The factor of 2 should be your smallest worry! Your equation (2) is plain wrong (in one space dimension)!

You must express dv in terms of dw, plus a prefactor, to get from equation (1) to a correct version of equation (2). No integration is required for this step. But you may want to verify that both forms of the density lead to the same value (n) when integrated over. This is elementary calculus.

Hi Harikrishnan,

the answer by Kåre should be suffcient, however let me stress a very important fact, that the exponential function depends on all momenta  p = (p1,p2, p3) (their number equals the dimention of the confiquration space of the particle), and only after suitably performed change of the independent variables one gets the density dependent on the the velocity   v,  the energy    e,  or any other observable uniquely determined by the  momenta, as indicated by Kåre.

Additionally, what is  $\Phi$  in your formula?. If this is the potential energy, then your problem is related to a more general case of  the Gibbs density, which -consequently- should be dependent on  both - the spatial  AND  the momentum variables (for 3-dimensional problem this leads to  6  indendent variables). Moreover, then the normalizing factor will depend on the spatial region of integration (no word about this in your problem).  Did  I  missunderstand something basic for the problem?

Dear Kare,  Thank you for your answer.