# Salpeter mass function - past paper question [duplicate]

I'm doing a past paper and this is the question I'm struggling with (it's a standalone question with no other information given outside of this screenshot):

To answer it, I used the Salpeter mass function $$\psi$$ to be (as I've been taught it):

$$\psi (m) = Cm^{-2.35}$$

where C is a normalising constant and m is mass.

Using this, I got 0.043, whereas the answer given is 0.05. To get to my answer, I ended up ignoring C because I'm not sure how to calculate it/what to do with it. I tried to use this to calculate it: $$\int_{0}^{\infty} \psi (m) dm = 1$$

but I couldn't get to a solution.

Does anyone know how I would go about calculating C? Or am I going wrong and there's actually no need to calculate C for this question because it can be cancelled out in some way?

(This is a self-study exercise/past paper, not something I will be graded on. But I'm not sure if a full solution to any past paper question is allowed to be given, so even a hint as to how I could work it out would be so helpful.)

The number of all stars with a mass between $$m_1$$ and $$m_2$$ will be$$\int_{m_1}^{m_2} \psi(m) \, dm.$$
The mass of all stars with a mass between $$m_1$$ and $$m_2$$ will be$$\int_{m_1}^{m_2} m \,\psi(m) \, dm.$$ Note the additional $$m$$ in the integral. This can be seen by noting that the mass of the stars in a range $$[m, m+\Delta m]$$ is approximately $$m \psi(m) \Delta m$$, summing over many "bins" of width $$\Delta m$$, and then taking the limit as the width of the bins goes to zero.
• Yeah, it's a little confusing at first. Particularly you can't actually calculate $C$ from the information given in the problem: any power-law distribution $x^{-\alpha}$ with $\alpha \geq 1$ can't be normalized without imposing a cutoff at some $x_\text{min} > 0$. Commented Aug 21, 2023 at 13:38