# 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!

• I think it's a direct application of the Fundamental Theorem of Calculus, where we have $\int_a^b \partial_t f(t) dt= f(b)-f(a)$ and assuming the function $f$ vanishes at the boundaries, we have $f(a) = f(b)=0$ and hence the integral vanishes. – wilsonw May 28 at 16:47
• Thanks for your comment! However, in this textbook we're considering these matrix integrals to be formal, in the sense that they're generating functions for maps, and it is mentioned in the book that they need not be convergent. Therefore I think it is not the case that the integrand vanishes at infinity. – Victor Eduardo May 31 at 11:08

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.
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.