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.

I was evaluating $$\int_0^{\frac{\pi}{2}}x\ln \cos x \, \text{d}x$$ I like to try with the fourier series $$\int_0^{\frac{\pi}{2}}\left(\sum_{k=1}^\infty\frac{(-1)^{k-1}\cos (2kx)}{k}x-x\ln 2\right) \, \text{d}x$$ But I dont know how to do the series part.

share|improve this question
add comment

1 Answer

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 $$-\dfrac{7}{16} \zeta(3) - \dfrac{\pi^2}8 \ln2$$

share|improve this answer
    
Oh Thx! I was not confident with these series :) –  Ryan Feb 13 '13 at 3:58
add comment

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.