(3-11)  As before we have: 

The difference this time is that we have to put in the relativistic momentum, which makes things trickier because it depends on v (and therefore on x) in a more complicated way. 

①  Start by re-writing the integral above with p replaced by its explicit relativistic definition (Chapter 3-3:  equation III-1).

②  As you did with the classical derivation, recognise that dx = vdt.  Rewrite and simplify your expression accordingly.  The result looks a bit strange, because it is an integration of a simple function (v) with respect to a complicated differential, d(v/√(1 – v2/c2)) . This is the kind of situation which might make your “integrate by parts” alarm light up.

③  If you have two functions of variable v, which we’ll call y(v) and z(v), then the general expression for an integration by parts is often written as:

Now, looking at your expression from ②, you can see that if we set y(v) = v and z(v) = v/√(1 – v2/c2), we can use integration by parts to rewrite the kinetic energy expression in a form that looks a bit more capable of being integrated. 

See if you can do this now.

④   The second part of this expression (the integral) still looks a bit tricky. Let’s just focus on this part for now:  what we need is a substitution to make the integration more manageable. Try introducing  z =  (1 – v2/c2) , and see where that takes you. Be careful with the limits of integration when you make the substitution.

⑤   By the miracle of integration by substitution, you should now have an easy integration to complete. Do this now. Apply the limits to work out the definite integral in terms of z, and then convert the answer back so it’s in terms of v again.

⑥  Re-insert the expression you’ve derived back into your kinetic energy equation from step ③.  You will see that the resulting expression is very interesting because it is a sum of terms, two of which depend on the velocity, v, while one does not. This is where the hairs on the back of your neck start to stand on end: we have an expression for kinetic energy which depends partly on a term that does not depend on the velocity. Spooky. Now separate this term out from the rest.

⑦ Next, tidy up the velocity-dependent part: get the two terms over the same denominator and see how your expression then simplifies.