin{equation} c_1:=sum_{j=1}^inftyBig(ln(1+1/j)+EX_jBig) \ =sum_{j=1}^inftyBig(ln(1+1/j)-frac{j}{(j+1/b)(j+1/2)}Big)inR end{equation} (in fact, $c_1=(ga b-2 b+b ln4+2 psileft(1+1/bright))/(2-b)$, where $ga=0.577dots$ is the Euler constant and $psi:=Ga'/Ga$; however, the actual value of $c_1$ does not matter here).

So, $R$ is the c.f. of the r.v. $T:=hc_1+S$. Also, by Lemma 1, $R$ is in $L^1$. It now follows that the function $I$, defined by (1), is $2pi$ times the density of the r.v. $T$. Thus, $I(a)ge0$ for all real $a$, as desired.

It remains to provide

Proof of Lemma 1: By (3), begin{equation*} frac{|Ga(s+it)|}{Ga(s)}=prod_{j=0}^inftyfrac{j+s}{|j+s+it|} =exp{-Si_{s,t}/2}, end{equation*} where begin{equation} Si_{s,t}:=sum_{j=0}^inftylnBig(1+frac{t^2}{(j+s)^2}Big). end{equation} Since $lnbig(1+frac{t^2}{(j+s)^2}big)$ is nonincreasing in $j$, we have begin{equation} J_{s,t}leSi_{s,t}le J_{s,t}+lnbig(1+frac{t^2}{s^2}big), end{equation} where begin{equation} J_{s,t}:=int_0^inftylnbig(1+frac{t^2}{(x+s)^2}big),dxsimpi|t| end{equation} as $|t|toinfty$, which completes the proof of Lemma 1 and the entire answer. (In fact, integrating by parts, for $tne0$ we find begin{equation} J_{s,t}=pi|t|-s ln left(s^2+t^2right)-2 t arctan(s/t)+2 s ln ssimpi|t|.) end{equation} The proof of Lemma 1 and the entire answer are now complete.