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By considering the Fourier sine series on the interval $[0,pi]$ for $f(x)=1$, show that $$frac{pi^2}{8}=1+frac{1}{3^2}+frac{1}{5^2}+frac{1}{7^2}+...$$

I am having trouble computing the Fourier sine series, which will have the form $$Sf(x)=frac{a_0}{2}+sum_{k=1}^{infty} a_kcosleft(frac{kpi x}{L}right), kgeq 1.$$
I have computed that $a_0=1$ and $a_n=0$, using $$a_0=frac{1}{L}int_{-L}^{L} f(x) dx, a_k=frac{1}{L}int_{-L}^{L} f(x)cosleft(frac{kpi x}{L}right) dx,$$ where $L$ denotes half a period. Once the correct Fourier series is calculated, showing the result using $$Vert{f}Vert^2_{w}=sum_{j=1}^{infty} A^2_jVertphi_jVert^2_w,$$
should not be so hard.

I have found the Fourier sine series, $$Sf(x)=frac{4}{pi}sum_{j=1}^{infty} frac{sin((2j-1)x)}{(2j-1)}.$$ But I having trouble proving the identity in the title. Using Parseval's identity provided above, I get that $$Vert fVert^2_w=1, sum_{j=1}^{infty} A^2_jVertphi_jVert^2_w=frac{8}{pi}sum_{j=1}^{infty}frac{1}{(2j-1)^2},$$ which does not show the required result (out by a factor of $pi$). Note: I took the weight function, $w$, to be $1$. Please help.

As mentioned in the comments, you're looking for a series of sines, not cosines here. I'm guessing you went for cosines because the function is even, like cosine. But the interval you're looking at here is $[0,pi]$, not $[-pi,pi]$, and in fact in this interval it is the sine function that is symmetric (because $sin(pi-x)=sin x$).

Your Fourier series should then look something like $displaystyle sum a_ksinleft(frac{kpi x}{L}right)$, where the $a_k$ are instead given by
$$a_k=frac{2}{pi}int_0^pi f(x)sinleft(frac{kpi x}{L}right),dx$$
(so integrating over just half of the interval, from $0$ to $pi$).

Another way to get this same series is to instead extend $f$ by defining $$f(x)=begin{cases}1&x>0\-1&x<0end{cases}$$
on the entire interval $[-pi,pi]$. Now you can use the form of the Fourier series you have in your question; since the function is odd, its series will have only sines. On the interval $[0,pi]$ it is just equal to $1$ still, though.