  We prove that the auxiliary function $\mathop{\mathcal R}(s)$ has the
integral representation \[\mathop{\mathcal R}(s)=-\frac{2^s \pi^{s}e^{\pi i
s/4}}{\Gamma(s)}\int_0^\infty y^{s}\frac{1-e^{-\pi y^2+\pi \omega y}}{1-e^{2\pi
\omega y}}\,\frac{dy}{y},\qquad \omega=e^{\pi i/4}, \quad\Re s>0,\] valid for
$\sigma>0$. The function in the integrand $\frac{1-e^{-\pi y^2+\pi \omega
y}}{1-e^{2\pi \omega y}}$ is entire. Therefore, no residue is added when we
move the path of integration.
