I know how to integrate, but I can't understand how the integral of this fourier serie is calculated. my problem is with integral of the sigma.
fourier:
integral:
Can anyone say me how this integral is calculated?
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1You need to change the order of integration and summation. – copper.hat Apr 21 '13 at 15:41
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Can you explain more please? – hpn Apr 21 '13 at 15:44
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2$\int \sum = \sum \int$ under certain conditions. See: http://math.stackexchange.com/questions/48119/criteria-for-swapping-integration-and-summation-order – Argon Apr 21 '13 at 15:44
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Since $|\cos((2n-1)x)|\le 1$, the series is uniformly convergent and so you can exchange the order of integration and summation.
\begin{eqnarray} \int_0^t (\frac{\pi}{2}-\frac{4}{\pi} \sum_{n=1}^\infty \frac{\cos((2n-1)x)}{(2n-1)^2} )dx &=& \int_0^t\frac{\pi}{2} dx-\int_0^t \frac{4}{\pi} \sum_{n=1}^\infty \frac{\cos((2n-1)x)}{(2n-1)^2} dx \\ &=& \frac{\pi}{2}t - \frac{4}{\pi} \sum_{n=1}^\infty \int_0^t \frac{\cos((2n-1)x)}{(2n-1)^2} dx \\ &=& \frac{\pi}{2}t - \frac{4}{\pi} \sum_{n=1}^\infty \frac{\sin((2n-1)t)}{(2n-1)^3} \end{eqnarray}
copper.hat
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thanks. my problem is with the 3rd line. why (2n-1)^2 became (2n-1)^3 ? according to the integration rules I expect to see [(2n-1)^3]/3 in the integral – hpn Apr 21 '13 at 15:58
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