Evaluating $\sum\limits_{k=1}^{n-1} k\cos (k x)$

Question

My intent is to find an expression for the following summation, $$\sum_{k=1}^{n-1} k\cos (k x).$$ We know that (cfr. Sines and Cosines of Angles in Arithmetic Progression by M. P. Knapp) $$\sum_{k=0}^{n-1} \cos (k x) = \frac{\sin(nx/2)}{\sin(x/2)} \cos\left(\frac{(n-1)x}{2}\right) $$ and I was wondering if something similar can be found if there is a and additional $k$-term in the summation. Thank you for your help.

Answer 1

Hints: Step 1: $\sum_{k=1}^{n-1}e^{ikx}$ is a geometric sum. Write down its value. Step 2: differentiate w.r.t. $x$. Step 3: divide by $i$. Step 4: take real part on both sides.

Answer 2

Using the above suggestion from User Simply Beautiful Art, that is, by writing the expression as a derivative of a sine function, we finally get: \begin{align} \sum_{k=1}^{n-1} k \cos (kx) &= \sum_{k=1}^{n-1} \frac{d}{dx} \sin{kx} \\ &= \frac{d}{dx} \left(\frac{\sin(kx/2)}{sin(x/2)} \sin\left(\frac{n-1}2 x\right)\right) \\ &= \frac 1 2 \left(k \frac{\cos(kx/2)}{\sin{x/2}}-\frac{\sin(kx/2)}{\sin^2(x/2)} \cos(x/2)\right)\sin\left( \frac {n-1} 2 x\right) \\ &+ \frac{n-1}2 \frac{\sin(kx/2)}{\sin(x/2)} \cos\left(\frac{n-1}2 x\right) \end{align} where in the second row we use again the formula from the reference given in the opening post. Thank you for your suggestions!