I'm working through Evans's proof of the solution for the nonhomogenous heat equation (Theorem 2.3.2) and I'm getting stuck on a certain part. He starts with the proposed solution: $$ u(x,t) = \int^t_0 \int_{\Bbb{R}^n} \Phi(y,s)f(x-y,t-s)dyds $$ In the middle of the proof, he calculates $$ u_t(x,t)-\Delta u(x,t) = \int^t_0 \int_{\Bbb{R}^n}\Phi(y,s)(\frac{\partial}{\partial t}-\Delta_x)f(x-y,t-s)dyds + \int_{\Bbb{R}^n}\Phi(y,t)f(x-y,0)dy = \quad \int^t_\varepsilon \int_{\Bbb{R}^n}\Phi(y,s)(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)dyds + \int^\varepsilon_0\int_{\Bbb{R}^n}\Phi(y,s)(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)dyds + \int_{\Bbb{R}^n}\Phi(y,t)f(x-y,0)dy$$. So far so good. Evans then denotes $$I_\varepsilon = \int^t_\varepsilon \int_{\Bbb{R}^n}\Phi(y,s)(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)dyds $$ and then argues that, by using integration by parts, we obtain: $$ I_\varepsilon = \int^t_\varepsilon \int_{\Bbb{R}^n}(\frac{\partial}{\partial s}-\Delta_y)\Phi(y,s)f(x-y,t-s)dyds + \int_{\Bbb{R}^n}\Phi(y,\varepsilon)f(x-y,t-\varepsilon)dy - \int_{Bbb{R}^n}\Phi(y,t)f(x-y,0)dy$$
This is where I am confused. I tried working out the details of this integration by parts by first looking at the $\frac{\partial}{\partial s} $ part of the integral first. Just looking at that part separately, we have $$ \int^t_\varepsilon \int_{\Bbb{R}^n}\Phi(y,s)[-\frac{\partial}{\partial s}f(x-y,t-s)]dyds = \int_{\Bbb{R}^n}\int^t_\varepsilon\Phi(y,s)[-\frac{\partial}{\partial s}f(x-y,t-s)]dsdy = \quad \int_{\Bbb{R}^n}\int^t_\varepsilon\frac{\partial}{\partial s}\Phi(y,s)f(x-y,t-s)dsdy - \int_{\Bbb{R}^n} [\Phi(y,s)f(x-y,t-s)]^t_\varepsilon dy = \int^t_\varepsilon\int_{\Bbb{R}^n}\frac{\partial}{\partial s}\Phi(y,s)f(x-y,t-s)dsdy - \int_{\Bbb{R}^n} \Phi(y,t)f(x-y,0) dy + \int_{\Bbb{R}^n} \Phi(y,\varepsilon)f(x-y,t-\varepsilon)dy$$
So that explains where the last two terms come from. However, I'm confused about how to handle the Laplacian part of this integral. Using Green's third identity, we have $$ \int^t_\varepsilon \int_{\Bbb{R}^n}\Phi(y,s)(-\Delta_y f(x-y,t-s))dyds = \int^t_\varepsilon \int_{\Bbb{R}^n}-\Delta_y \Phi(y,s)f(x-y,t-s)dyds + \int^t_\varepsilon \int_{\partial \Bbb{R}^n} [\Phi(y,s)\nabla f(x-y,t-s) \cdot \nu - f(x-y,t-s) \nabla \Phi(y,s) \cdot \nu]dS $$
Here, $\nu$ is the outward facing unit normal vector. Based on the result that Evans gets, I assume that $$(\ast) \quad \int^t_\varepsilon \int_{\partial \Bbb{R}^n} [\Phi(y,s)\nabla f(x-y,t-s) \cdot \nu - f(x-y,t-s) \nabla \Phi(y,s) \cdot \nu]dS = 0 $$ but I'm not sure how to verify this. I'm also not sure what it means to integrate a function over $\partial \Bbb{R}^n$ (I don't know what the boundary of $\Bbb{R}^n$ would be. So my question is, is my work thus far correct, and if so, how do I verify that ($\ast$) is correct? Any help would be appreciated.