= Solution
The one-loop <one-particle irreducible> correction is a <tadpole diagram>: two external lines meet one <quartic interaction> and the other two legs are joined. Interchanging the two ends of its loop gives the <symmetry factor> $2$.
\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2023-scalar-diagrams.png]
{title=Original scalar diagrams: the two-dimensional tadpole, charge-conserving light-scalar exchange, and heavy-scalar box amplitudes. Dashed lines carry charge two; solid lines carry charge one.}
Define the <self-energy> by an insertion $-i\Sigma(p^2)$ and a full <scalar propagator> $i/[p^2-m^2-\Sigma(p^2)+i\epsilon]$ before <counterterms>. The supplied <cutoff regularization> integral gives
$$
T_\Lambda:=\int_{|k_E|<\Lambda}\frac{d^2k_E}{(2\pi)^2}\frac1{k_E^2+m^2}
=\frac1{4\pi}\log\left(1+\frac{\Lambda^2}{m^2}\right),
\qquad
-i\Sigma_1=\frac{-ig_4}{2}\int\frac{d^2k}{(2\pi)^2}\frac{i}{k^2-m^2+i\epsilon}.
$$
Thus
$$
\boxed{\Sigma_1(p^2)=\frac{g_4}{2}T_\Lambda
=\frac{g_4}{8\pi}\log\left(1+\frac{\Lambda^2}{m^2}\right).}
$$
It has no external <momentum> dependence. The correction to the unamputated <two-point correlation function> in <momentum space> is
$$
\Delta G(p)=\frac{i}{p^2-m^2+i\epsilon}[-i\Sigma_1]
\frac{i}{p^2-m^2+i\epsilon}
=\frac{i\Sigma_1}{(p^2-m^2+i\epsilon)^2}.
$$
The positive sign of $\Sigma_1$ raises the unrenormalized squared <pole mass> for $g_4>0$.
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