<s>[INST] Sign of integral of $\frac{ 2 ^{\frac{it}{2/3}} \Gamma ( \frac{it +1}{2/3}) }{ 2 ^{\frac{it}{1.5}} \Gamma ( \frac{it +1}{1.5}) } \frac{1}{(a+it)^k}$ Can we determine the sign of the following function
\begin{align}
f(a,k)=\int_{-\infty}^\infty  \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right)  }{  2 ^{\frac{it}{3/2}}  \Gamma \left( \frac{it +1}{3/2}\right) }   \frac{1}{(a+it)^k} dt,
\end{align} 
where $a\neq 0$ and $k \ge 1$ is some positive integer. 
The conjecture is that the sign of the integral is equal to
 \begin{align}
{\rm sign } (f(a,k))={\rm sign}(a)^k.
\end{align} 
Perhpas the following limit can be usefull. By using a  method  in this question  it is not difficult to see that
\begin{align}
\left |  \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right)  }{  2 ^{\frac{it}{3/2}}  \Gamma \left( \frac{it +1}{3/2}\right) }  \right| \to O( e^{- (\frac{3}{2}-\frac{2}{3}) t}) \text{ as } t \to \infty.
\end{align}
Thanks
 [/INST] Not quite an answer, but a good start. Let's look at
$$f(a,k)=\int_{-\infty}^\infty  \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right)  }{  2 ^{\frac{it}{3/2}}  \Gamma \left( \frac{it +1}{3/2}\right) }   \frac{1}{(a+it)^k} dt$$
which can be rewritten as
$$f(a,k)= (-i) \int_{-i\infty}^{i\infty}   2 ^{\frac{5z}{6}}\frac{ \Gamma \left( \frac{3(z +1)}{2}\right)  }{ \Gamma \left( \frac{2(z +1)}{3}\right) }   \frac{1}{(a+z)^k} dz$$
Observe that $\Gamma \left( \frac{3(z +1)}{2}\right)$ has poles in the left half plane when $\frac{3(z+1)}{2} = n$ and $ n = 0, -1, -2, -3,....$, so when $z = \frac{2}{3}n - 1$ we have a pole. The principal part is
$$\frac{(-1)^n}{n!(\frac{3(z+1)}{2} +n)} = \frac{2}{3}\frac{(-1)^n}{n!(z + 1 + \frac{2}{3}n)}$$
Take a semicircle contour that grows in the left half plane. A simple exercise in Mellin transforms gives that, if $a < 0$
$$f(a,k) = \frac{4\pi}{3}\sum_{n=0}^\infty \frac{(-1)^n2^{-\frac{5}{6}(1+\frac{2}{3}n)}}{n!\Gamma(-\frac{4}{9}n)(a-1-\frac{2}{3}n)^k}$$
Now showing that $\text{sign}(f(a,k)) = \text{sign}(a)^k$ involves talking about this series. Note that some of the terms disappear if $n = 0 \,\mod 9$ (because the Gamma function on the bottom vanishes there).  Not sure how you would really approach this, but probably discussing that this series oscillates wildly has something to do with it. It seems obvious though that if $a<0$ then $\text{sign}(f(a,k))^k = \text{sign}(a)^k$.
EDIT: If $a > 0$ there's another term added because $\frac{1}{(a+z)^k}$ has a pole in the left half plane. This needs to be handled in cases, because when $a = -\frac{2}{3}n -1$ the residues get all wonky. I'll leave it to you to find that extra term, which isn't too hard to get at. It just involves taking the $k$'th derivative of the rest of the integrand (something I'm not in the mood to do).
PS: I may have screwed up some arithmetic; the amount of fractions I just crossed out and rearranged in my head had me blurry eyed.
</s>