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


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.

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    $\begingroup$ 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. $\endgroup$ – Conrad Turner Oct 8 '15 at 2:53
  • $\begingroup$ 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. $\endgroup$ – 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 $(1+\alpha)L^*$

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    $\begingroup$ Not sure your maths is correct there as you get a $L_{\star}$ to disappear in your final expression for $N_{tot}$. $\endgroup$ – MichaelJRoberts Oct 8 '15 at 8:56
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    $\begingroup$ Thinking it was just a typo. $\endgroup$ – MichaelJRoberts Oct 8 '15 at 8:57
  • $\begingroup$ Hi Conrad, could you explain or derive the last sentence where you state that the Gamma functions cancel...could you show me how this is done exactly... $\endgroup$ – MichaelJRoberts Oct 8 '15 at 13:11
  • $\begingroup$ 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. $\endgroup$ – Conrad Turner Oct 8 '15 at 13:31
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    $\begingroup$ $\int_0^{\infty}\exp(-x)dx=\left[-\exp(-x) \right]_0^{\infty}=-\exp(-\infty)+\exp(-0)=0+1=1$ $\endgroup$ – Conrad Turner Oct 9 '15 at 5:27

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