Integral of Bolzmann distribution function for dark matter (modern cosmology)

Question

In Dodelson's textbook modern cosmology ch4.5, I wonder if anyone knows how he did the integration by part over the phasespace on the distribution function of Eq(4.72) and Eq(4.78).

Here is part of Eq(4.72) $$\int \frac{d^3p}{(2\pi)^3}p\frac{\partial f_{dm}}{\partial p} = \frac{4\pi}{(2\pi)^3}\int_0^\infty dpp^3\frac{\partial f_{dm}}{\partial p}$$ $$= -3 \frac{4\pi}{(2\pi)^3}\int_0^{\infty}dpp^2f_{dm}$$

I don't understand why he assumes this in the last step. $$p^3f_{dm}|_0^{\infty} = 0.$$

Here is part of Eq(4.78) $$\int_0^\infty dp \dfrac{p^4}{E}\dfrac{\partial f_{dm}}{\partial p}=\int_0^\infty dp f_{dm} \left( \dfrac{p^5}{E^3} - \dfrac{4p^3}{E}\right)$$

Similarly, he also used integration by part but why is this correct. $$\frac{p^4}{E} f_{dm}|_0^{\infty} = \dfrac{p^5}{E^3}$$

Dodelson mentions that we don't need an explicit expression of the zero-order distribution function for dark matter, $f_{dm}$, during the derivation, but then I could not follow his integration in this case.

Any help is appreciated.

Answer 1

Actually, the end points of the integral vanishes ($f_{dm}$ goes like $e^{-p}$ for large $p$, so it goes to zero faster than the power law in $p$ diverges). We get:
$$\frac{p^4}{E} f_{dm}|_0^{\infty} = 0.$$

The two terms come from the derivative of $\frac{p^4}{E}$: $$ \frac{\partial}{\partial p} \left(\frac{p^4}{E}\right) = \frac{4p^3}{E} - \frac{p^4}{E^2}\frac{dE}{dp} = \frac{4p^3}{E} - \frac{p^5}{E^3}.$$