 Solution

Use the normalized Gaussian integral with action $\mathrm{tr}(M^2)/2$. Its Wick contraction isThere is no factor $1/N$ here: this part uses $V$, whereas the following part uses $NV$. Expanding the normalized Hermitian matrix model integral to first order givesThe subtracted term removes Vacuum Feynman diagrams disconnected from the external pair. Since the one-point function vanishes by $M\mapsto -M$, the resulting two-point function is a connected correlation function.

Write the vertex as $M^a{}_b{M}^b{}_c{M}^c{}_d{M}^d{}_a$. Each external field must contract with a different vertex field, leaving the other two to form a tadpole diagram. Eight of the twelve connected pairings attach the external fields at adjacent cyclic positions. The remaining index loop gives $N{\delta }^i{}_l{\delta }^k{}_j$ in each case. The other four attach them at opposite positions and give ${\delta }^i{}_j{\delta }^k{}_l$, with no free index loop. ThereforeBoth index structures are required at finite $N$. For $N=1$ the answer is $1-3g+O(g^2)$, agreeing with the ordinary zero-dimensional quartic integral. This is a formal perturbative quantum field theory expansion; the real integral is convergent for $g\ge 0$.