When integrating functions $\dfrac {\sin x}{\sin x}$, as well as $\dfrac {\sin3x}{\sin x}$ , $\dfrac {\sin 5x}{\sin x}$, $\dfrac {\sin 7x}{\sin x}$ etc. on interval from $0$ to $\dfrac {\pi}{2}$, the answer turns out to be $\dfrac {\pi}{2}$. How can this be proven?
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1Have you seen the Chebyshev polynomials, by any chance? – J. M. ain't a mathematician May 30 '13 at 16:59
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No, I haven't, I came across this question as an extra credit problem on an old exam where I used to study. I tried Induction, but the term aren't getting easier. I tried sum formula to "peel" of a sine to take out denom, but this isn't getting anywhere. Series wouldn't do it either. I feel I spent enough time on it in the sens I need some direction. I do want to have a method that falls within Calculus BC material – imranfat May 30 '13 at 17:03
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Be aware of false generalizations http://math.stackexchange.com/a/111499/312 – leonbloy May 30 '13 at 17:07
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1Three answers and I'm the only one who's up-voted the question so far. This often seems to get neglected. – Michael Hardy May 30 '13 at 18:03
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Hint:
$$1+2 \cos{2 x} + 2 \cos{4 x} + \ldots+2 \cos{2 n x} = \frac{\sin{(2 n+1) x}}{\sin{x}} $$
Ron Gordon
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Well, that does it, isn't it? And I think I have seen this identity somewhere in my old notes too! But I don't think my posted problem (the general case) is suitable for Calc BC puppies anymore. Thanks! – imranfat May 30 '13 at 17:28
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The function on both sides of this trigonometric multiple-angle identity is the Dirichlet kernel. Its convolution with any $2\pi$-periodic function is that function's $n$th-degree Fourier approximation. – Michael Hardy May 30 '13 at 18:02
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\pi/2}{% \sin\pars{\bracks{2n + 1}x} \over \sin\pars{x}} \,\dd x\,\right\vert_{\,n\ \in\ \mathbb{N}_{\,\geq\ 0}}} \\[5mm] = &\ \Im\int_{0}^{\pi/2}{\expo{\ic\pars{2n + 1}x}\,\,\, -\ 1 \over \sin\pars{x}}\,\dd x \\[5mm] = &\ \left.\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\,\, {z^{2n + 1} -1 \over \pars{z - 1/z}/\pars{2\ic}} \,{\dd z \over \ic z}\,\right\vert_{\,z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left.2\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\,\, {1 - z^{2n + 1} \over 1 - z^{2}} \,\dd z\,\right\vert_{\,z\ =\ \exp\pars{\ic x}} \end{align} Hereafter, I'll "close" the integration around a quarter circle in the complex plane first quadrant: Thetre isn't any pole inside such contour.
Namely, \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\pi/2}{% \sin\pars{\bracks{2n + 1}x} \over \sin\pars{x}} \,\dd x\,\right\vert_{\,n\ \in\ \mathbb{N}_{\,\geq\ 0}}} \\[5mm] = &\ -2\,\Im\int_{1}^{0}\,\, {1 - \ic^{2n + 1}\,\,y^{2n + 1}\,\, \over 1 - \ic^{2}\,y^{2}\,} \,\ic\,\dd y\ -\ \pars{\substack{\mbox{An integral, along the}\ \ds{x}\mbox{-axis} \\[1mm] \mbox{segment}\ \ds{\pars{0,1}},\ \mbox{which} \\[1mm] \ds{vanishes\ out}\ \mbox{because} \\[1mm]\mbox{the integrand}\ \ds{\in \mathbb{R}.}}} \\[5mm] = &\ 2\int_{0}^{1}{\dd y \over 1 + y^{2}} = \bbx{\pi \over 2} \\ & \end{align}
Felix Marin
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I recommend you to prove by induction with all odd integers since $$\int \frac{\sin3x}{\sin x}dx$$ will depend on $$\int \frac{\sin x}{\sin x}dx$$
using integration by parts.
hyg17
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