# Is there a closed-form of $\sum_{n=1}^{\infty} \frac{\sin(n)}{n^4}$

Is there a closed-form summation result for Fourier series:

$$\sum_{n=1}^{\infty}\frac{\sin(n)}{n^4}?\tag{1}$$

I tried using available result of the following (odd) function :

$$\frac{\pi-x}{2}=\sum_{n\geq 1}\frac{\sin(nx)}{n}.\tag{2}$$

However I couldn't get the series to agree because when I integrate the formula to evaluate coefficients, the powers of $n$ become odd..

• I would try to work with the imaginary part of the series $\sum_{n=1}^\infty\frac{e^{inx}}{n^4}$, and afterwards to evaluate at $x=1$...I am lazy now, sorry. – Matemáticos Chibchas Apr 3 '15 at 13:28
• Sorry but what did you fail to understand from the answers received there, precisely? – Did Apr 3 '15 at 14:00
• @Did Sorry, I cannot understand what you mean – E.H.E Apr 3 '15 at 14:04
• That you are reposting again and again questions that previous answers you received on the site provide powerful tools to attack. Thus there are two possibilities: either you did understand these previous answers and then I wonder why you post this, or you did not and then the game can go on forever. – Did Apr 3 '15 at 14:07

Consider the series \begin{align} f(x) = \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{4}} \end{align} for which upon differentiation the following is obtained. \begin{align} f'(x) &= \sum_{n=1}^{\infty} \frac{\cos(n x)}{n^{3}} \\ f''(x) &= - \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{2}} \\ f'''(x) &= - \sum_{n=1}^{\infty} \frac{\cos(n x)}{n} = \frac{1}{2} \left( \ln(1 - e^{i x}) + \ln(1 - e^{- i x}) \right). \end{align} By integration the following is obtained \begin{align} f''(x) &= \frac{i}{2} \left( Li_{2}(e^{i x}) - Li_{2}(e^{-i x}) \right) + c_{2} \\ f'(x) &= \frac{1}{2} \left( Li_{3}(e^{i x}) - Li_{3}(e^{- i x}) \right) + c_{2} x + c_{1} \\ f(x) &= \frac{-i}{2} \left( Li_{4}(e^{i x}) - Li_{4}(e^{-i x}) \right) + \frac{c_{2} x^{2}}{2} + c_{1} x + c_{0} \end{align} where $Li_{m}(x)$ is the polylogarithm function. In order to determine the constants of integration evaluation of the series is required. This is determined by setting $x = 0$. \begin{align} f(0) &= 0 = c_{0} \\ f'(0) &= \zeta(3) = c_{1} \\ f''(0) &= 0 = c_{2}. \end{align} The result is then \begin{align} \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{4}} &= \frac{-i}{2} \left( Li_{4}(e^{i x}) - Li_{4}(e^{-i x}) \right) + \zeta(3) \, x \end{align}