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Prove that : $$\displaystyle \int_0^{\frac{\pi}{2}}p(x)\cot x\text{d}x=2\sum_{k=1}^{\infty}\int_0^{\frac{\pi}{2}}p(x)\sin (2kx)\text{d}x$$ where $\displaystyle p(x)=x^n$

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up vote 3 down vote accepted

Using the formula for the sum of a geometric series: $$ \begin{align} 2\sum_{k=1}^{m-1}\sin(2kx) &=-i\sum_{k=1}^{m-1}(e^{i2kx}-e^{-i2kx})\\ &=-i\left(\frac{e^{i2x}-e^{i2mx}}{1-e^{i2x}}-\frac{e^{-i2x}-e^{-i2mx}}{1-e^{-i2x}}\right)\\ &=-i\left(\frac{e^{ix}-e^{i(2m-1)x}}{e^{-ix}-e^{ix}}-\frac{e^{-ix}-e^{-i(2m-1)x}}{e^{ix}-e^{-ix}}\right)\\ &=i\frac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}-i\frac{e^{i(2m-1)x}+e^{-i(2m-1)x}}{e^{ix}-e^{-ix}}\\[6pt] &=\frac{\cos(x)}{\sin(x)}-\frac{\cos((2m-1)x)}{\sin(x)}\\[9pt] &=\cot(x)-\frac{\cos((2m-1)x)}{\sin(x)} \end{align} $$ Therefore, if $p(x)/x$ is integrable, then by the Riemann-Lebesgue Lemma $$ \begin{align} 2\sum_{k=1}^\infty\int_0^{\pi/2}p(x)\sin(2kx)\,\mathrm{d}x &=\lim_{m\to\infty}2\sum_{k=1}^{m-1}\int_0^{\pi/2}p(x)\sin(2kx)\,\mathrm{d}x\\ &=\lim_{m\to\infty}\int_0^{\pi/2}p(x)\left(\cot(x)-\frac{\cos((2m-1)x)}{\sin(x)}\right)\,\mathrm{d}x\\ &=\int_0^{\pi/2}p(x)\cot(x)\,\mathrm{d}x-{\small\lim_{m\to\infty}\int_0^{\pi/2}\frac{p(x)}{\sin(x)}\cos((2m-1)x)\,\mathrm{d}x}\\ &=\int_0^{\pi/2}p(x)\cot(x)\,\mathrm{d}x \end{align} $$

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Thx, how we know this converge? : ) – gauss115 Jan 22 '13 at 8:59
@gauss115: I have muddied up the argument with the grimy details :-) – robjohn Jan 22 '13 at 10:16
Thank you so much :) – gauss115 Jan 22 '13 at 12:06

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