# Integration by parts - weird choosing of factors

I have come across a weird integration where author states that i can get RHS out of LHS using integration by parts:

$$\int\limits_0^x \! \frac{d}{d t}\Big[ mv \gamma\Big]\,\, d x = v \!\cdot\! mv \gamma - \int\limits_0^v \! m v \gamma\, dv$$

Formula for integration by parts goes like this:

$$\int\! \frac{dg}{dx} f\,\, d x = f \!\cdot\! g -\!\! \int\! \frac{df}{dx}~g\,\, d x$$

and i don't know how to choose $\frac{dg}{dx}$ or $f$ in my equation.

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It's a computation of the work done by a relativistic force. It's a "physicists calculation", so it might not be very mathematically rigourous, but it is nevertheless correct. – Raskolnikov Mar 21 '13 at 20:08
Yes but i stick to the philosophy: "Ask mathematician about mathematics". Physics is all about interpreting the result u get at the end and rest is math. So i chose you guys here over guys at physics stackoverflow. – 71GA Mar 21 '13 at 20:50

Try to make the LH into an integral with respect to $t$ by noting that $$v=\frac{dx}{dt}$$ From there on you can apply partial integration straightforwardly.
EDIT: This works because $$D_t(v\cdot mv\gamma)=D_t(v)\cdot mv\gamma + v \cdot D_t (mv\gamma) \; .$$ Integrating with respect to $t$ we get $$v\cdot mv\gamma = \int_{t_0}^{t_f}D_t(v)\cdot mv\gamma \; dt + \int_{t_0}^{t_f} v D_t (mv\gamma) \; dt \; .$$ Then, in the first integral, we make a substitution so that we integrate over $v$. In the second integral, we make a substitution so that we integrate over $x$. These substitution rules are just $dv=\frac{dv}{dt}dt$ and $dx=\frac{dx}{dt}dt$ in differential notation, but they are completely legitimate.
But then i get: $\int\limits_{0}^{x} \frac{d}{dt}\Big[(mv\gamma)\Big]\, dx = v \!\cdot\! mv\gamma - \int\limits_{0}^{t} \frac{dv}{dt}\cdot mv\gamma\, \, dt$ This would be OK if i can do this: $\frac{dv}{dt} dt = dv$ and then i have to change the limits from $\int\limits_{0}^{t}$ to $\int\limits_{0}^{v}$... But can i do this? Is this allowed? – 71GA Mar 21 '13 at 20:44