= Solution
The positive denominators in this part are \b[Euclidean]. Starting with the <Minkowski spacetime> <Feynman rules>, first perform a <Wick rotation>; a bare <Minkowski metric> integral without a pole prescription would not equal the displayed positive expression. Work initially with $m^2>0$ and $\Re n>D/2$, where the <Euclidean massive loop integral> converges.
Use <Schwinger parameterization> followed by the <Gaussian integral>:
$$
\frac1{(k^2+m^2)^n}=\frac1{\Gamma(n)}\int_0^\infty ds\,s^{n-1}e^{-s(k^2+m^2)},
\qquad
\int\frac{d^Dk}{(2\pi)^D}e^{-sk^2}=\frac1{(4\pi s)^{D/2}}.
$$
The remaining <Gamma function> integral gives
$$
\begin{aligned}
I_{D,n}(m^2)
&=\frac1{(4\pi)^{D/2}\Gamma(n)}\int_0^\infty ds\,s^{n-D/2-1}e^{-sm^2}\\
&=\boxed{\frac{\Gamma(n-D/2)}{(4\pi)^{D/2}(n-1)!}(m^2)^{D/2-n}}.
\end{aligned}
$$
The factorial form is for positive integral $n$. <Dimensional regularization> defines the result elsewhere by <analytic continuation>.
For the <scalar bubble integral>, combine the two denominators with a <Feynman parameter> $x$ and shift the integration variable to $\ell=k-(1-x)p_E$. Then
$$
B_D(p_E)=\int_0^1dx\,I_{D,2}(M_x^2),
\qquad M_x^2=m^2+x(1-x)p_E^2.
$$
For $D=4-\epsilon$ this is
$$
B_D(p_E)=\frac{\Gamma(\epsilon/2)}{(4\pi)^{2-\epsilon/2}}\int_0^1dx\,(M_x^2)^{-\epsilon/2}.
$$
The <Gamma function> recurrence gives $\Gamma(\epsilon/2)=2/\epsilon+O(1)$, and the parameter integral is $1+O(\epsilon)$. Consequently
$$
\boxed{B_D(p_E)=\frac1{8\pi^2\epsilon}+O(1).}
$$
This <ultraviolet divergence> is independent of the external <momentum>. The assumption $m^2>0$ avoids an <infrared divergence>. In particular the completely massless zero-momentum integral is a <scaleless integral> and vanishes in <dimensional regularization>; its ultraviolet and infrared poles must be separated before assigning it the displayed ultraviolet residue.
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