Expand $\cos(x)$ in sinus-series in the interval $(0,\pi/2)$ and use the result to calculate $$\sum_{n=1}^{\infty}\frac{n^2}{(4n^2-1)^2}.$$

Since we only need the odd terms in the Fourier expansion, we know that $a_n=0$, so we only need $b_n$ in order to find the sine expansion.

In my book, there is a chapter on "Fourier series on intervals", and there they state that:

If $f$ is a piecewise smooth function on $[0,l]$, then the Fourier sine expansion is given by

$$f(x)=\sum_{n=1}^{\infty}b_n\sin\left(\frac{\pi n x}{l}\right),\tag{1}$$


$$b_n=\frac{2}{l}\int\limits_{0}^{l}f(x)\sin\left(\frac{\pi n x}{l}\right).\tag{2}$$

So we should be able to use only $(1)$ and $(2)$ with $l=\pi/2$ to solve this problem. So

\begin{align} b_n=\frac{4}{\pi}\int\limits_{0}^{\pi/2}\cos(x)\sin(2nx)dx = \frac{4(\sin(\pi n)-2n)}{\pi(1-4n^2)}\tag{3}. \end{align}

This gives

$$\cos(x)=\frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n}{4n^2-1}\sin(2nx), \quad \forall \ n\in\mathbb{N}\tag{4}$$

2 questions remain:

  1. I used wolfram alpha to calculate the integral in $(3)$. What methods do I use to do this? Repeated integration by parts and solve for the integral?
  2. How do I calculate the desired sum?
  • $\begingroup$ The function to expand is probably not the regular $\cos$. Instead, consider the function that coincides with $\cos$ on $(0, \frac \pi 2)$, that is odd and is $\pi$ periodic. $\endgroup$ Jan 25, 2019 at 16:30
  • $\begingroup$ But they explicitly state that I'm to expand $\cos(x)$. So I am pretty convinced that $\cos(x)$ is the regular cos that is to be expanded. I might have misunderstood your comment. $\endgroup$
    – Parseval
    Jan 25, 2019 at 17:30

2 Answers 2


As posted by the OP, two steps remain:

  1. Perform the integral:

$$ b_n=\frac{4}{\pi}\int\limits_{0}^{\pi/2}\cos(x)\sin(2nx)dx \\ = \frac{1}{i \pi}\int\limits_{0}^{\pi/2}e^{ix(1+2n)}- e^{ix(1-2n)}+e^{-ix(1-2n)}-e^{-ix(1+2n)}dx \\ = \frac{1}{i \pi} \left(\frac{e^{ix(1+2n)}}{i(1+2n)}|_{0}^{\pi/2} - \frac{e^{ix(1-2n)}}{i(1-2n)}|_{0}^{\pi/2} + \frac{e^{-ix(1-2n)}}{-i(1-2n)}|_{0}^{\pi/2} -\frac{e^{-ix(1+2n)}}{-i(1+2n)}|_{0}^{\pi/2} \right)\\ = \frac{1}{i \pi} \left(\frac{i(-1)^n - 1}{i(1+2n)} - \frac{i(-1)^n - 1}{i(1-2n)} + \frac{-i(-1)^n - 1}{-i(1-2n)} - \frac{-i(-1)^n - 1}{-i(1+2n)} \right)\\ = \frac{1}{- \pi (1 - 4 n^2)} \left( -4n(i(-1)^n - 1) - 4n (-i(-1)^n - 1) \right)\\ = \frac{-8n}{\pi(1-4n^2)} \\ = \frac{4(\sin(\pi n)-2n)}{\pi(1-4n^2)} $$ where the last step only indicates the "desired" solution by WolframAlpha, where anyway $\sin(\pi n) = 0$ .

2 Calculating the desired sum.

From OP's last result, $$ \cos(x)=\frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n}{4n^2-1}\sin(2nx), $$ multiply this formula with $\cos(x) $ and integrate, using the integral which was just derived:

$$ \frac{\pi}{4} = \int\limits_{0}^{\pi/2}\cos^2(x) dx = \frac{8}{\pi} \sum_{n=1}^\infty \frac{n}{4n^2 -1} \int\limits_{0}^{\pi/2}\cos(x)\cdot\sin(2nx) dx = \\ \frac{8}{\pi} \sum_{n=1}^\infty \frac{n}{4n^2 -1} \cdot \frac{2 n}{4n^2 -1}\\ = \frac{16}{\pi} \sum_{n=1}^\infty \frac{n^2}{(4n^2 -1)^2} $$ so finally

$$ \sum_{n=1}^\infty \frac{n^2}{(4n^2 -1)^2} = \frac{\pi^2}{64} $$

  • $\begingroup$ Wonderful, thanks a lot! $\endgroup$
    – Parseval
    Jan 28, 2019 at 12:46

Consider the function $f(x) = \cos(x) \cdot {\rm{sign}}(x)$ in the interval $(-\pi, \pi)$. This function is identical to $\cos(x)$ in the interval $(0, \pi/2)$ so the expansion will be correct in that interval.

You want $f(x) = \sum_{m=1}^\infty b_m \sin(mx)$.

Integrating with $\sin(nx)$ gives

$$ \int\limits_{-\pi}^{\pi}\sin( n x)\sin(nx)dx = \pi $$ $$ \int\limits_{-\pi}^{\pi}\sin( m x)\sin(nx)dx = 0 \quad (m\ne n) $$ and

$$ \int\limits_{-\pi}^{\pi}\cos(x)\cdot {\rm{sign}}(x) \cdot \sin(nx)dx = 2 \int\limits_{0}^{\pi}\cos(x)\cdot \sin(nx)dx = 2 \frac{n (1+\cos(\pi n))}{n^2 - 1} \tag{*} $$

So you have $$ \cos(x) = \frac{2}{\pi} \sum_{n=1}^\infty \frac{n (1+\cos(\pi n))}{n^2 - 1} \sin(nx) $$ where the coefficients are nonzero for even n.

Substituting $n$ with $2 m$ gives $$ \cos(x) = \frac{8}{\pi} \sum_{m=1}^\infty \frac{m}{4m^2 -1} \sin(2 m x) $$

No use the integral in (*) again: multiply the last formula with $\cos(x) $ and integrate:

$$ \frac{\pi}{2} = \int\limits_{0}^{\pi}\cos^2(x) dx = \frac{8}{\pi} \sum_{m=1}^\infty \frac{m}{4m^2 -1} \int\limits_{0}^{\pi}\cos(x)\cdot\sin(2mx) dx = \frac{32}{\pi} \sum_{m=1}^\infty \frac{m^2}{(4m^2 -1)^2} $$ so finally

$$ \sum_{m=1}^\infty \frac{m^2}{(4m^2 -1)^2} = \frac{\pi^2}{64} $$

  • $\begingroup$ Thanks for a nice solution and I'll give you an upvote. However this answer fails to provide an answer that aids in helping me to understand my mistake. I'd rather learn where I went wrong in my own attempt then see a customized soluton that would only work for this particular problem. For example, I've never had to deal with the $sgn(x)$ function and there is not a single example in the book where they use a method like this to solve a similar problem. This problem is solveable using standard given formulae and I want to know why it did not work this time. $\endgroup$
    – Parseval
    Jan 25, 2019 at 19:29
  • $\begingroup$ If you want it without that symmetry condition, you would have to set $\cos(x) = \sum_{m=1}^\infty b_m \sin(2mx)$ which we know (in hindsight) from the solution above. It is not clear to me a priori that the odd terms $\sin((2m+1)x)$ should not be present in the series at all - do you have an argument? Then, you can integrate $\int_0^{\pi/2} \sin(2mx) \sin(2nx) dx = 0$ for $m \ne n$ and everything works as above. (continued) $\endgroup$
    – Andreas
    Jan 25, 2019 at 19:59
  • $\begingroup$ (continued) The point is that in general, $\int_0^{\pi/2} \sin(mx) \sin(nx) dx $ is not zero and therefore the calculation of the coefficients does not work straightforward if you cannot exclude the odd values of m right from the start. $\endgroup$
    – Andreas
    Jan 25, 2019 at 19:59
  • $\begingroup$ I've been working on this problem a bit further now, I'll edit the answer and show you what I mean by applying the formulas straight on. Stand by. $\endgroup$
    – Parseval
    Jan 25, 2019 at 20:52
  • $\begingroup$ Please see my edit now Andreas. Thanks for the time and effort you put in, I apreciate it. Oh and the answer $\pi^2/64$ is indeed correct. $\endgroup$
    – Parseval
    Jan 25, 2019 at 21:13

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