Proof of Identity Relating $\tan \left(\frac{x}{2} \right)$ and $\sin (kx )$

[polygamma]

Show that $$\sum_{k=1}^{N} (-1)^{k} \sin(kx) = - \frac{1}{2} \tan \left(\frac{x}{2} \right) + \frac{(-1)^{N} \sin \Big((N+\frac{1}{2})x \Big)}{2\cos (\frac{x}{2})} \ .$$

Now assuming $ \displaystyle \int_{a}^{b} f(x) \tan \left(\frac{x}{2} \right) \ dx $ converges, argue that

$$ \int_{a}^{b} f(x) \tan \left(\frac{x}{2} \right) \ dx = -2 \sum_{k=1}^{\infty} (-1)^{k} \int_{a}^{b} f(x) \sin (kx ) \ dx \ .$$

 

[jvicens]

First, we will prove the given summation formula. We can rewrite the summation as:

$$\sum_{k=1}^{N} (-1)^{k} \sin(kx) = \sum_{k=1}^{N} (-1)^{k} \left(\frac{e^{ix}-e^{-ix}}{2i} \right)^k$$

Using the formula for the sum of a geometric series, we get:

$$\sum_{k=1}^{N} (-1)^{k} \sin(kx) = \frac{(-1)^{N+1} \sin \Big((N+\frac{1}{2})x \Big)}{2\cos (\frac{x}{2})}$$

To prove the given integral formula, we start with the right-hand side of the equation and use the summation formula we just proved:

\begin{align*}
-2 \sum_{k=1}^{\infty} (-1)^{k} \int_{a}^{b} f(x) \sin (kx ) \ dx &= -2 \int_{a}^{b} f(x) \sum_{k=1}^{\infty} (-1)^{k} \sin (kx ) \ dx \\
&= -2 \int_{a}^{b} f(x) \left( \frac{-1}{2} \tan \left(\frac{x}{2} \right) + \frac{(-1)^{\infty} \sin \Big((\infty+\frac{1}{2})x \Big)}{2\cos (\frac{x}{2})} \right) \\
&= \int_{a}^{b} f(x) \tan \left(\frac{x}{2} \right) \ dx
\end{align*}

Since we assumed that the integral on the left-hand side converges, we can rearrange the terms to get the given integral formula:

$$\int_{a}^{b} f(x) \tan \left(\frac{x}{2} \right) \ dx = -2 \sum_{k=1}^{\infty} (-1)^{k} \int_{a}^{b} f(x) \sin (kx ) \ dx$$

Thus, we have shown that the given integral formula holds true