# Luminosity Schechter function for galaxies

Just a question I am having trouble understanding. I have the Schechter luminosity function for galaxies, given as:

$$\Phi(L)dL=\Phi_{0}\left({\frac{L}{L_{\star}}}\right)^{\alpha}e^{-\frac{L}{L_{\star}}}\frac{dL}{L_{\star}}$$

I need to consider the case when $\alpha=-1$. And then show that the average luminosity of a galaxy is exactly $L_{\star}$. Could somebody explain how I could go about doing this and perhaps a hint or some part of a setup would be excellent. I really need to understand this.

Another part, which is related to the above question, is asking me to explain why the total luminosity is a finite number, whereas the total number of galaxies diverges. By this does it mean that the total number is infinite? Any extra comments on this would also be really appreciated.

• The second part of this is easy enough, it just involves a couple of integrals, the first converges (for the total luminosity per unit volume) to $\Phi_0 L^*$ and the second diverges at the low luminosity limit (for the total number of galaxies per unit volume). My problem is that this implies that the average luminosity of a galaxy is zero. Only if we take $\Phi_0$ as some sort of "typical" space number density for galaxies do we have that $L^*$ is the corresponding "typical" luminosity. – Conrad Turner Oct 8 '15 at 2:53
• Yeh, I understand if you times by $L$ then integrate the LF with respect to $L$ you get total luminosity, and then integrating the LF with respect to $L$ you get the number density. Divide one by the other to get the average. But I come out with some very problematic Gamma functions. – MichaelJRoberts Oct 8 '15 at 8:27

Total Luminosity with $\alpha=-1$: \begin{aligned} L_{tot}&=\int_0^{\infty}L \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\ &=\int_0^{\infty}\Phi_0 e^{-\frac{L}{L^*}}dL \end{aligned} Now put $L'=L/L^*$ and you get: $$L_{tot}=\Phi_0L^* \int_0^{\infty}e^{-L'}dL'=\Phi_0L^*$$ Total number: \begin{aligned} N_{tot}&=\int_0^{\infty} \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\ &=\int_0^{\infty} \Phi_0 \frac{1}{L}e^{-\frac{L}{L^*}}dL \end{aligned} which diverges at the lower limit.
Observe with $\alpha=-1$ we have no gamma functions.
Also IIRC with $\alpha \in (-1,0)$ when computing $L_{tot}/N_{tot}$ the gamma functions cancel leaving $(1+\alpha)L^*$
• Not sure your maths is correct there as you get a $L_{\star}$ to disappear in your final expression for $N_{tot}$. – MichaelJRoberts Oct 8 '15 at 8:56
• You end up with something like $L_{tot}/N_{tot}=L^* \times \frac{\Gamma(\alpha+2)}{\Gamma(\alpha+1)}$ but $\Gamma(\alpha+2)=(\alpha+1)\Gamma(\alpha+1)$. Sorry I can't be more precise the derivation is at home and I'm not at present. – Conrad Turner Oct 8 '15 at 13:31
• $\int_0^{\infty}\exp(-x)dx=\left[-\exp(-x) \right]_0^{\infty}=-\exp(-\infty)+\exp(-0)=0+1=1$ – Conrad Turner Oct 9 '15 at 5:27