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Put $I_0=\sqrt{2\pi /m^2}$ and assume $m^2>0$. The normalized free Gaussian integral has covariance $\Delta =1/m^2$. Expanding the interaction exponential and using Wick's theorem gives the Feynman rules: a quartic vertex contributes $-\lambda$, each internal propagator contributes $\Delta$, and each vacuum diagram is divided by its Feynman-diagram symmetry factor. Multiply the resulting sum by $I_0$. Because we are expanding $I$, rather than $\log I$, disconnected vacuum diagrams must also be included.

At order ${\lambda }^2$ there are two quartic vertices and four internal propagators. Let $r$ be the number of edges joining the two vertices. The remaining half-edges must pair at their own vertex, so $r$ is even. The possibilities $r=0,2,4$ exhaust the vacuum diagrams:

For $r=0$, each vertex has three possible pairings, giving $N_0=9$ Wick contractions. For $r=2$, choose the two half-edges at each vertex, then pair the chosen half-edges across the vertices: $N_2={\left(\genfrac{}{}{0ex}{}{4}{2}\right)}^{2}2!=72$. The remaining two half-edges at each vertex form a self-loop. For $r=4$, all half-edges pair across the vertices, giving $N_4=4!=24$. The common denominator from the expansion is $2!(4!{)}^2=1152$, so the Feynman-diagram symmetry factors areEach vacuum diagram has the same factor $I_0{\lambda }^2{\Delta }^4$ before division by its Feynman-diagram symmetry factor. Consequently the coefficient of ${\lambda }^2$ isAs an independent check, the eighth moment obtained from the Gaussian moment pairing theorem is $⟨x^8{⟩}_0=7!!/m^8=105/m^8$, and $105/[2!(4!{)}^2]=35/384$. The first terms of the zero-dimensional quartic perturbation series are thereforeStrictly, the printed phrase “Taylor expansion” means a formal expansion or a right-sided asymptotic expansion. The general coefficient has absolute successive ratio $(4v+3)(4v+1)/[24(v+1)m^4]\to \infty$, so the power series has zero radius of convergence. Nevertheless, for $\lambda \ge 0$ the remainder after any fixed truncation of $e^{-\lambda x^4/4!}$ is bounded by the next absolute term, and its Gaussian integral is finite. This proves the displayed asymptotic expansion as $\lambda \downarrow 0$. For negative real $\lambda$ the defining integral diverges.