6

Calculate$$\int_{0}^{\pi/3}\ln^4\left(\frac{\sin x}{\sin(x+\pi/3)}\right)\mathrm dx$$

My try: $$\sin(x+\pi/3)=\sin x\cos(\pi/3)+\sin(x+\pi/3)\cos x=\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x$$

$$\int_{0}^{\pi/3}\ln^4\left(\frac{2\sin x}{\sin x+\sqrt{3}\cos x}\right)\mathrm dx$$

$$\int_{0}^{\pi/3}\ln^4\left(\frac{2}{1+\sqrt{3}\cot x}\right)\mathrm dx$$

$u=\frac{2}{1+\sqrt{3}\cot x}$

$u^{'}=\frac{2\sqrt{3}}{\sin^2 x(1+\sqrt{3}\cot x)^2}$

$$\frac{1}{2\sqrt{3}}\int_{0}^{1}\sin^2 x(1+\sqrt{3}\cot x)^2\ln^4(u)\mathrm du$$

$$\frac{2}{\sqrt{3}}\int_{0}^{1}\sin^2 x u^{-2}\ln^4(u)\mathrm du$$

$\cot^2 x=\frac{(2-u)^2}{3u^2}$

using $1+\cot^2 x=\frac{1}{\sin^2 x}$

$$\frac{2}{\sqrt{3}}\int_{0}^{1}\frac{3u^{2}}{3u^2+(2-u)^2} u^{-2}\ln^4(u)\mathrm du$$

$$\frac{6}{\sqrt{3}}\int_{0}^{1}\frac{1}{3u^2+(2-u)^2}\ln^4(u)\mathrm du$$

$$\frac{3}{\sqrt{3}}\int_{0}^{1}\frac{1}{u^2-u+1}\ln^4(u)\mathrm du$$

I am not sure what to do next...

Zacky
  • 27,674
  • I doubt you'd be able to find the indefinite integral, so you need to use the specific limits of integration you have. try some symmetry arguments or something – mathworker21 Jul 06 '19 at 15:29
  • At the end, instead of going from $\frac{6}{\sqrt{3}}$ to $\frac{3}{\sqrt{3}}$, you should be going to $\frac{3}{2\sqrt{3}}$, as you are multiplying by $4$ in the integral. – Varun Vejalla Jul 06 '19 at 15:41
  • Mathematica says the answer is $\frac{17\pi^{5}}{243}$. I don't know how to get there though. – Varun Vejalla Jul 06 '19 at 15:43
  • Didn't check the previous work, but by applying partial fractions and geometric series, the last integral boils down to computing $$\sum_{n=0}^\infty a_n\frac{\mathrm d}{\mathrm d\alpha}\int_0^1u^\alpha~\mathrm du\bigg|_{\alpha=n}$$ – Simply Beautiful Art Jul 06 '19 at 15:44

1 Answers1

8

$$I=\int_{0}^{\pi/3}\ln^4\left(\frac{\sin(x+\pi/3)}{\sin x}\right)dx=\int_0^\frac{\pi}{3}\ln^4 \left(\frac12 +\frac{\sqrt 3}{2}\cot x\right)dx=\frac{\sqrt 3}{2}\int_0^1 \frac{\ln^4 x}{x^2-x+1}dx$$ Above follows by the substitution $\frac12+\frac{\sqrt{3}}{2}\cot x\to x$. We also have for $t\in(0,\pi), x\in (-1,1)$: $$\frac{\sin t}{x^2-2x\cos t+1}=\sum_{n=1}^\infty x^{n-1}\sin(nt)$$ $$\Rightarrow I=\sum_{n=1}^\infty \sin\left(\frac{n\pi}{3} \right)\int_0^1 x^{n-1} \ln^4 x dx=24\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3} \right)}{n^5}$$

Similarly to here, we have for $x\in(0,2\pi)$: $$\frac{\pi-x}{2}=\sum_{n=1}^\infty\frac{\sin(nx)}{n}$$ Integrating the above with respect to $x$ gives: $$\sum_{n=1}^\infty \frac{\cos(nx)}{n^2}=\frac{(\pi-x)^2}{4}+C_1$$ Setting $x=\pi$ gives $C_1=-\frac{\pi^2}{12}$ and integrating again produces: $$\sum_{n=1}^\infty \frac{\sin(nx)}{n^3}=-\frac{(\pi-x)^3}{12}-\frac{\pi^2}{12}x+C_2$$ Putting $x=\pi $ yields $C_2= \frac{\pi^3}{12}$. One more time:

$$\sum_{n=1}^\infty \frac{\cos(nx)}{n^4}=-\frac{(\pi-x)^4}{48}+\frac{\pi^2}{24}x^2-\frac{\pi^3}{12}x+C_3$$ For $x=\pi \Rightarrow C_3=\frac{23\pi^4}{720}$. And finally, one more similar step gives for $x \in(0,2\pi)$: $$S(x)=\sum_{n=1}^\infty \frac{\sin(nx)}{n^5}=\frac{(\pi-x)^5}{240}+\frac{\pi^2}{72}x^3-\frac{\pi^3}{24}x^2+\frac{23\pi^4}{720}x-\frac{\pi^5}{240}$$

And well, the value of the integral is just $24S\left(\frac{\pi}{3}\right)$. Doing the algebra yields: $$\boxed{\int_{0}^{\pi/3}\ln^4\left(\frac{\sin x}{\sin(x+\pi/3)}\right)dx=\frac{17\pi^5}{243}}$$

Community
  • 1
Zacky
  • 27,674
  • Why are you allowed to interchange the sum and the integral to get $I = \sum_{n=1}^\infty \sin(\frac{n\pi}{3})\int_0^1 x^{n-1}\ln^4 x dx$? – mathworker21 Jul 07 '19 at 13:12
  • Well, can't we integrate term by term? Since in the radius of convergence is fine to interchange them. See a https://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged?noredirect=1&lq=1 – Zacky Jul 07 '19 at 14:28
  • 1
    The link you sent gives general conditions for integrating term by term, general conditions I am very familiar with. I don't see what in the link you sent justified what you did. In any event, I think what you did would be fine if you replaced the integral from $0$ to $1$ by an integral from $\epsilon$ to $1-\epsilon$ and then took $\epsilon \to 0$ at the end of the day. – mathworker21 Jul 07 '19 at 23:12
  • Thanks for that! Usually I don't do things rigorously (since I actually don't have the knowledge to do so). I just enjoy solving the integrals, but can't provide perfect solutions. // By the way, since you are familiar with it, can you tell me if the conditions from that link apply to double integrals too? I assumed it does, here: https://math.stackexchange.com/a/3284584/515527 – Zacky Jul 07 '19 at 23:23
  • 1
    Yea, it is usually more fun to be non-rigorous. I don't think it's helpful to think as generally as "does it apply to double integrals". It's very dependent on the setup. Something you should take away from the first link you sent, though, is that you can do $\sum_n \int a_n(x)dx = \int \sum_n a_n(x)dx$ if $a_n(x) \ge 0$ for each $n$ or $x$, or if $\sum_n \int |a_n(x)|dx < \infty$ (which is equivalent to $\int \sum_n |a_n(x)|dx < \infty$ since $|a_n(x)| \ge 0$ for each $n,x$!). The reason I bring this up is because it is the same commonly used sufficient conditions for interchanging a double.. – mathworker21 Jul 07 '19 at 23:27
  • 1
    sum. That is, $\sum_n \sum_m a_{n,m} = \sum_m \sum_n a_{n,m}$ if $a_{n,m} \ge 0$ for each $n,m$ or if $\sum_n \sum_m |a_{n,m}| < \infty$ (which, once again, is equivalent to $\sum_m \sum_n |a_{n,m}| < \infty$). – mathworker21 Jul 07 '19 at 23:28
  • 1
    Now, unfortunately, I don't think either of these two sufficient conditions apply to the answer we're commenting on now, or the answer involving the double integral you just linked to (actually, I think the setups are the same, it's just in that answer, you are interchanging a sum with an integral twice!). In either case, I think you might need to truncate the limits of integration to make some sort of dominated convergence theorem argument work, which is something more complicated. I haven't thought it through though. In any event, very nice answers :) – mathworker21 Jul 07 '19 at 23:31