*Warning: The following contains formal manipulations that ignore convergence.*

**Proposition:**

\begin{align}
\mathrm{regularized} \int_0^\infty \mathrm{d}x^s = 0
\end{align}
for all $s$ such that $\Re(s) \neq 0$.

**“Proof” 1:**

\begin{align}
\int_0^\infty \mathrm{d}x^s
&= \int_0^1 \mathrm{d}x^s + \int_1^\infty \mathrm{d}x^s \\
&= 1 - 1 \\
&= 0
\end{align}

where we simultaneously assumed $\Re(s) > 0$ and $\Re(s) < 0$ for the first and second integrals, respectively.

**“Proof” 2:**

Let
\begin{align}
f_\pm(\eta) = \mathrm{e}^{-\eta} \left( 1 \pm \frac{\eta}s \right)
\end{align}

Case $\Re(s) > 0$:
\begin{align}
\int_0^\infty x^{s-1} \mathrm{d}x &= \int_0^\infty \lim_{\varepsilon \downarrow 0} x^{s-1} f_-(\varepsilon x) \mathrm{d}x \\
&\stackrel{!}{=} \lim_{\varepsilon \downarrow 0} \int_0^\infty x^{s-1} f_-(\varepsilon x) \mathrm{d}x \\
&= \lim_{\varepsilon \downarrow 0} 0 & \Re(s) > 0, \Re(\varepsilon) > 0 \\
&= 0
\end{align}

Case $\Re(s) < 0$:
\begin{align}
\int_0^\infty x^{s-1} \mathrm{d}x
&= \int_0^\infty \lim_{\varepsilon \downarrow 0} x^{s-1} f_+(\varepsilon x^{-1}) \mathrm{d}x \\
&\stackrel{!}{=} \lim_{\varepsilon \downarrow 0} \int_0^\infty x^{s-1} f_+(\varepsilon x^{-1}) \mathrm{d}x \\
&= \lim_{\varepsilon \downarrow 0} 0 & \Re(s) < 0, \Re(\varepsilon) > 0 \\
&= 0
\end{align}

See here for more context about this regulator-based approach.

Applying linearity of integration yields your equalities.