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I was evaluating $$ \int_{0}^{^\pi/_2}x\ln\left(\vphantom{\large A}\cos\left(x\right)\right)\,{\rm d}x $$ I like to try with the fourier series $$ \int_{0}^{^\pi/_2} \left[\,\,\sum_{k = 1}^{\infty} {\left(-1\right)^{k - 1}\cos\left(2kx\right) \over k}\,x - x\ln\left(2\right)\right] \,{\rm d}x $$ But I dont know how to do the series part.

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up vote 1 down vote accepted

Note that $$I_k = \int_0^{\pi/2} x \cos(2kx) dx = \dfrac{\cos(\pi k) -1}{4k^2} = \begin{cases} 0 & \text{if }k \text{ is even}\\ - \dfrac1{2k^2} & \text{if }k \text{ is odd}\end{cases}$$ Hence, $$\int_{0}^{\pi/2} \left(\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}k x \cos(2kx) - x \ln 2\right) dx = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}k I_k - \dfrac{\pi^2/4}2 \ln2$$ $$\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}k I_k = - \dfrac12\sum_{k \text{ is odd}} \dfrac1{k^3} = -\dfrac{7}{16} \zeta(3)$$ Hence, the answer is $$-\dfrac{7}{16} \zeta(3) - \dfrac{\pi^2}8 \ln2$$

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Oh Thx! I was not confident with these series :) – Ryan Feb 13 '13 at 3:58

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