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I'm asked to find the fourier series of the $2 pi $ periodic function f(x) which is $sin(x)$ between $0$ and $pi$ and $0$ between $pi$ and $2pi$

I use the complex form to proceed and get $$frac{1}{2pi}int_{0}^{pi}sin(x)e^{-ikx}dx$$. The complex coefficient $c_k$ I get as result is $frac{-1}{pi(k^2-1)}$ where $k=2n$ (k even) which is also correct according to WolframAlpha.

But then, I'm asked to use this result to evaluate $sum_{n=1}^{infty}frac{1}{(4n^2-1)^2}$. For that, I switch to the real coefficients using $a_k=c_k+c_{-k}, b_k=i(c_k-c_{-k})$. I get: $a_k=frac{-2}{pi(k^2-1)}, a_0=frac{2}{pi}$ and $b_k$ is $0$.

So $f(x)=frac{a_0}{2}+sum_{k=1}^{infty}a_kcos(kx)$

I then use Parseval's theorem to evaluate the sum we are looking for, remembering that k is even, i.e. $k=2n$ and that the function is $0$ between $pi$ and $2pi$:

$$frac{1}{pi}int_{0}^{2pi}|f(x)|^2 dx=frac{a_0^2}{2}+sum_{k=1}^{infty}a_k^2$$ (1)

$$frac{1}{pi}int_{0}^{pi}sin^2(x)dx=frac{frac{2^2}{pi^2}}{2}+sum_{n=1}^{infty}frac{4}{pi^2(4n^2-1)^2}$$ (2)
$$frac{1}{2}-frac{2}{pi^2}=sum_{n=1}^{infty}frac{4}{pi^2(4n^2-1)^2}$$ (3)

So finally I get$frac{pi^2}{8}-frac{1}{2}=sum_{n=1}^{infty}frac{1}{(4n^2-1)^2}$ (4)

However, WolframAlpha gets $frac{pi^2}{16}-frac{1}{2}$ so I must somehow have forgotten a factor of $frac{1}{2}$ or put an extra factor of $2$ by $frac{pi^2}{8}$.

Logically, this missing/extra factor must have happened while I was evaluating $frac{1}{pi}int_{0}^{pi}sin^2(x)dx$ because the $-frac{1}{2}$ at the end that came from the right side of the equality is correct according to WolframAlpha. But even when I evaluate $frac{1}{pi}int_{0}^{pi}sin^2(x)dx$ in WolframAlpha, I get the $frac{1}{2}$ from step (3) which finally becomes $frac{pi^2}{8}$ and again, an factor of $frac{1}{2}$ is missing, so I'm a bit perplex about what's wrong.

Thanks for your help !

Edit: the strange thing is that when I evaluate this sum using Parseval but with the fourier series of |sin(x)| between $0$ and $2pi$, I get the correct result.

I can't check your calculations since you haven't included them, but it is clear that the Fourier series you found is not the Fourier series of $f$. Your function is not even, so it cannot have a Fourier cosine series. For example,

$$ b_1 = frac{1}{pi} int_0^{2pi} f(x) sin(x) , dx = frac{1}{pi} int_0^{pi} sin^2(x) , dx = frac{1}{2}. $$

My guess is that you haven't been careful in checking the special case when integrating the complex form (that is, $int e^{ikx} , dx = frac{e^{ikx}}{ik} + C$ only when $k neq 0$). In fact, the Fourier series of $f$ is given by

$$ sum_{k = 1}^{infty} frac{2}{pi(1-4k^2)} cos(2kx) + frac{1}{pi} + frac{1}{2} sin(x) $$

and you'll get your missing factor from the extra $sin$ term.

The complex coefficient $c_1$ is given by

$$ c_1 = frac{1}{2pi} int_0^{2pi} f(x) e^{-ix} , dx = frac{1}{2pi} int_0^{pi} frac{e^{ix} - e^{-ix}}{2i} e^{-ix} , dx = frac{1}{4pi i} int_0^{pi} (1 - e^{-2ix}) dx = frac{1}{4pi i} left[x - frac{e^{-2ix}}{-2i} right]_{x = 0}^{x = pi} = -frac{1}{4}i. $$
Similarly,

$$ c_{-1} = frac{1}{2pi} int_0^{2pi} f(x) e^{ix} , dx = frac{1}{2pi} int_0^{pi} frac{e^{ix} - e^{-ix}}{2i} e^{ix} , dx = frac{1}{4pi i} int_0^{pi} (e^{2ix} - 1) dx = frac{1}{4pi i} left[frac{e^{2ix}}{2i} - x right]_{x = 0}^{x = pi} = frac{1}{4}i. $$

Hence,

$$ b_1 = i(c_1 - c_{-1}) = i(-frac{1}{4}i - frac{1}{4}i) = frac{1}{2}. $$