= Solution
Assume a finite optimum with $X>0$ and continuous price near its completion time. With <costate> $p^*$ for the remaining-file state, the minimization <Hamiltonian of an optimal-control problem> is $H=u(p(t)-p^*)$. It is independent of $x$, so $p^*$ is constant. By the <Pontryagin maximum principle>, pointwise minimization over $[0,1]$ gives
$$
\boxed{u(t)=1\text{ if }p(t)<p^*,\qquad u(t)=0\text{ if }p(t)>p^*.}
$$
This is a <bang-bang control> away from equality. At equality partial transmission or either endpoint rate is allowed. This also follows by exchanging a small amount of transmission from a more expensive time to a cheaper unused time. The <free-terminal-time transversality condition> is $\min_uH(T,u)+2\gamma T=0$. Since $T>0$, it forces $p^*>p(T)$ and terminal transmission at rate one, yielding
$$
\boxed{p^*=p(T)+2\gamma T.}
$$
Without regularity/existence hypotheses an arbitrary known price need not have a finite differentiable optimum; the equality is the interior terminal-time condition used here.
For $p(t)=t+1/t$, $X=1$, feasible completion requires $T>1$ for finite cost. The price decreases to its minimum at one and then increases. Its cheapest unrestricted interval of length one has endpoints $a,a+1$ of equal price, giving $a(a+1)=1$ and $a+1=\varphi=(1+\sqrt5)/2$. For $1<T\leq\varphi$, the cheapest length-one subset of $[0,T]$ is the terminal interval $[T-1,T]$: its left endpoint has at least the price of its right endpoint, and the earlier times are still more expensive. For $T\geq\varphi$ the transmission minimum is unchanged from that unrestricted interval, while the delay cost increases, so no optimum lies beyond $\varphi$.
For the terminal-interval policy the total cost is
$$
J(T)=T-\frac12+\log\frac T{T-1}+\gamma T^2,\qquad
J'(T)=1-\frac1{T(T-1)}+2\gamma T.
$$
Its <derivative> is strictly increasing on $T>1$, since $J''(T)=(2T-1)/[T^2(T-1)^2]+2\gamma>0$. It is negative near one and positive at $\varphi$, so the unique global optimum is
$$
\boxed{u=1\text{ on }[T-1,T],\quad u=0\text{ earlier},\qquad\frac1{(T-1)T}=1+2\gamma T.}
$$
There is exactly one <positive root>: none lies in $(0,1)$ because the left side is negative, and the strictly monotone comparison gives exactly one in $(1,\varphi)$. Its threshold is $p(T-1)=p(T)+2\gamma T$.
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