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Question Number 33894 by math khazana by abdo last updated on 26/Apr/18

$$1)letfR\to C2\pi periodiceven/f\left(x\right)=x$$$$\mathrm{\forall }x\in [0,\pi [developpfatfourierserie$$$$2)calculate\sum _{p=0}^{\mathrm{\infty }}\frac{1}{{(2p+1)}^2}.$$

Commented by abdo imad last updated on 28/Apr/18

$$fisevensof\left(x\right)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cos\left(nx\right)with$$$$a_n=\frac{2}{T}{\int }_{\left[T\right]}f\left(x\right)cos\left(nx\right)dx=\frac{2}{2\pi }{\int }_{-\pi }^{\pi }xcos\left(nx\right)dx$$$$=\frac{2}{\pi }{\int }_0^{\pi }xcos\left(nx\right)dx\Rightarrow \frac{\pi }{2}{a}_n={\int }_0^{\pi }xcos\left(nx\right)dxletintegrate$$$$byparts{u}^{}=xand{v}^{{}^{\prime }}=cos\left(nx\right)\Rightarrow$$$$\frac{\pi }{2}{a}_n={\left(\frac{x}{n}sin\left(nx\right)\right]}_0^{\pi }-{\int }_0^{\pi }\frac{1}{n}sin\left(nx\right)dx$$$$=-\frac{1}{n}{\int }_0^{\pi }sin\left(nx\right)=\frac{1}{n^2}{\left[cos\left(nx\right)\right]}_0^{\pi }=\frac{1}{n^2}({(-1)}^n-1)\Rightarrow$$$$a_n=\frac{2}{\pi n^2}({(-1)}^n-1)\Rightarrow a_{2n}=0and{a}_{2n+1}=\frac{-4}{\pi {(2n+1)}^2}$$$$\frac{\pi }{2}{a}_0={\int }_0^{\pi }xdx=\frac{{\pi }^2}{2}\Rightarrow a_0=\pi \Rightarrow$$$$f\left(x\right)=\frac{\pi }{2}-\frac{4}{\pi }\sum _{n=0}^{\mathrm{\infty }}\frac{cos(2n+1)x}{{(2n+1)}^2}\left(d\right)$$$$2)lettakex=0in\left(d\right)weget$$$$0=\frac{\pi }{2}-\frac{4}{\pi }\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}\Rightarrow \frac{4}{\pi }\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}=\frac{\pi }{2}\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}=\frac{{\pi }^2}{8}.$$

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