Why is $u(z) = \frac {z^{1-p}}{1-p}$ taken as $log(z)$ when $p=1$?

Question

We want our function $u(z)$ to have constant $-\frac{zu''(z)}{u'(z)}$.

Let $u(z) = \frac {z^{1-p}}{1-p}$ when $p$ is not 1, and $u(z) = log(z)$ when $p=1$.

Why do we take it as $log(z)$? How does this keep the $-\frac{zu''(z)}{u'(z)}$ constant?

Answer 1

Well, you run into this impasse when you try to find an antiderivative for $x^{-1}$ (because using the power rule, we get stuck with $\dfrac{x^0}0$, which doesn't work). Thus one (cleverly) defines and studies $\ln x:=\int^x_1\dfrac1t\operatorname{dt}$.

Then it turns out that this function is the inverse of $e^x$.

Answer 2

You have $$ \lim_{p\rightarrow 1}\Big(\frac{z^{1-p}}{1-p}-\frac{1}{1-p}\Big) = \lim_{p\rightarrow 1}\Big(\frac{e^{(1-p)\ln z} -1}{1-p}\Big) = \lim_{x\rightarrow 0}\frac{e^{x\ln z} -1}{x} =\ln z$$