<p>Total Luminosity with $\alpha=-1$:
$$
\begin{aligned}
L_{tot}&amp;=\int_0^{\infty}L \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\
&amp;=\int_0^{\infty}\Phi_0 e^{-\frac{L}{L^*}}dL
\end{aligned}
$$
Now put $L'=L/L^*$ and you get:
$$
L_{tot}=\Phi_0L^* \int_0^{\infty}e^{-L'}dL'=\Phi_0L^*
$$
Total number:
$$
\begin{aligned}
N_{tot}&amp;=\int_0^{\infty} \Phi_0 \frac{L^*}{L}e^{-\frac{L}{L^*}}\frac{dL}{L^*}\\
&amp;=\int_0^{\infty} \Phi_0 \frac{1}{L}e^{-\frac{L}{L^*}}dL
\end{aligned}
$$
which diverges at the lower limit.</p>

<p>Observe with $\alpha=-1$ we have no gamma functions.</p>

<p>Also IIRC with $\alpha \in (-1,0)$ when computing $L_{tot}/N_{tot}$ the gamma functions cancel leaving $(1+\alpha)L^*$</p>

CC BY-SA 3.0
2015-10-08T08:56:00.177Z
Not sure your maths is correct there as you get a $L_{\star}$ to disappear in your final expression for $N_{tot}$.
MichaelJRoberts
CC BY-SA 3.0
2015-10-08T08:57:51.400Z
Thinking it was just a typo.
MichaelJRoberts
CC BY-SA 3.0
2015-10-08T13:11:46.640Z
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...
MichaelJRoberts
CC BY-SA 3.0
2015-10-08T13:31:01.863Z
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
CC BY-SA 3.0
2015-10-08T13:32:54.340Z
When you get back to your derivation at home could you share it with me. Still not sure how it shows that the average luminosity is $L_{\star}$ tho...
MichaelJRoberts
CC BY-SA 3.0
2015-10-08T13:34:42.093Z
It doesn't, what you have is it typically is $L^*$ as per my first comment on the original question.
Conrad Turner
CC BY-SA 3.0
2015-10-08T18:22:34.807Z
Just a quick question, on the third line how did you get the definite integral to be 1? Could you show those steps for me?
MichaelJRoberts
CC BY-SA 3.0
2015-10-09T05:27:30.923Z
$\int_0^{\infty}\exp(-x)dx=\left[-\exp(-x) \right]_0^{\infty}=-\exp(-\infty)+\exp(-0)=0+1=1$
Conrad Turner
CC BY-SA 3.0
2015-10-09T08:26:30.877Z
Of course, yes. Simple.
MichaelJRoberts
CC BY-SA 3.0
2015-10-08T08:36:10.280Z
2015-10-08T10:12:17.060Z
2015-10-08T10:12:17.060Z
Conrad Turner
<p>Just a question I am having trouble understanding. I have the Schechter luminosity function for galaxies, given as:</p>

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

<p>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. </p>

<p>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.</p>

CC BY-SA 3.0
2015-10-08T02:53:37.587Z
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
CC BY-SA 3.0
2015-10-08T08:27:41.403Z
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
<galaxy><luminosity>
Luminosity Schechter function for galaxies
MichaelJRoberts
# Luminosity Schechter function for galaxies

(Asked by: username_0 on 2015-10-06)

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.

**Tags:** galaxy, luminosity

> **Comments:**
> **username_1**: 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.  
> **username_0**: 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.  

---

## 1 answers

### Answer by username_1 on 2015-10-08. Score: 2 | Accepted answer

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 $(1+\alpha)L^\*$

> **Comments:**
> **username_0**: Not sure your maths is correct there as you get a $L_{\star}$ to disappear in your final expression for $N_{tot}$.  
> **username_0**: Thinking it was just a typo.  
> **username_0**: 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...  
> **username_1**: 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.  
> **username_0**: When you get back to your derivation at home could you share it with me. Still not sure how it shows that the average luminosity is $L_{\star}$ tho...  
> **username_1**: It doesn't, what you have is it typically is $L^*$ as per my first comment on the original question.  
> **username_0**: Just a quick question, on the third line how did you get the definite integral to be 1? Could you show those steps for me?  
> **username_1**: $\int_0^{\infty}\exp(-x)dx=\left[-\exp(-x) \right]_0^{\infty}=-\exp(-\infty)+\exp(-0)=0+1=1$  
> **username_0**: Of course, yes. Simple.  

---

