Integrals of Power This question has two parts: a question about a specific expression and a question about a family of expressions. In what follows I will omit all constants of integration.
If we integrate force with respect to time we get momentum: $\int F \, \operatorname{d}\!t = mv$. If an object accelerates along a line then we can find its acceleration at any given point and write force as a function of distance. Doing so, and integrating gives kinetic energy (or work done): $\int F \, \operatorname{d}\!x = \frac{1}{2}mv^2$.
I decided to try the same thing with power using the definition $P = Fv$. When I integrated with respect to time I got, as expected, $\frac{1}{2}mv^2$ which is work done. If an object accelerates along a line then we can measure the power at regular distances and write power as a function of distance. Then we may integrate power with respect to distance:
$$\int P \, \operatorname{d}\!x = \int mav \, \operatorname{d}\!x = m\int\frac{\operatorname{d}\!v}{\operatorname{d}\!t}v\, \operatorname{d}\!x = m\int\frac{\operatorname{d}\!x}{\operatorname{d}\!t}\frac{\operatorname{d}\!v}{\operatorname{d}\!x}v\,\operatorname{d}\!x = m\int v^2\operatorname{d}\!v = \tfrac{1}{3}mv^3$$
Here is the first question: What is $\frac{1}{3}mv^3$? Does it have a name? How can I understand it conceptually?
Here is the second question: Momentum is $mv$, kinetic energy is $\frac{1}{2}mv^2$, this new quantity $\frac{1}{3}mv^3$ has turned up. It seems that $\frac{1}{k}mv^k$ for positive whole numbers $k$ plays a role. These can all be combined into one power series. Provided $v$ is very small, we have
$$mv + \tfrac{1}{2}mv^2 + \tfrac{1}{3}mv^3 + \cdots + \tfrac{1}{k}mv^k + \cdots \equiv m\ln\left(\frac{1}{1-v}\right)$$
Do the expressions $\frac{1}{k}mv^k$ have any general meaning; and does this logarithmic expression have any use? (I realise that the units in the sum are all different.)
 A: Comments to the question (v2): OP's observation can be summarized as follows.


*

*In $D$ dimensions: Let $g=g(v^2)$ be a function of the square $v^2$ of the speed $v=|{\bf v}|$. Let $G=G(v^2)$ be a primitive integral (aka. antiderivative or indefinite integral) of $g$, i.e. $G^{\prime}(v^2) = g(v^2)$. Then in analogy with the work-energy theorem for a non-relativistic particle with mass $m$, one has
$$\int_{{\bf r}_i}^{{\bf r}_f} \! g(v^2)~
{\bf F}\cdot {\rm d}{\bf r}
~=~\int_{t_i}^{t_f} \! g(v^2)~
m\frac{{\rm d}{\bf v}}{{\rm d}t}\cdot{\bf v}~ {\rm d}t
~=~\frac{m}{2}\int_{t_i}^{t_f} \! g(v^2)~
\frac{{\rm d}(v^2)}{{\rm d}t}{\rm d}t$$
$$~=~\frac{m}{2}\int_{v_i^2}^{v_f^2} \! g(v^2) ~ {\rm d}(v^2)
~=~\frac{m}{2} G(v_f^2) -\frac{m}{2} G(v_i^2). $$
The standard work-energy theorem corresponds to $g(v^2)=1$ and $G(v^2)=v^2$. OP's first non-standard example corresponds to $g(v^2)=v$ and $G(v^2)=\frac{2}{3}v^3$.

*In $D=1$ dimension: We use boldface to denote signed 1D quantities, i.e. vectors in 1D. Let $h=h({\bf v})$ be a function of the 1D velocity ${\bf v}$. Let $H=H({\bf v})$ be a primitive integral of ${\bf v}h({\bf v})$, i.e. $H^{\prime}({\bf v}) = {\bf v}h({\bf v})$. Then the generalized work-energy theorem read
$$\int_{{\bf x}_i}^{{\bf x}_f} \! h({\bf v})~{\bf F}~{\rm d}{\bf x}
~=~\int_{t_i}^{t_f} \! h({\bf v})~
m\frac{{\rm d}{\bf v}}{{\rm d}t} {\bf v} ~{\rm d}t$$
$$~=~m\int_{{\bf v}_i}^{{\bf v}_f} \! H^{\prime}({\bf v})~{\rm d}{\bf v}
~=~m H({\bf v}_i)-m H({\bf v}_f). $$
The standard work-energy theorem corresponds to $h({\bf v})=1$ and $H({\bf v})=\frac{1}{2}{\bf v}^2$. OP's first non-standard example corresponds to $h({\bf v})={\bf v}$ and $H({\bf v})=\frac{1}{3}{\bf v}^3$.
I don't know if the above generalizations of the work-energy theorem have a name, though. Usually these generalizations are not needed in order to solve a physical problem. I speculate that similar observations might be useful in integrable systems to keep track of an infinite hierarchy of higher conservation laws.
