Fixing the closed form of $\sum_{k=1}^nk\sin^2(kx).$

Question

I've been working on finding the closed form of this:$$\sum_{k=1}^nk\sin^2(kx).$$ Using the fact that:$$\sum_{k=1}^nku^k={u\over (1-u)^2}\bigg[nu^{n+1}-(n+1)u^n+1\bigg]\forall u\ge 1\quad (1)$$

I started by substituting $$cos(2kx)={exp(i(2kx))+exp(-i(2kx))\over 2}\quad \quad (2)$$

Which helped me simplify down into $${1\over4}\bigg[n(n+1)-\sum_{k=1}^nk(e^{i2x})^k-\sum_{k=1}^nk(e^{-i2x})^k\bigg]$$ Then I was able to apply $(1)$ to it along with a multiplication of a nifty one in the form of ${exp(i4x)\over exp(i4x)}.$

I then was able to simplify down into: $${1\over4}\bigg[n(n+1)+{1\over 2\cos(2x)-1}\bigg(n(2cos(2x))^{n+1}-(n+1)(2cos(2x))^n+2\bigg)\bigg]$$ Question:

I believed this was the closed form. I was told though that It wasn't complete enough because I didn't take into account when $u=1$ Why do I need to take that into account.

If you'd like me to expand my work just tell me and I will expand it.

EDIT:

So I have figured out that I need to consider when $u=1$ and my $u=e^{12x}$ and $u=e^{-i2x}$ So I found that both of my $u$'s equal $1$ when $x=n\pi$ now do I just go back and plug my $x$ into the original summation. Which since $n$ and $k$ are integers it's always going to be a multiple of $\pi$ so my $\sin$ is always going to be $0$ right?

Answer 1

You stated (1) for $u \ge 1$, but in your application $u = e^{2kxi}$ is complex. What you mean is $u \ne 1$. You then have to consider the case $u = 1$ separately. I don't see how you get a denominator of $2 \cos(2x)-1$: I think it should be $\cos(2x)-1$

Answer 2

Why not just use $\sin^2{k x} = \frac12 - \frac12 \cos{2 k x}$? The sum ends up being

$$\frac{n(n+1)}{4} - \frac12 \sum_{k=1}^n k \cos{2 k x}$$

To do the sum on the RHS, note that

$$\begin{align}\frac12 \sum_{k=1}^n k \cos{2 k x} &= \frac14 \frac{d}{dx}\sum_{k=1}^n \sin{2 k x} \\ &= \frac14 \frac{d}{dx}\Im{\sum_{k=1}^n e^{i 2 k x}}\\ &= \frac14\frac{d}{dx} \Im \left [e^{i 2 x} \frac{e^{i 2 n x}-1}{e^{i 2 x}-1} \right ] \\ &= \frac{d}{dx}\frac{\sin{n x} \sin{(n+1) x}}{4 \sin{x}}\\ &=\frac18 \frac{d}{dx} \left [\cot{x} - \frac{\sin{(2 n+1) x}}{\sin{x}} \right ]\end{align}$$

etc. You get the picture I hope.