Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 !

share|improve this question

3 Answers 3

up vote 13 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,$\displaystyle \int_0^{\frac\pi2} \frac{\sin (2013)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin x}{\sin x} dx=\frac{\pi}2$

share|improve this answer
    
nice solution there (+1) –  Chris's sis 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 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 $\displaystyle\int_0^{\frac{\pi}2}\frac{\sin x}{\sin x}dx=\frac{\pi}2$

share|improve this answer
    
@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! –  Chris's sis Dec 22 '12 at 16:49
    
I think you're wrong on the 3rd line (at a closer look). –  Chris's sis 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! –  Chris's sis 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}

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

share|improve this answer
    
Nice (+1). It seems you like my questions! (by the way, I created some thousands of such questions) –  Chris's sis May 7 at 11:01
    
You are always posting interesting questions. Thanks. –  Felix Marin May 7 at 19:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.