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I'm reading the David Tong's notes (p.111) from Cambridge and I'm facing an integral which I don't know how the author solve it.

$$\int_0^\infty \frac{2y^n}{e^{2y} - 1} dy = \frac{I_n}{2^n}$$

$$I_n = \int \frac{y^n}{e^y-1} dy$$

It might be trivial, but I don't see how the first integral becomes essentially: $$\frac{1}{2^n} \int \frac{y^n}{e^y-1} dy$$

pie
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epselonzero
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    Do you mean $(2y)^n$, not $2y^n$? Use parentheses to make your intentions clear. Also, you write $e^2 - 1$ rather than $e^y - 1$. – KCd Apr 15 '24 at 23:14
  • @amWhy Ok, I removed it. I do not think however that this is a duplicate. As I mentioned below the question is simply about the process of transforming the initial integral to $I_n/2^n$ and not about the evaluation of $I_n$ itself. – Gary Apr 16 '24 at 01:30
  • @KCd I fixed $e^2$. however, it is $2y^n$ and not $(2y)^2$.

    Also, this post is not a duplicate. Since As Gary says I'm not interested about the evaluation of $I_n$.

     – epselonzero Apr 16 '24 at 03:37
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    In the original integral, perform the change of variables from $y$ to $t$ by $t=2y$. What do you get? – Gary Apr 16 '24 at 03:40
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    Exactly. I saw my mistake at the same moment you wrote the solution. It was not that hard, but I couldn't find it by myself sadly. Thank you! – epselonzero Apr 16 '24 at 04:00

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