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I'm reading through the lecture notes of Wayne Hu regarding the Damping Scale of the CMB. He give the following steps to calculating the damping scale, $k_D$:$$k_D^{-2}=\int \frac{1}{6(1+R)}\left( \frac {16}{15}+ \frac{R^2}{(1+R)}\right)\frac{1}{\dot\tau} d\eta$$Limiting forms:$$\lim\limits_{R \to 0}k_D^{-2}= \frac{1}{6}\frac{16}{15}\int \frac{1}{\dot\tau} d\eta$$ $$\lim\limits_{R \to \infty}k_D^{-2}= \frac{1}{6}\int \frac{1}{\dot\tau} d\eta$$and finally$$k_D=\frac{\sqrt{6}}{\sqrt{\eta \dot\tau^{-1}}}$$I see roughly what he's doing, but my math is rusty. Could anyone explain in greater detail how he arrived at $k_D$? For example, why did he discard the limit as ${R \to 0}$? Why is only ${R \to \infty}$ used in the final formula? Why was he able to extract everything but $\frac{1}{\dot\tau}$ from the integral (since R is also a function of $\eta$)?

EDIT: While function $f(\eta)=\frac{1}{6(1+R(\eta))}\left( \frac {16}{15}+ \frac{R(\eta)^2}{(1+R(\eta))}\right)$ is relatively constant, it does change with $\eta$ and looks like this:

enter image description here

and the function $g(\eta)=\frac{1}{\dot\tau(\eta)}$ looks like this: enter image description here

The bottom axis is $\eta$ in seconds, $s$.

Quark Soup
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  • What are $R$ and $\tau'$ -- are they dependent on $\eta$? If $R$ is independent of $\eta,$ then you can factor out all the functions of $R,$ so he's just taken limits. Also, over what domain are you integrating? – Allawonder Sep 12 '19 at 18:00
  • I updated the post with graphs of the functions. Hopefully it will answer your questions. As far as the domain, the author shows it as an indefinite integral, but I've seen other sources show the domain from $(0,η_0)$ where $\eta_0$ is the current conformal time (a constant). – Quark Soup Sep 12 '19 at 19:46

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From the second limiting form $$\dfrac{1}{k_D^2}=\dfrac{1}{6}\int \dfrac{1}{\tau} d\eta \\ 1=\dfrac{1}{6}\int \dfrac{1}{\tau} d\eta \ k_D^2 \\ 1=\dfrac{1}{6} \tau^{-1}\eta \ k_D^2 \\ k_D=\sqrt{\dfrac{6}{\tau^{-1}\eta}} \\ k_D=\dfrac{\sqrt{6}}{\sqrt{\tau^{-1}\eta}} $$ And as you can see, this is the answer.

Novice
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  • @Quarkly I don't catch what you have in mind, this is indefinite integral, so there's no upper and lower limit. – Novice Sep 12 '19 at 18:44
  • I get the part about moving the variables around. I was hoping for more information about why the upper limit was taken and why the lower limit was discarded. Also, are 0 and $\infty$ assumed as the upper and lower limits on an indefinite integral? What justification did we get from the first equation to use these limits (again, my math is rusty. Sorry if it's a stupid question). – Quark Soup Sep 12 '19 at 18:44
  • @Quarkly 'Also, are 0 and ∞ assumed as the upper and lower limits when the integral has no limits given?' No. Not at all. As I said previously it's indefinite integral so there's no limits and such integrals we also can count. And answering your second question: about which derivative are you talking about? Because I don't see here even one. You don't need to apologize :) – Novice Sep 12 '19 at 18:48
  • I'm trying to understand the justification for taking the limits at 0 and $\infty$. They seem like arbitrary numbers since I didn't see them in the first integral. I'd also like to understand why the lower limit (as R goes to 0) was discarded in the final answer. – Quark Soup Sep 12 '19 at 18:51
  • @Quarkly these limits don't correspond to the first integral and its limits. I don't know what exactly R means in this case, but for sure the first formula describes a certain casus and the lecturer wanted to show how this equation behaves when R is close to 0 and ∞. And the reasons why the limit when R goes to zero was discarded may be a lot - maybe simply the case with ∞ was more proper for situation considered by the lecturer. – Novice Sep 12 '19 at 19:11