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How can I compute $$\int_{-\pi}^\pi\frac{\sin(13x)}{\sin x}\cdot\frac1{1+2^x}\mathrm dx?$$

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Hint 1: $$ \int\limits_{-a}^a f(x)dx=\int\limits_{-a}^a \frac{f(x)+f(-x)}{2}dx $$ Hint 2: $$ \frac{\sin (n x)}{\sin x}= \frac{(e^{ix})^n-(e^{-ix})^{n}}{e^{ix}-e^{-ix}}= \sum\limits_{k=0}^{n-1} (e^{ix})^{n-k-1}(e^{-ix})^k $$ Hint 3: $$ \int\limits_{-\pi}^{\pi} e^{ikx}dx= \begin{cases} 2\pi&\text{ if }\quad k=0\\ 0 &\text{ if }\quad k\in\mathbb{Z}\setminus\{0\} \end{cases} $$

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    @did Thanks!${}{}$2012-07-08
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    Sorry for the complexity in my previous comment. The sum in Hint 2 should be $$\sum_{k=0}^{n-1}e^{i(n-2k-1)x}$$2012-07-08
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    I was going to suggest the change you just made to Hint 1 :-)2012-07-08
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    @robjohn, thanks for your attention. You are always on the alert :)2012-07-08
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    You're quite welcome! (BTW Hint 2 has become even more hinky; see my revised comment above).2012-07-08
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    I think it is clear and follows from $$a^n-b^n=(a-b)\sum\limits_{k=0}^{n-1}a^{n-1-k}b^{k}$$2012-07-08
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    But you still have $e^{ikx}$ instead of $e^{ix}$ inside the first pair of parentheses.2012-07-08
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    I fixed it; I hope you don't mind. (+1)2012-07-08
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    @robjohn Thanks!2012-07-08