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How would you approach $$\int_0^{\pi/2}\frac{\sin 2013x }{\sin x} \ dx\space?$$ The way I see here involves Dirichlet kernel. I wonder what else can we do, maybe some easy/elementary approaching ways. Thanks !

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up vote 14 down vote accepted

Let $I=\displaystyle\int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx$

As $\displaystyle\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$

$\displaystyle I=\int_0^{\frac\pi2} \frac{\sin (2n+1)(\frac\pi2-x)}{\sin (\frac\pi2-x)} dx$ $\displaystyle =\int_0^{\frac\pi2} \frac{\sin \{n\pi+\frac\pi2-(2n+1)x\}}{\cos x} dx$

$\displaystyle =\int_0^{\frac\pi2} \frac{\cos (2n+1)x}{\cos x} dx$ if $n$ is even.

$\displaystyle =-\int_0^{\frac\pi2} \frac{\cos (2n+1)x}{\cos x} dx$ if $n$ is odd.

If $n$ is odd, $\displaystyle 2I=\int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx-\int_0^{\frac\pi2} \frac{\cos (2n+1)x}{\cos x} dx$ $\displaystyle =\int_0^{\frac\pi2} \frac{\sin (2n)x}{\sin x\cos x} dx$ $\displaystyle =2\int_0^{\frac\pi2} \frac{\sin (2n)x}{\sin 2x} dx$ $\displaystyle =2\frac12\int_0^{\pi} \frac{\sin ny}{\sin y} dy$ $\displaystyle =2\int_0^{\frac{\pi}2} \frac{\sin ny}{\sin y} dy$ as $\displaystyle\frac{\sin ny}{\sin y}$ is an even function.

So, $\displaystyle\int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx=I=\int_0^{\frac{\pi}2} \frac{\sin nx}{\sin x} dx$ if $n$ is odd.

Similarly, $\displaystyle \int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin (n+1)x}{\sin x} dx$ if $n$ is even.

If we put, $2n+1=2013, n=1006$ which is even.

S0, $\displaystyle \int_0^{\frac\pi2} \frac{\sin (2013)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin (1007)x}{\sin x} dx$

Now, if we put $2n+1=1007,n=503$ which is odd.

So, $\displaystyle \int_0^{\frac{\pi}2} \frac{\sin (1007)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin (503)x}{\sin x} dx$

Now, if $2n+1=503,n=251$

The reduction of $n$ will follow :$2013,1007,503,251,125,63,31,15,7,3,1$

So,$\displaystyle \int_0^{\frac\pi2} \frac{\sin (2013)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin x}{\sin x} dx=\frac{\pi}2$

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nice solution there (+1) – user 1618033 Dec 22 '12 at 14:44
@I must thank you for the verification. – lab bhattacharjee Dec 22 '12 at 14:46
This is very much related to the other question...good solution...but I like the next solution more. – Hawk Feb 8 '14 at 4:58

$\sin(m+2)x-\sin mx=2\sin x\cos(m+1)x$

So, $\displaystyle\frac{\sin(m+2)x}{\sin x}-\frac{\sin mx}{\sin x}=2\cos (m+1)x$

$\displaystyle\int_0^{\frac{\pi}2}\frac{\sin(m+2)x}{\sin x}dx-\int_0^{\frac{\pi}2}\frac{\sin mx}{\sin x}dx=2\int_0^{\frac{\pi}2}\cos (m+1)xdx=\frac2{m+1}\sin(m+1)\frac{\pi}2$

Putting $m=2n-1$

$\displaystyle\int_0^{\frac{\pi}2}\frac{\sin(2n+1)x}{\sin x}dx-\int_0^{\frac{\pi}2}\frac{\sin (2n-1)x}{\sin x}dx=\frac2{m+1}\sin(2n-1+1)\frac{\pi}2=0$

So, $$\int_0^{\frac{\pi}2}\frac{\sin(2n+1)x}{\sin x}dx=\int_0^{\frac{\pi}2}\frac{\sin (2n-1)x}{\sin x}dx$$ and so on up to $\displaystyle\int_0^{\frac{\pi}2}\frac{\sin x}{\sin x}dx=\frac{\pi}2$

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@Chris'ssister, would you please have a look into this? – lab bhattacharjee Dec 22 '12 at 16:46
I did it in the first 30 seconds after you posted the answer and upvoted you! :) Thanks! – user 1618033 Dec 22 '12 at 16:49
I think you're wrong on the 3rd line (at a closer look). – user 1618033 Dec 22 '12 at 17:01
@Chris'ssister, thanks for your observation. Would you please have a look into the rectified version? – lab bhattacharjee Dec 22 '12 at 17:36
that's fine now! Thank you! – user 1618033 Dec 22 '12 at 17:40

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\pi/2}{\sin\pars{2013 x} \over \sin\pars{x}}\,\dd x:\ {\large ?}}$

We'll consider $\ds{\pars{~\mbox{with}\ \color{#c00000}{\large n=\ {\tt 2013}}~}}$: \begin{align} &\color{#c00000}{\int_{0}^{\pi/2}{\sin\pars{nx} \over \sin\pars{x}}\,\dd x}= \half\int_{-\pi/2}^{\pi/2}{\sin\pars{nx} \over \sin\pars{x}}\,\dd x = \half\int_{0}^{\pi}{-\cos\pars{nx} \over -\cos\pars{x}}\,\dd x = {1 \over 4}\int_{-\pi}^{\pi}{\cos\pars{nx} \over \cos\pars{x}}\,\dd x \\[3mm]&={1 \over 4}\Re \oint_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}} {z^{n} \over \pars{z^{2} + 1}/\pars{2z}}\,{\dd z \over \ic z} \end{align}

$$ \color{#c00000}{\int_{0}^{\pi/2}{\sin\pars{nx} \over \sin\pars{x}}\,\dd x} =\half\Im \oint_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}} {z^{n} \over z^{2} + 1}\,\dd z\tag{1} $$

The only contribution to the integration arises from two 'small' semicircles around $\ds{\pm\ic}$: \begin{align} &\color{#c00000}{\int_{0}^{\pi/2}{\sin\pars{nx} \over \sin\pars{x}}\,\dd x} \\[3mm]&=\half\Im\lim_{\epsilon \to 0^{+}}\bracks{% -\int_{2\pi}^{\pi}{\pars{\ic + \epsilon\expo{\ic\theta}}^{n} \over \pars{\ic + \epsilon\expo{\ic\theta}}^{2} + 1}\, \epsilon\expo{\ic\theta}\ic\,\dd\theta -\int_{\pi}^{0}{\pars{-\ic + \epsilon\expo{\ic\theta}}^{n} \over \pars{-\ic + \epsilon\expo{\ic\theta}}^{2} + 1}\, \epsilon\expo{\ic\theta}\ic\,\dd\theta} \\[3mm]&=\half\pars{{\pi \over 2} + {\pi \over 2}}.\qquad\qquad \mbox{Note that}\quad \pars{\pm \ic}^{2013} = \pm\ic. \end{align}

$$ \color{#00f}{\Large\int_{0}^{\pi/2}{\sin\pars{2013 x} \over \sin\pars{x}}\,\dd x ={\pi \over 2}} $$

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Nice (+1). It seems you like my questions! (by the way, I created some thousands of such questions) – user 1618033 May 7 '14 at 11:01
You are always posting interesting questions. Thanks. – Felix Marin May 7 '14 at 19:15

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