Why does this integral term vanishes in Pauli's proof that Coulomb's Law follows from $\nabla\times\mathbf{E}=0,\;\nabla\cdot\mathbf{E}=4\pi\varrho$

The following is from Electrodynamics Volume 1 of Pauli Lectures on Physics. My question is regarding the final sentence and is: why does the second term not contribute?

We have derived the field equations [2.4] , [2.3], or [4.4] , [2.5], \begin{equation*} \operatorname{curl} \mathbf{E} = 0, \qquad \operatorname{div} \mathbf{E} = 4\pi \rho, \end{equation*} or \begin{equation*} \mathbf{E} = -\operatorname{grad} \varphi, \qquad \nabla^2 \varphi = -4\pi \rho, \end{equation*} from Coulomb's law. We now wish to show that, conversely, Coulomb's law follows from these equations. In order to demonstrate this we must, in addition, require that $\varphi \to 0$ at least as fast as $1/r$ as $r \to \infty$.

\begin{align*} \left(\varphi(\mathbf{r}) = \mathscr{O} \frac{1}{r} \; \backepsilon\; r \to \infty\right) &\iff \left(\limsup_{r \to \infty} \left| r \, \varphi(\mathbf{r})\right| < \infty\right) \iff \exists_{C,R\in\mathbb{R}^{+}}\forall_{r\geq R}\left|r\varphi(\mathbf{r})\right|\leq C \end{align*}

For simplicity we will perform the proof only for volume charges. For this, we will need Green's theorem. If $\mathbf{A} = \varphi \operatorname{grad} \psi$ is substituted into Gauss's theorem, \begin{equation*} \oint_F A_n \, df = \oint_F \mathbf{A} \cdot \mathbf{n} \, df = \int_V \operatorname{div} \mathbf{A} \, dV, \end{equation*} then, because \begin{equation*} \operatorname{div} (\varphi \operatorname{grad} \psi) = \varphi \nabla^2 \psi + \operatorname{grad} \varphi \cdot \operatorname{grad} \psi, \tag{6.1}\label{eq:6.1} \end{equation*} we obtain Green's first identity: \begin{equation*} \oint_F \varphi \frac{\partial \psi}{\partial n} \, df = \int_V \varphi \nabla^2 \psi \, dV + \int_V \operatorname{grad} \varphi \cdot \operatorname{grad} \psi \, dV. \tag{6.2}\label{eq:6.2} \end{equation*} Interchanging $\varphi$ and $\psi$ in \eqref{eq:6.1} we obtain, upon subtraction, Green's second identity: \begin{align*} \operatorname{div} (\varphi \operatorname{grad} \psi - \psi \operatorname{grad} \varphi) &= \varphi \nabla^2 \psi - \psi \nabla^2 \varphi,\ \oint_F \left( \varphi \frac{\partial \psi}{\partial n} - \psi \frac{\partial \varphi}{\partial n} \right) df &= \int_V (\varphi \nabla^2 \psi - \psi \nabla^2 \varphi) \, dV. \tag{6.3}\label{eq:6.3} \end{align*} These results are valid for any arbitrary volume in which $\varphi$ and $\psi$ are regular. Green's second identity can be applied to the integration of Poisson's equation by considering $\varphi$ as the sought potential function and letting $\psi_{P'} = 1/r_{PP'}$. In order that the integrands be regular, we limit our integration to the region between a small sphere $K_P$ about the fixed point $P$ and a large sphere $K$ (see Fig. 6.1). Because \begin{equation*} \nabla^2 \psi = 0 \quad \text{for} \quad P \neq P', \end{equation*} we have \begin{equation*} \oint_K \left( \varphi \frac{\partial (1/r)}{\partial n} - \frac{1}{r} \frac{\partial \varphi}{\partial n} \right) df_{P'} + \oint_{K_P} \left( \varphi \frac{\partial (1/r)}{\partial n} - \frac{1}{r} \frac{\partial \varphi}{\partial n} \right) df_{P'} = 4\pi \int \frac{\rho_{P'}}{r_{PP'}} \, dV_{P'}. \end{equation*} Now, the integral over the sphere $K$ vanishes because of our assumption about the behavior of $\varphi$ at infinity. In addition, the second term in the $K_P$ integral does not contribute.

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Elsewhere Pauli uses a $K_P$ sphere to excise a point charge. But if there were a point charge at $P$ the integral term would not vanish. It would be the electric flux over $K_P$ divided by the fixed radius. If we had a finite volume charge density the term would again not vanish for the same reason. What additional assumptions, if any, are required to explain the vanishing of this term?