Integrating $\sum_{n=1}^{\infty} \frac{\sin nx }{n^4}$

Consider : $\displaystyle f(x)= \sum_{n=1}^{\infty} \frac{\sin nx }{n^4}$

Find : $\displaystyle \int_0^{x} f(t)\ \mathrm{d}t$.

• Can you find $\int_0^x{\sin nt\over n^4}\,dt$? – Gerry Myerson Feb 8 '13 at 11:28
• If $f_N(t)=\sum_{n=1}^N \sin(nt)/n^4$, then $f_N:[0,x]\to\mathbb{R}$ is converges uniformly to $f$ for all $x$. So you can integrate this series term-by-term. – Hanul Jeon Feb 8 '13 at 11:52
• @GerryMyerson : yes I can but how to calculate this : $\displaystyle \sum \frac{\cos {nx} }{n^5}$ – aziiri Feb 8 '13 at 11:55
• @aziiri - if only it were an odd power instead of a power of 4.. – nbubis Feb 8 '13 at 11:58
• Good question. Fourier series? – Gerry Myerson Feb 8 '13 at 12:30

Note that, $$\int_{0}^{x}f(t)\ \mathrm dt=\sum_{n=1}^{\infty}\frac{1}{n^4}\int_{0}^{x}\sin(nt) \ \mathrm dt=\sum_{n=1}^{\infty}\frac{1}{n^4}\left( \frac{1}{n}-\frac{\cos(nx)}{n} \right)$$

$$\implies \int_{0}^{x}f(t) \ \mathrm dt=\sum_{n=1}^{\infty}\frac{1}{n^5}-\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^5}$$

$$\implies f(x)=\zeta(5)-\frac{1}{2}\sum_{n=1}^{\infty} \frac{e^{inx} }{n^5} - \frac{1}{2}\sum_{n=1}^{\infty} \frac{e^{-inx} }{n^5}$$

$$\implies f(x)=\zeta(5)-\frac{1}{2}( \operatorname{Li}_{5}(e^{ix})+ \operatorname{Li}_{5}(e^{-ix}) ),$$

where $\operatorname{Li}_{s}(z)$ is the Polylogarithm function.

• this was my first thought, but I assumed that there is a solution without polylogarithms. – aziiri Feb 8 '13 at 14:42
• @aziiri: I do not think you can get it in terms of simpler functions. – Mhenni Benghorbal Feb 8 '13 at 14:51
• OK, thank you anyway. – aziiri Feb 8 '13 at 15:14
• @aziiri: You are welcome. – Mhenni Benghorbal Feb 8 '13 at 15:16
• @aziiri: By the way, since it was your first thought, why you did not mention it, so we know what we should post. – Mhenni Benghorbal Feb 8 '13 at 15:28