ポアソン核$\displaystyle\sum_{n=1}^{+\infty} r^n cos(nx)$を求めます．
部分和
$$\displaystyle\sum_{n=1}^{\zeta} r^n cos(nx)=\displaystyle\sum_{n=1}^{\zeta} r^n \displaystyle\frac{e^{inx}+e^{-inx}}{2}$$
$$=\displaystyle\frac{1}{2}\displaystyle\sum_{n=1}^{\zeta}[r^n e^{inx}+r^n e^{-inx}] =\displaystyle\frac{1}{2} [\displaystyle\frac{re^{ix}(1-r^\zeta e^{i\zeta x})} {1-re^{ix}}+\displaystyle\frac{re^{-ix}(1-r^\zeta e^{-i\zeta x})}{1-r e^{-ix}}]$$
極限をとる．ただし，$0<r<1$である．
$$\displaystyle\frac{1}{2} \displaystyle\lim_{\zeta \to \infty} [ \displaystyle\frac{re^{ix}(1-r^\zeta e^{i\zeta x})} {1-re^{ix}}+\displaystyle\frac{re^{-ix}(1-r^\zeta e^{-i\zeta x})}{1-r e^{-ix}}]$$
$$=\displaystyle\frac{1}{2}[\displaystyle\frac{re^{ix}} {1-re^{ix}}+\displaystyle\frac{re^{-ix}} {1-re^{-ix}}]=\displaystyle\frac{1}{2}[\displaystyle\frac{r(e^{ix}+e^{-ix})-2r^2} {1-r(e^{ix}+e^{-ix})+r^2}]$$
$$=\displaystyle\frac{1}{2}[\displaystyle\frac{2rcosx-2r^2}{1-2rcosx+r^2}]=\displaystyle\frac{r(cosx-r)}{1-2rcos x+r^2}$$