Integral of total derivative vanishes I am reading Bertrand Eynard's book on Counting Surfaces.
In this book he mentions that the integral of a total derivative vanishes. What does he mean by this?
Basically, I am trying to understand how the formula 2.5.1 follows.

I know that the total derivative of a function is the best linear approximation of that function at a point, as explained in the Wikipedia article. However, I don't see how this total derivative vanishes upon integration.
Thanks!
 A: In this setting total derivative is the divergence of a "vector" field: the divergence of $G(M)_{i,j} e^{-\frac{N}{t} \operatorname{Tr} V(M)}$. By the divergence theorem we have 
\begin{equation}
\int_V (\nabla \cdot F) \, dV = \oint_{\partial V} F \cdot dS,
\end{equation}
where the second integral is the flux of F trough the boundary of V. If $F$ vanishes on the boundary, the LHS will vanish as well. For example with a vector field in two variables $F(x,y) = (f(x,y),g(x,y))$, we get
\begin{equation}
\int_V \frac{\partial f(x,y)}{\partial x} + \frac{\partial g(x,y)}{\partial y} = 0.
\end{equation}
 If we move on to complex variables, $F(w,z) = (f(w,z),g(w,z))$, by the definition of complex derivative in terms of real and imaginary part, we have
\begin{equation}
\int_V \frac{\partial f(w,z)}{\partial \text{Re}{\,w}} -i \frac{\partial f(w,z)}{\partial \text{Im}{\,w}} +
 \frac{\partial g(w,z)}{\partial \text{Re}{\,z}} -i \frac{\partial g(w,z)}{\partial \text{Im}{\,z}}= 0
\end{equation}
Now you can translate this to the matrix case and obtain equation (2.5.1). In this case since entries of $M$ on the diagonal are real, and only entries with $i\leq j$ are independent variables, we get exactly what's in the text.  
Finally we must justify that the flux vanishes at infinity. Note that you can still have vanishing integrals in the formal sense (all the coefficients of the formal series vanishes). As explained in remark 2.5.1, the statement holds for both convergent and formal matrix integrals.
In the convergent case, this is because the potential V(M) goes to infinity and so the weight $e^{-\frac{N}{t} \operatorname{Tr}V(M)}$ kills $G(M)$ (we take $t_3,...,t_d$ negative). In the formal case remember that we first expand the exponential and then perform gaussian integrations with the remaining $e^{-\frac{N}{2t} \operatorname{Tr} M^2}$, and this will also kill any monomial of $G(M)$ together with monomials from the expansion of the other traces.
