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$ I have the following function
$$
\int_{-\infty}^{\infty} \left\{{2^{1/\beta-1/2} \over v}\right\}^{it}
{ \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} }dt
$$
where $1<\beta<2$, $v>0$. Need to show it is positive.
The inverse Mellin transform of
$$
 \left\{2^{1/\beta-1/2} \right\}^{it}
{ \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} }
$$
is
$$
{C \over v}\int_{-\infty}^{\infty} \left\{{2^{1/\beta-1/2} \over v}\right\}^{it}
{ \Gamma\{(it+1)/\beta\}\over \Gamma\{(it+1)/2\} }dt
$$
 A: $\newcommand\Ga\Gamma
\newcommand{\R}{\mathbb{R}}
\newcommand{\de}{\delta}
\newcommand{\ga}{\gamma}
\newcommand{\Si}{\Sigma}$
We have to show that for $a:=-\ln(2^{1/b-1/2}/v)\in\R$ and $b:=\beta\in(1,2)$,
\begin{equation*}
I(a):=\int_{-\infty}^{\infty} e^{-iat}R(t)\,dt>0,   \tag{1}
\end{equation*}
where
\begin{equation*}
    R(t):=\frac{\Ga\big((1+it)/b\big)}{\Ga\big((1+it)/2\big)}.  \tag{2}
\end{equation*}
The key is Euler's product formula
\begin{equation*}
    \Ga(z)=\frac1z\,\prod_{j=1}^\infty\frac{(1+1/j)^z}{1+z/j}
\end{equation*}
for $z\in\mathbb C\setminus\{0,-1,-2,\dots\}$, which yields
\begin{equation*}
    \frac{\Ga(s+it)}{\Ga(s)}=\prod_{j=1}^\infty(1+1/j)^{it}
    \Big/\prod_{j=0}^\infty\Big(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 $|t|\to\infty$.

The proof of Lemma 1 will be given at the end of this answer.
It also follows from (3) that
\begin{equation*}
    R(t)=c\prod_{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.) $X_j\sim p_j\de_0+(1-p_j)Exp(-1/(1+bj))$, 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 $Var\,X_j\le1/(bj)^2\le1/j^2$. 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}^\infty\Big(\ln(1+1/j)+EX_j\Big) \\ 
    =\sum_{j=1}^\infty\Big(\ln(1+1/j)-\frac{j}{(j+1/b)(j+1/2)}\Big)\in\R
\end{equation}
(in fact, $c_1=(\ga  b-2 b+b \ln4+2 \psi\left(1+1/b\right))/(2-b)$, where $\ga=0.577\dots$ 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 $2\pi$ 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}^\infty\frac{j+s}{|j+s+it|}
    =\exp\{-\Si_{s,t}/2\}, 
\end{equation*}
where
\begin{equation}
    \Si_{s,t}:=\sum_{j=0}^\infty\ln\Big(1+\frac{t^2}{(j+s)^2}\Big). 
\end{equation}
Since $\ln\big(1+\frac{t^2}{(j+s)^2}\big)$ is nonincreasing in $j$, we have
\begin{equation}
    J_{s,t}\le\Si_{s,t}\le J_{s,t}+\ln\big(1+\frac{t^2}{s^2}\big), 
\end{equation}
where
\begin{equation}
    J_{s,t}:=\int_0^\infty\ln\big(1+\frac{t^2}{(x+s)^2}\big)\,dx\sim\pi|t|
\end{equation}
as $|t|\to\infty$, which completes the proof of Lemma 1 and the entire answer.
(In fact, integrating by parts, for $t\ne0$ we find
\begin{equation}
    J_{s,t}=\pi|t|-s \ln \left(s^2+t^2\right)-2 t \arctan(s/t)+2 s \ln s\sim\pi|t|.) 
\end{equation}
The proof of Lemma 1 and the entire answer are now complete.
