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Show that $$ \int_0^{\pi\over 2}\frac{\sin(2nx)}{\sin^{2n+2}(x)}\frac{1}{e^{2\pi \cot x}-1}dx =(-1)^{n-1}\frac{2n-1}{4(2n+1)} $$

My attempt

Lemma-1 \begin{align*} \frac{\sin(2nx)}{\sin^{2n}(x)}&=\sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r-1}\cot^{2n-2r+1}(x) \\ \end{align*}

proof

\begin{align*} \frac{\sin(2nx)}{\sin^{2n}(x)} &= \text{Im}\frac{e^{2inx}}{\sin^{2n}(x)} \\ &= \text{Im} \;\left( \frac{e^{ix}}{\sin x}\right)^{2n} \\ &= \text{Im} \; \left( \frac{\cos x+i\sin x}{\sin x}\right)^{2n} \\ &= \text{Im} \; (\cot x+i)^{2n} \\ &= \text{Im} \sum_{r=0}^{2n}\binom{2n}{r}i^r \cot^{2n-r}(x) \\ &= \sum_{r=0}^{2n}\binom{2n}{r}\sin\left(\frac{\pi r}{2} \right) \cot^{2n-r}(x) \\ &= \sum_{r=1}^n \binom{2n}{2r-1} (-1)^{r-1}\cot^{2n-2r+1}(x) \end{align*}

Let $I$ denote the integral. Now using the identity above I mentioned

$$I = \int_0^{\pi\over 2}\frac{\sin(2nx)}{\sin^{2n+2}(x)}\frac{1}{e^{2\pi \cot x}-1}dx $$

$$ = \int_0^{\pi\over 2}\left( \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r-1}\cot^{2n-2r+1}(x)\right)\frac{\csc^2(x)}{e^{2\pi \cot x}-1}dx $$

$$ = \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r-1} \int_0^{\pi \over 2}\cot^{2n-2r+1}(x)\frac{\csc^2(x)}{e^{2\pi \cot x}-1}dx $$

$$ = \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r-1} \int_0^\infty \frac{t^{2n-2r+1}}{e^{2\pi t}-1}dt \quad \color{blue}{(t=\cot x)} $$

$$ = \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r-1}\left\{ \frac{(2n-2r+1)!\zeta(2n-2r+2)}{(2\pi)^{2n-2r+2}}\right\} $$

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  • $\begingroup$ And then...? Why do you always end your attempt half the way in? You seem to know already the answer to many of the integrals you post, but still don't provide any context for them. When asked about it you ignore it. I don't understand the purpose of your questions honestly. $\endgroup$
    – Zima
    Feb 21 at 19:27
  • $\begingroup$ Have you tried expanding $\zeta(2n-2r+2)$? If so, what problem have you encountered? $\endgroup$
    – Zima
    Feb 21 at 19:29
  • $\begingroup$ @Zima I don't know the answer to any of the questions; I only know half of the answer to the integral I last posted. $\endgroup$
    – user1285841
    Feb 21 at 19:40
  • $\begingroup$ Now explain to me why I would post only half a solution if I already have the full solution. Uploading the question to the site would be a lot of work for me(don’t take it on a wrong way) $\endgroup$
    – user1285841
    Feb 21 at 19:42
  • $\begingroup$ @KStarGamer math.meta.stackexchange.com/q/20447/1285841 and math.meta.stackexchange.com/a/23082/1285841 $\endgroup$
    – user1285841
    Feb 24 at 5:24

1 Answer 1

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(Assuming $n$ is a positive integer.) As you have noted $$\frac{\sin 2nx}{\sin^{2n}x}=\Im(\cot x+i)^{2n}=\frac1{2i}\left((\cot x+i)^{2n}-(\cot x-i)^{2n}\right),$$ the given integral, after the substitution $t=\cot x$, equals $$I=\frac1{2i}\int_0^\infty\frac{(t+i)^{2n}-(t-i)^{2n}}{e^{2\pi t}-1}\,dt.$$

Now, for $0<r<1$, consider the contour $\lambda_r$ that is the boundary of $$\{z\in\mathbb{C} : \Re z>0 \land |\Im z|<1\}\setminus\{z\in\mathbb{C} : |z|<r\}$$ (goes along $+\infty-i\to-i\to i\to+\infty+i$ with a notch around $z=0$). Then \begin{align} 0=\int_{\lambda_r}f(z)\,dz &=\int_0^\infty\big(f(x+i)-f(x-i)\big)\,dx\\ &+i\int_r^1\big(f(ix)+f(-ix)\big)\,dx\\ &+\int_{-\pi/2}^{\pi/2}f(re^{it})\,rie^{it}\,dt, \end{align} where we set $f(z)=\displaystyle\frac{z^{2n}-i^{2n}}{e^{2\pi z}-1}$.

The first of the three terms is $2iI$; as $r\to 0$, the second one tends to $i^{2n+1}$ times $$\int_0^1\left(\frac{x^{2n}-1}{e^{2i\pi x}-1}+\frac{x^{2n}-1}{e^{-2i\pi x}-1}\right)dx=\int_0^1(1-x^{2n})\,dx=\frac{2n}{2n+1},$$ and the last one tends to $i\pi\displaystyle\operatorname*{Res}_{z=0}f(z)=-i^{2n+1}/2$. The expected result follows.

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