I quote Life Insurance Mathematics (Gerber, 1997).
Let us consider a person aged $x$ years, denoted by $(x)$. We denote his/her future lifetime by $T$ or, more explicitly, by $T(x)$. Thus $x+T$ will be the age at death of the person.
The future lifetime $T$ is a random variable with a probability distribution function \begin{equation} G(t)=\mathbb{P}\left(T\leq t\right)\text{, }t\geq0\tag{1} \end{equation} Define $_t p_x$ as follows: \begin{equation} _t p_x=1-G(t)=\mathbb{P}\left(T>t\right)\text{, }t\geq0\tag{2} \end{equation} Define $\mu_{x+t}$ as follows: \begin{equation} \mu_{x+t}=-\frac{d}{dt}\ln\hspace{0,1cm} _{t}p_{x}\tag{3} \end{equation}
For $0<u<1$, let us assume that $\mu_{x+u}$ is a constant. Let us denote the constant value of $\mu_{x+u}$ by $\mu_{x+\frac{1}{2}}$.
$\color{red}{\text{Using }(3)\text{ one finds:}}$ \begin{equation} \color{red}{\mu_{x+\frac{1}{2}}=-\ln p_x\tag{4}} \end{equation}
I cannot really understand why applying $(3)$, one would get $(4)$ for $t=\frac{1}{2}$. Could you please help me understand that?
I am not sure what was meant to be understood by "assume that $\mu_{x+u}$ is a constant." I will suppose it means that for $0 < t < 1$ and a fixed $x$ we are to assume that $\ln {}_tp_x$ has a constant derivative with respect to $t$. (This is the application of Equation $(3)$.)
We then find that $\ln {}_tp_x = at + b$, that is, ${}_tp_x = e^{at + b},$ for some constants $a$ and $b$, provided that $0 < t < 1.$ If we also make the reasonable assumption (perhaps already implied in the textbook) that ${}_tp_x = \mathbb P(T>t)$ is a continuous function of $t,$ we find that ${}_0p_x = \lim_{t\to 0} e^{at + b} = e^b$ and ${}_1p_x = \lim_{t\to 1} e^{at + b} = e^{a+b}.$
Presumably $\mathbb P(T>0) = 1$, so $$ e^b = {}_0p_x = \mathbb P(T>0) = 1, $$ which implies $b = 0,$ so $\ln {}_tp_x = at,$ and in particular $\ln {}_1p_x = a.$ Then (for $0 < t < 1$) $$ \mu_{x+t} = -\frac{\mathrm d}{\mathrm dt}\ln {}_tp_x = -\frac{\mathrm d}{\mathrm dt}at = -a = -\ln {}_1p_x. $$