"Summing" the series $\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-...$

Question

"Summing" the series $\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...$

Pose $$S=\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...$$ $$C=\cos(x)-\dfrac{1}{2}\cos(2x)+\dfrac{1}{3}\cos(3x)-\dfrac{1}{4}\cos(4x)+...$$

$$C+iS = e^{ix}-\dfrac{1}{2}(e^{ix})^2+\dfrac{1}{3}(e^{ix})^3-\dfrac{1}{4}(e^{ix})^4+...$$

Let $t$ = $e^ix$

Then we have a series of $t-\dfrac{t}{2}+\dfrac{t}{3}-\dfrac{t}{4}+...=\log(1+t)$

Which is $\log(1+e^{ix})=\log(1+\cos(x)+i\sin(x))$, use the formula $\log(A+iB)=\dfrac{1}{2}\log(A^2+B^2)+\arctan\left(\dfrac{B}{A}\right)$

$\log([1+\cos(x)]^2+i\sin(x))=\dfrac{1}{2}\log(\log([1+\cos(x)]^2+i\sin^2(x))+\arctan\left(\dfrac{i\sin(x)}{1+\cos(x)}\right)$. Since we are interested only in the imaginary part, we have the sum for $S$ is:

$$\arctan\left(\dfrac{i\sin(x)}{1+\cos(x)}\right)$$

I don't know what to do next.

In his paper "Subsidium Calculi Sinuum", Euler wrote the series has the "sum"

$$\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...=\dfrac{x}{2}$$

We obtained this by multiplying $dx$ integrating (his words is Illa autem series per $dx$ multiplicata et integrata dat:"

$$\cos(x)-4\cos(2x)+9\cos(3x)-16\cos(4x)...=0$$

I don't know how he obtains $\dfrac{x}{2}$, since the left hand side is $0$, how can it be $x/2$. Returning to my own "sum", how can I obtain the $1/2$

This series is quite important because it appears in Fourier "Analytical Theory of Heat".

Answer 1

You are almost there! We have:

$$\ln(1 + \cos(x) + i\sin(x))= \ln(A + iB) = \frac{1}{2}\ln(A^{2}+B^{2})+i\arctan\bigg(\frac{B}{A}\bigg)$$

$$=\frac{1}{2}\ln\big((1 + \cos(x))^{2} + \sin^{2}(x)\big)+i\arctan\bigg(\frac{\sin(x)}{1 + \cos(x)}\bigg)$$

Then, the imaginary part is:

$$\arctan\bigg(\frac{\sin(x)}{1 + \cos(x)}\bigg)$$

Using the identity $\displaystyle\tan\bigg(\frac{x}{2}\bigg) = \frac{\sin(x)}{1 + \cos(x)}$:

$$ = \arctan\bigg(\tan\bigg(\frac{x}{2}\bigg)\bigg) = \boxed{\frac{x}{2}}$$

Of course, with this approach you will run into the usual domain problems with inverse trig functions and complex logarithms.

Answer 2

Note

$$S’(x) = \cos x - \cos2x +\cos 3x -\cos 4x+ \cdots\\ = Re(e^{i x} - e^{i 2x} + e^{i 3x} -e^{i 4x} + \cdots) = Re \frac 1{1+ e^{-i x}}= \frac 12 $$

Thus

$$S(x) = \int_0^x S’(t) dt = \frac12x$$