A sine integral

Question

The integral \begin{align} \int_{0}^{\pi/2} \frac{ \sin(n\theta) }{ \sin(\theta) } \ d\theta \end{align} is claimed to not have a closed form expression. In this view find the series solution of the integral as a series involving of $n$.

Editorial note: As described in the problem several series may be obtained, of which, all seem to hold validity. As a particular case, from notes that were made a long while ago, the formula \begin{align} \int_{0}^{\pi/2} \frac{ \sin(n\theta) }{ \sin(\theta) } \ d\theta = \sum_{r=1}^{\infty} (-1)^{r-1} \ \ln\left(\frac{2r+1}{2r-1}\right) \ \sin(r n \pi) \end{align} is stated, but left unproved. Can this formula be proven along with finding other series dependent upon $n$?

Answer 1

$$I_n=\int_0^{\pi/2} \frac{\sin(n\theta)}{\sin\theta}d\theta$$

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Case i): When $n$ is odd i.e $n=2k+1$. $$I_{2k+1}-I_{2k+3}=\int_0^{\pi/2} \frac{\sin((2k+1)\theta)-\sin((2k+3)\theta)}{\sin\theta}d\theta=-2\int_0^{\pi/2} \cos(2(k+1)\theta)\,d\theta$$ $$\Rightarrow I_{2k+1}-I_{2k+3}=0 \Rightarrow I_{2k+3}=I_{2k+1}=\cdots=I_1=\frac{\pi}{2}$$ i.e whenever $n$ is odd, the value of the integral is always $\pi/2$

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Case ii): When $n$ is even i.e $n=2k$. $$I_{2k}-I_{2k+2}=\int_0^{\pi/2} \frac{\sin(2k\theta)-\sin((2k+2)\theta)}{\sin\theta}d\theta=-2\int_0^{\pi/2}\cos((2k+1)\theta)d\theta$$ $$\Rightarrow I_{2k}-I_{2k+2}=\frac{(-1)^k2}{2k+1} \Rightarrow I_n-I_{n+2}=\frac{2i^n}{n+1}$$ And I doubt the above recursive relation has a nice solution. :P

Answer 2

In the case that $n=2k$ is an even positive integer we have the identity $$\frac{\sin 2kx}{\sin x} = 2 \cos(2k-1)x + 2 \cos(2k-3)x + \ldots + 2 \cos x.$$

Therefore,

$$ \begin{align} \int_{0}^{\pi /2} \frac{\sin 2kx}{\sin x} \ dx &= 2 \int_{0}^{\pi/2} \Big(\cos x + \cos 3x + \ldots + \cos(2k-1) \ x \Big) \ dx \\ &= 2 \left(\sin x + \frac{1}{3} \sin 3x + \ldots + \frac{1}{2k-1} \sin(2k-1) x \right)\Bigg|^{\pi/2}_{0} \\ &= 2 \left(1- \frac{1}{3} + \ldots + \frac{(-1)^{k+1}}{2k-1} \right). \end{align}$$

EDIT:

To prove that $$ \frac{\sin 2kx}{\sin x} = 2 \cos(2k-1)x + 2 \cos(2k-3)x + \ldots + 2 \cos x,$$ notice that

$$\begin{align} \frac{\sin kx}{\sin x} &= \frac{e^{ikx}-e^{-ikx}}{e^{ix}-e^{-ix}} \frac{e^{-ix}}{e^{-ix}} \\ &= \frac{e^{i(k-1)x} - e^{i(-k-1)x}}{1-e^{-2ix}} \\ &= e^{i(k-1)x} \frac{1-e^{-2ikx}}{1-e^{-2ix}} \\ &= e^{i(k-1)x} \left(1 + e^{-2ix} + e^{-4ix} + \ldots + e^{-2i(k-1)x} \right) \\ &= e^{i(k-1)x} + e^{i(k-3)x} + \ldots + e^{i(-k+1)x}. \end{align}$$

Therefore,

$$ \frac{\sin 2kx}{\sin x} = e^{i(2k-1)x} + e^{i(2k-3)x} + \ldots + e^{ix} + e^{-ix} + \ldots + e^{i(-2k+1)x}$$

$$ = 2 \cos(2k-1)x + 2 \cos(2k-3)x + \ldots + 2 \cos x.$$

Answer 3

Let $U_n(x)$ be the Chebyshev's polynomial of the second kind. We know

$$U_{n}(\cos\theta) = \frac{\sin(n+1)\theta}{\sin\theta}$$ and it has a generating function of the form $$\sum_{n=0}^\infty U_n(x)t^n = \frac{1}{1-2xt+t^2}$$ Substitute $x$ by $\cos\theta$ in above expression, integrate $\theta$ over $[0,\frac{\pi}{2}]$ and uses a relatively easy to prove identity

$$\int_0^{\frac{\pi}{2}} \frac{d\theta}{K - \cos\theta} = \frac{2}{\sqrt{K^2-1}}\tan^{-1}\sqrt{\frac{K+1}{K-1}}\quad\text{ for }\quad K > 1$$ We have

$$\begin{align} \sum_{n=0}^\infty t^n \int_0^{\frac{\pi}{2}} U_n(\cos\theta)d\theta &= \frac{1}{2t}\int_0^{\frac{\pi}{2}}\frac{d\theta}{\frac{1+t^2}{2t} - \cos\theta}\\ &= \frac{1}{2t}\frac{2}{\sqrt{\left(\frac{1+t^2}{2t}\right)^2-1}}\tan^{-1}\sqrt{\frac{\frac{1+t^2}{2t}+1}{\frac{1+t^2}{2t}-1}}\\ &= \frac{2}{1-t^2}\tan^{-1}\left(\frac{1+t}{1-t}\right)\\ &= \frac{2}{1-t^2}\left(\frac{\pi}{4} + \tan^{-1} t\right)\\ &= \frac{1}{1-t^2}\left(\frac{\pi}{2} + 2\sum_{k=0}^\infty \frac{(-1)^k t^{2k+1}}{2k+1}\right) \end{align} $$ Compare the coefficients of both sides, we get

$$\int_0^{\frac{\pi}{2}}\frac{\sin((n+1)\theta)}{\sin\theta}d\theta = \int_0^{\frac{\pi}{2}} U_n(\cos\theta)d\theta = \begin{cases} \frac{\pi}{2},& n = 2k\\ \displaystyle\;2\sum\limits_{j=0}^k \frac{(-1)^j}{2j+1},& n = 2k+1 \end{cases}$$

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I_{n}\equiv \int_{0}^{\pi/2}{\sin\pars{n\theta} \over \sin\pars{\theta}}\,\dd\theta: \ {\large ?}}$

Since $\ds{I_{n} = -I_{-n}}$, we'll study the case $\ds{n > 0}$: \begin{align} I_{n\ >\ 0}&=\Im\int_{0}^{\pi/2}{\expo{\ic n\theta} - 1\over \sin\pars{\theta}}\,\dd\theta =\Im \int_{\verts{z}\ =\ 1 \atop {\phantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi/2}} {z^{n} - 1 \over \pars{z^{2} - 1}/\pars{2\ic z}}\,{\dd z \over \ic z} \\[3mm]&=2\,\Im \int_{\verts{z}\ =\ 1 \atop {\phantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi/2}} {1 - z^{n} \over -z^{2} + 1}\,\dd z \\[3mm]&=2\,\Im\bracks{% -\int_{1}^{0}{1 - \expo{\ic\pi n/2}y^{n} \over y^{2} + 1}\,\ic\,\dd y -\int_{0}^{1}{1 - x^{n} \over -x^{2} + 1}\,\dd x} =2\,\Re\int_{0}^{1}{1 - \expo{\ic\pi n/2}y^{n} \over y^{2} + 1}\,\dd y \\[3mm]&=\underbrace{2\int_{0}^{1}{\dd y \over y^{2} + 1}} _{\ds{=\ {\pi \over 2}}}\ -\ 2\cos\pars{n\pi \over 2}\int_{0}^{1}{y^{n}\,\dd y \over y^{2} + 1} \end{align}

$$ I_{n\ >\ 0}={\pi \over 2} -2\cos\pars{n\pi \over 2}\int_{0}^{1}{y^{n}\,\dd y \over y^{2} + 1} \,,\qquad\qquad I_{n\ <\ 0} = -I_{-n} $$

\begin{align} &\color{#c00000}{\int_{0}^{1}{y^{n}\,\dd y \over y^{2} + 1}} =\int_{0}^{\infty}{\expo{-\pars{n + 1}t} \over 1 + \expo{-2t}}\,\dd t =\sum_{\ell = 0}^{\infty}\pars{-1}^{\ell} \int_{0}^{\infty}\expo{-\pars{2\ell + n + 1}t}\,\dd t =\sum_{\ell = 0}^{\infty}{\pars{-1}^{\ell} \over 2\ell + n + 1} \\[3mm]&={1 \over 4}\bracks{\Psi\pars{n + 3 \over 4} - \Psi\pars{n + 1 \over 4}} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function ${\bf\mbox{6.3.1}}$.

$$\color{#44f}{% I_{n}=\left\lbrace\begin{array}{lcl} -I_{-n} & \mbox{if} & n < 0 \\[1mm] 0 & \mbox{if} & n = 0 \\[3mm] \color{#c00000}{\left.\begin{array}{lcl} {\pi \over 2} & \mbox{if} & n\ \mbox{is odd} \\ {\pi \over 2} - \half\,\pars{-1}^{n/2} \bracks{\Psi\pars{n + 3 \over 4} - \Psi\pars{n + 1 \over 4}} & \mbox{if} & n\ \mbox{is even} \end{array}\right\rbrace} & \mbox{if} & n > 0 \end{array}\right.} $$