Hi I am trying to understand the concept of dex and how to use it in calculations. The usual definition is that it is the order of magnitude, so $10^{0.1}$ is $0.1$ dex.

I want to do a simple exercise of calculating the value of the RHS of Eqn 4 in this paper arxiv paper, the gammas are incomplete gamma functions and the value of $\alpha$, $\phi_∗$ and $M_∗$ are given in table 2. The units of $\phi_∗$ is $h_{70}^3$ Mpc$^{−3}$ dex$^{−1}$. The value of the integral is given in the following paragraph as $5.06×10^5 h_{70}^3$ $M_⊙$ Mpc$^{−3}$ (let's take only the first one). If I substitute the values from table 2 to the RHS of Eq 4 and then calculate I get $7.58 × 10^5$ $h_{70}^3$ Mpc$^{−3}$ dex$^{−1}$ . What do I do with the dex$^{-1}$? Is it because of the dex$^{-1}$ I am a little off?

EDIT 1: Does the $d(\log(M_{bh}))$ in the integral take care of the $\rm{dex}^{-1}$, and my little off answer is due to rounding errors?


1 Answer 1


The "Schechter-like" Equation 3 in that paper is meant to give the value of black-hole mass density in units of "number of black holes of a particular mass per volume per log black hole mass", with the volume in units of megaparsecs and the "log black hole mass" in units of 0.1 log solar-mass units -- i.e., 0.1 dex.

Since the integral involves multiplying by $d \log M_{\rm BH}$, you are multiplying $(\log M_{\rm BH})^{-1}$ by $\log M_{\rm BH}$, and thus the "dex" units do indeed cancel out.

As for the exact value, you're going to have trouble reproducing that. First of all, the value from the text that you quoted ("5.06") is the result when they integrate over all BH masses, not just the $10^6$ to $10^{10}$ limits in Equation 4. From their Table 3, the relevant comparison value would be 4.9 if you're just doing the integration in Equation 4.

Second, the values they quote in Table 3 (and in the text for the full integration) are apparently not straightforward computations from their Equation 4 (or from the full integral). They go through a complicated exercise of computing this value 10001 times, each time tweaking the input coefficients of Equation 1, then applying some more using random Gaussian deviations, so they end up with a distribution of BH density values, not just a single one. They say

The final SMBH density is defined as the median of this distribution with the 1σ error given by 68 percentile range


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