$\mathcal{X}\left(f\right):$

$\zeta \left(-2\pi i\alpha +\frac{1}{2}\right)={\displaystyle {\sum }_{\left|x\right|=1}^{\left|x\right|=\infty }{\left|x\right|}^{2\pi i\alpha -\frac{1}{2}}}$

$\stackrel{U_x^{\xi ,r}:}{\overbrace{\to }}$

$U_f^{\xi ,r}\mathcal{X}\left(f\right)=a^{r+1}{\text{e}}^{-2\pi i\xi \left(1-a\right)f}\mathcal{X}\left(af\right):$

${\displaystyle {\sum }_{\left|x\right|=1}^{\left|x\right|=\infty }{\text{e}}^{-2\pi i\alpha \left|x\right|}}$

$\downarrow$

(warping operator, $r=-\frac{1}{2}$ , $\xi =0$ , $a={\text{e}}^{-x}$ ; $U_x^{\xi =0,r=-\frac{1}{2}}=U_x$ )

$\downarrow$

${\mathcal{M}}_f^{\xi }\left[\mathcal{X}\left(f\right)\right]=\mathcal{F}{U}_n^{\xi ,r}\left[\mathcal{X}\left(f\right)\right]:$

$\mathcal{F}{\displaystyle {\sum }_{\left|x\right|=1}^{\left|x\right|=\infty }{\text{e}}^{-2\pi i\alpha \left|x\right|}}={\displaystyle {\sum }_{\left|x\right|=1}^{\left|x\right|=\infty }\delta \left(\left|x\right|-\alpha \right)}$

with $\mathcal{F}$ the Fourier Transform with ordinary unitary frequency

$\underset{{\mathcal{M}}_x}{\underbrace{\to }}$

$\begin{array}{l}{\mathcal{M}}_f^{\xi }\left[U_x^{\xi ,r}\mathcal{X}\right]\left(\beta \right)\\ =a^{-2\pi i\beta }{\mathcal{M}}_f^{\xi }\left[\mathcal{X}\right]\left(\mu \right)\cdot {\displaystyle {\int }_0^{\infty }{\displaystyle {\sum }_{\left|x\right|=1}^{\left|x\right|=\infty }{\text{e}}^{-2\pi i\beta \left|x\right|}{\left|x\right|}^{\mu -1}\text{d}x}}\\ ={\displaystyle {\int }_0^{\infty }{\displaystyle {\sum }_{\left|x\right|=1}^{\left|x\right|=\infty }\delta \left(\left|x\right|-\beta \right){\left|x\right|}^{\mu -1}\text{d}x}}\\ =\zeta \left(1-\mu \right)\end{array}$

The result is equivalent with $-1^{-\mu }{i}^{\mu }\frac{\Gamma \left(\mu +\alpha \right)}{\Gamma \left(\alpha \right)}\varsigma \left(\mu +\alpha ,\beta \right)$ for $\xi \ne 0$ , $\mu =2\pi i\beta +r$ , $a={\text{e}}^{-x}$ and initial input $\varsigma \left(\alpha \right)$ . Hence the factor $a^{-2\pi i\beta }$ may be moved to the integrand of the found Fourier Transform $\begin{array}{l}{\displaystyle {\int }_{-\infty }^{\infty }{\displaystyle {\sum }_{\left|f\right|=1}^{\left|f\right|=\infty }{\left|f\right|}^{-\alpha }{\text{e}}^{-2\pi i\left(\xi +\beta \right)f}{f}^{\mu }\text{d}f}}\\ =2\pi i^{-\mu }{\displaystyle {\sum }_{\left|\xi \right|=1}^{\left|\xi \right|=\infty }\frac{\Gamma \left(\mu +\alpha \right)}{\Gamma \left(\alpha \right)}{\left|\xi +\beta \right|}^{-\mu -\alpha }}\\ =2\pi i^{-\mu }\frac{\Gamma \left(\mu +\alpha \right)}{\Gamma \left(\alpha \right)}\left\{\varsigma \left(\mu +\alpha ,\beta \right)-{\left|\beta \right|}^{-\mu -\alpha }\right\}\end{array}$ (71),

with $\varsigma \left(s,\lambda \right)$ the Hurwitz Zeta function (100).