I am trying to understand a derivation in Ma and Chen (2013), in particular one step in their example 3.1 on page 3177. The step that I am having problems understanding is part of Equation 24 on page 9 of the PDF: $$\int_0^\infty e^{-ix} x^{-1/2-it} dx = e^{-i\pi(1/2-it)/2} \int_0^\infty e^{-x} x^{-1/2-it} dx$$ where $t\in \mathbb{R}$.
If I rewrite $$\int_0^\infty e^{-ix} x^{-1/2-it} dx = \int_0^\infty e^{-ix} x^{(1/2-it)-1} dx,$$ then I get the Mellin transform of the function $e^{-ix}$, which is $$i^{-(1/2-it)}\Gamma(1/2-it)$$ However, looking at their answer and mine, there is a discrepancy in the factors multiplying the Gamma function. I get $i^{-(1/2-it)}$ while they get $e^{-i\pi(1/2-it)/2}$.
To reconcile the two factors, let $a \equiv 1/2-it$, so that $$i^{-a} = (-1)^{-a/2} = \left( \frac{1}{e^{i\pi}} \right)^{a/2} = e^{-i\pi a/2}.$$
Is this correct? or am I missing something obvious and going the Mellin transform route is just a booby trap that my brain decided to set-up for me?
The reason for which I would like to understand this is that I would like to check the conditions of their main theorem for something like $$\int_\mathbb{R} f(k) \left( \int_0^\infty e^{-ikx} x^{-1/2-it} dx \right) dk$$ where $k \in \mathbb{R}$ where $f$ is an $L_1$ positive function.
Any suggestions for understanding their example would be greatly appreciated. If suggestions for how to approach the second equality above are given, that would be extra fantastic :) Thank you!
The Mellin transform of $e^{-i x}$ is as follows:
$$\int_0^\infty e^{-i x} x^{s-1}\,dx=e^{-\frac{1}{2} i \pi s}\ \Gamma(s)\ ,\quad 0<\Re(s)<1\tag{1}$$
Evaluating at $s=\frac{1}{2}-i t$ leads to
$$\int_0^\infty e^{-i x} x^{-\frac{1}{2}-i t}\,dx=\int_0^\infty e^{-i x} x^{(\frac{1}{2}-i t)-1}\,dx=e^{-\frac{1}{4} \pi (2 t+i)} \Gamma\left(\frac{1}{2}-i t\right)\tag{2}$$