A sine integral 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 $n$? 
 A: 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,
$$ \frac{\sin 2kx}{\sin x} = e^{i(2k-1)x} + e^{i(2k-3)x} + \ldots + e^{ix} + e^{-ix} + \ldots + e^{i(-2k+1)x}$$
$$ = 2 \cos(2k-1)x + 2 \cos(2k-3)x + \ldots + 2 \cos x.$$
A: 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
$$\int_0^{\frac{\pi}{2}}\frac{\sin((n+1)\theta)}{\sin\theta}d\theta
= \int_0^{\frac{\pi}{2}} U_n(\cos\theta)d\theta 
= \begin{cases} 
\frac{\pi}{2},& n = 2k\\
\displaystyle\;2\sum\limits_{j=0}^k \frac{(-1)^j}{2j+1},& n = 2k+1
\end{cases}$$
A: $\newcommand{\+}{^{\dagger}}
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$\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
  ${\bf\mbox{6.3.1}}$.

$$\color{#44f}{%
I_{n}=\left\lbrace\begin{array}{lcl}
-I_{-n} & \mbox{if} & n < 0
\\[1mm]
0 & \mbox{if} & n = 0
\\[3mm]
\color{#c00000}{\left.\begin{array}{lcl}
{\pi \over 2} & \mbox{if} & n\ \mbox{is odd}
\\
{\pi \over 2} - \half\,\pars{-1}^{n/2}
\bracks{\Psi\pars{n + 3 \over 4} - \Psi\pars{n + 1 \over 4}}
& \mbox{if} & n\ \mbox{is even}
\end{array}\right\rbrace} & \mbox{if} & n > 0
\end{array}\right.}
$$
A: $$I_n=\int_0^{\pi/2} \frac{\sin(n\theta)}{\sin\theta}d\theta$$

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 $\pi/2$

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
