# Integration by parts to derive relativistic kinetic energy

I have come across a weird integration during derivation of relativistic kinetic energy. Our professor states that i can get RHS out of LHS using integration by parts:

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

I know a standard formula for integration by parts which i wrote below, but I don't know how to choose $\frac{dg}{dx}$ or $f$ in my case.

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

How did my profesor do this? Any help will be appreciated.

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You set the dot $\cdot$ yourself. Plug in

$\text d x=\frac{\text d x}{\text d t} \text d t = v\ \text d t,$

$\text d v=\frac{\text d v}{\text d t}\text d t$

on the left and right hand side, respectively, and identify what is been differentiated with respect to $t$.

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What follows is the quick and dirty way, which will be applauded by most physicists, scorned at by most mathematicians, and used wisely by us physicist-mathematicians who know what we're secretly doing when we abuse notation in this way...

Note that you have a $\mathrm{d}x$ over a $\mathrm{d}t$ on the left. Well, $$\frac{\mathrm{d}x}{\mathrm{d}t} = v,$$ so you can replace that. Now you have an integral of $v$ with respect to $mv\gamma$ (as opposed to the usual $f$ with respect to $x$). The right hand side follows immediately from integration by parts. Don't forget the limits are now $v(x = 0)$ (presumably equal to $0$) and $v(x)$ (written just $v$ for short). Actually the most confusing thing about this problem is using the same symbol for both the limits of integration and the dummy integration variable.

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A little trick required here. Perform a substitution first: $$mv \gamma(v) = u ,$$ hence your integral becomes $$\int \frac{du}{dt}dx ,$$ but notice that $$\frac{du}{dt} = \frac{du}{dx}\frac{dx}{dt}dx ,$$ but $\frac{dx}{dt} = v$ - the definition of velocity! So the integral becomes: $$\int \frac{du}{dt}dx = \int \frac{du}{dx}vdx .$$ So now you can just apply your general formula with no changes! I.e. $$\int \frac{du}{dx}vdx = uv - \int udv .$$ Substituting for $u$ get $$\int \frac{du}{dx}vdx = mv\gamma(v)v - \int mv\gamma(v)dv ,$$ QED.

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Also don't forget about the limits like I just did :-) –  Eugene B Apr 29 '13 at 18:23