令 $\varphi(x) =\dfrac {\mathrm e x}{\mathrm e-x}$，则有$$\varphi(\mu)\cdot \varphi(\mathrm e-\mu)=\dfrac {\mathrm e\mu}{\mathrm e-\mu}\cdot \dfrac {\mathrm e(\mathrm e-\mu)}{\mathrm e-(\mathrm e-\mu)}=\mathrm e^2,$$从而有\[\begin{split}\sum \limits_{k=1}^{2010}f\left(\dfrac {k\mathrm e}{2011}\right)&=\sum \limits_{k=1}^{2010}\ln \varphi \left(\dfrac {k\mathrm e}{2011}\right)\\ &=\ln \prod \limits_{k=1}^{2010}\varphi\left(\dfrac {k\mathrm e}{2011}\right)\\ &=\ln\prod \limits_{k=1}^{1005}\varphi\left(\dfrac {k\mathrm e}{2011}\right)\cdot \varphi\left[\dfrac {(2011-k)\mathrm e}{2011}\right]\\ &=\ln\prod \limits_{k=1}^{1005}\mathrm e^2=\ln \mathrm e^{2010}=2010.\end{split}\]