# How does one integrate a time derivative with respect displacement

How does one evaluate the following integral? $$\frac{1}{2}\frac{d}{dt}\int\limits_{-\infty}^\infty \left[ (v_t)^2 + c^2(v_x)^2 \right]dx - \frac{d}{dx}\int\limits_{-\infty}^\infty c^2 v_tv_xdx=-\nu\int\limits_{-\infty}^\infty(v_t)^2dx$$ where $c$ and $\nu$ are constants, and where any derivatives of $v$ vanish whenever $\left| x \right|\to \infty$.

I have no clue how to integrate a partial derivative with respect to one variable over another variable. Some hints would be appreciated.

• I suspect you have made a mistake, since the second integral on the left is constant in $x$ so its $d/dx$ is 0 – Calvin Khor Jun 2 '16 at 21:30

$$\frac{1}{2}\frac{d}{dt}\left[v_t^2+c^2 v_x^2\right]-\frac{d}{dx}c^2 v_t v_x = v_t v_{tt}+ c^2 v_x v_{xt}-c^2 v_x v_{tx}-c^2 v_t v_{xx}$$ but assuming that Schwarz' condition $v_{xt}=v_{tx}$ holds, the RHS equals: $$v_t v_{tt}-c^2v_t v_{xx} = \frac{1}{2}\frac{d}{dt} v_t^2-c^2 v_t v_{xx}=v_t\color{red}{(v_{tt}-c^2 v_{xx})}.$$ The red term has something to do with the Minkowski metric.