It does not really matter, if you interpret the formula I did post above correctly. Given 2 identical cores, one runs with higher frequency and voltage, it will never get the efficiency back by finishing faster. Because you only gain time back, which is linear to frequency, while the power shows a cubic increase. Pdyn = Cdyn * V^2 * f
With other words, you always loose efficiency if you increase voltage - independent how you calculate.
Another conclusion from this formula is, that power is always increasing faster than frequency (and performance does not increase faster than frequency) - hence the lower the voltage the higher the efficiency.