Sum of $\sin(kx)/k = \pi-x/2$

Question

i need to show this equality: $$ \sum_{k=1}^n \frac{\sin(kx)}{k} = \frac{\pi - x}{2}$$

I should use that $\displaystyle\frac{\sin(kx)}{k} = \int_\pi ^x \cos(kt)\,\mathrm dt$.

I tried many times to solve this, but I just got stuck. Is there a trick or anything I don't see?

Do you know what the sum of $\cos(kx)$ is with respect to $k$? — Chinny84, May 16 '17 at 20:53
That holds only for $x\in(0,2\pi)$ when you consider the limit as $n\to +\infty$. It is a standard exercise in Fourier series than can be simply tackled through $\sin(z)=\text{Im},e^{iz}$ and $\sum_{n\geq 1}\frac{z^n}{n}=\log(1-z)$ for any $z$ such that $|z|<1$. — Jack D'Aurizio, May 16 '17 at 20:55
well ..i got as far as $\int_\pi ^x \frac{sin(nt/2)* cos(tn+t))/2)}{sin(t/2) }dt $ do you mean that? — wondering1123, May 16 '17 at 20:57

score 2 · Answer 1 · edited Dec 26 '19 at 19:24

2

Hint:

$\displaystyle \sum_{k=1}^n \dfrac{\sin(kx)}{k} = \int_\pi ^x \sum_{k=1}^n \cos(kt)\,\mathrm dt$ by linearity of integral then write $\displaystyle\sum_{k=1}^n \cos(kt)$ as the real part of $\displaystyle\sum_{k=1}^n e^{ikt}$ which is a geometric sum.

edited Dec 26 '19 at 19:24

an4s

3,716

answered May 16 '17 at 20:55

Hugh the Thistle

106

score 1 · Answer 2 · answered May 16 '17 at 21:53

HINT

$$\sum\limits_{k=1}^n{\sin(kx)\over k} = -\sum\limits_{k=1}^n\int\limits_0^x\cos(kx)dx = -\Re\int\limits_0^x\left(\sum\limits_{k=1}^ne^{ikx}\right)dx = -\Re\int\limits_0^x{e^{ikx}\over1-e^{ikx}}dx$$ $$ = \Re\left({1\over ik}\int\limits_0^x{d\left(1-e^{ikx}\right)\over1-e^{ikx}}dx\right) = \Re\left({1\over ik}\log(1-ikx)\right) = {1\over k}\Im\log(1-ikx) = {1\over k}\left(\arg(1-ikx)+2\pi m\right) = {1\over k}\left(-\arctan kx+2\pi m\right)$$

Sum of $\sin(kx)/k = \pi-x/2$

2 Answers2