# How might I evaluate $\sum_{n=1}^{\infty}\frac{\sin(nx+c)}{n}$?

I was trying to evaluate the sums:

$$S_1=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}$$ $$S_2=\sum_{n=1}^{\infty}\frac{\sin(nx+c)}{n}$$

I've found that $$S_1=\tan^{-1}\left(\cot\left(\frac{x}{2}\right)\right)$$, however I can't seem to evaluate $$S_2$$. The problem is the $$+c$$ making it hard to simplify $$\Im\left(e^{ixn}e^{ic}\right)$$ with a similar technique.

To evaluate $$S_1$$, I said:

$$S_1=\sum_{n=1}^{\infty}\frac{\Im\left(e^{ix\cdot n}\right)}{n}=\Im\left(\sum_{n=1}^{\infty}\frac{\left(e^{ix}\right)^n}{n}\right)$$ Now, using the taylor series expansion of $$\ln(1-a)$$: $$-\ln(1-a)=\sum_{n=1}^{\infty}\frac{a^n}{n}$$ We have: $$S_1=\Im\left(-\ln\left(1-e^{ix}\right)\right)=\Im\left(-\ln\left(1-\cos(x)-i\sin(x)\right)\right)$$ Converting $$1-\cos(x)-i\sin(x)$$ into modulus-argument form: $$1-\cos(x)-i\sin(a)=\left(\sqrt{(1-\cos(x))^2+\sin^2(x)}\right)e^{i\tan^{-1}\left(\frac{-\sin(x)}{1-\cos(x)}\right)}$$ Hence: $$-\ln\left(1-\cos(x)-i\sin(x)\right)=-\ln\left(\sqrt{(1-\cos(x))^2+\sin^2(x)}\right)-i\tan^{-1}\left(\frac{-\sin(x)}{1-\cos(x)}\right)$$ So: $$S_1=\Im\left(-\ln\left(1-\cos(x)-i\sin(x)\right)\right)=-\tan^{-1}\left(\frac{-\sin(x)}{1-\cos(x)}\right)$$ Simplifying with trig identities yields: $$S_1=\tan^{-1}\left(\cot\left(\frac{x}{2}\right)\right)$$

• Use $\sin(nx +c) =\cos(c)\sin(nx)+\sin(c)\cos(nx)$ Commented Mar 26, 2019 at 22:53
• Hmm, your Taylor series expansion for $\ln(1-a)$ actually only holds for $|a|<1$. Commented Mar 26, 2019 at 23:07
• Also note that $\arctan(\cot(\tfrac x2)) = \tfrac{\pi-x}{2}$ for $x\neq 2k\pi$. Commented Mar 26, 2019 at 23:13
• @amsmath Does that make my derivation invalid? Commented Mar 26, 2019 at 23:20
• Hey Daniel. I'm not sure, to be honest. Commented Mar 26, 2019 at 23:34

$$\text{Im}(e^{inx}e^{ic}) =\cos(c)\text{Im}(e^{inx})+\sin(c)\text{Re}(e^{inx})$$
• Thanks - I got $\cos(c)\tan^{-1}\left(\cot\left(\frac{x}{2}\right)\right)-\frac{1}{2}\sin(c)\ln(2-2\cos(x))$ Commented Mar 26, 2019 at 23:07