\int_{\mathbb{R}^n\setminus B_\delta(x)}\ K(x,y,t)\ dy&=& \int_{\mathbb{R}^n\setminus B_\delta(x)}\ (4\pi t)^{-n/2}e^{-|x-y|^2/4t}\ dy\\ Then u(x,t) given by Poisson's formula (\ref{poisson1}) is ...\int_{\mathbb{R}^n\setminus B_\delta(x)}\ K(x,y,t)\ dy&=& \int_{\mathbb{R}^n\setminus B_\delta(x)}\ (4\pi t)^{-n/2}e^{-|x-y|^2/4t}\ dy\\ Then u(x,t) given by Poisson's formula (\ref{poisson1}) is in C^{\infty}(\mathbb{R}^n\times\mathbb{R}^1_+), continuous on \mathbb{R}^n\times[0,\infty) and a solution of the initial value problem (6.2), (6.3). u(x,t)-\phi(\xi)=\int_{\mathbb{R}^n}\ K(x,y,t)\left(\phi(y)-\phi(\xi)\right)\ dy.
