How to show that $\sum_{n=1}^{\infty} \sin(nx)=\frac{1}{2}\cot(\frac{x}{2})$ in the sense of distributions?

Question

Show that in the sense of distributions $$\sum_{n=0}^{\infty}\sin(nx)=\frac{1}{2}\cot(\frac{x}{2})$$

---

I know how to derive the result if this summation is finite, but how could I get this infinite sum from the finite case?

Answer 1

In the sense of Distributions, the statement

$$\sum_{n=1}^\infty \sin(nx)\sim \frac12\cot(x/2)$$

is interpreted to mean that for any suitable test function $\phi(x)$ we have

$$\lim_{N\to \infty}\sum_{n=1}^N \int_{-\infty}^\infty \phi(x)\sin(nx)\,dx=\int_{-\infty}^\infty \phi(x)\left(\frac12\cot(x/2)\right)\,dx \tag1$$

Evidently, the space of suitable test functions is restricted such that

$$\int_{-\infty}^\infty \phi(x)\left(\frac12\cot(x/2)\right)\,dx<\infty\tag2$$

---

To see that $(1)$ is correct subject to $(2)$, we first note that

$$\sum_{n=1}^N \sin(nx)= \frac12\cot(x/2)-\frac12 \csc(x/2)\cos((N+1/2)x)$$

Then upon applying the Riemann-Lebesgue Lemma we find immediately that

$$\begin{align} \lim_{N\to \infty}\sum_{n=1}^N \int_{-\infty}^\infty \phi(x)\sin(nx)\,dx&=\int_{-\infty}^\infty \phi(x)\left(\frac12\cot(x/2)\right)\,dx\\\\ &-\lim_{N\to \infty}\int_{-\infty}^\infty \phi(x)\left(\frac12\csc(x/2)\cos((N+1/2)x)\right)\,dx\\\\ &=\int_{-\infty}^\infty \phi(x)\left(\frac12\cot(x/2)\right)\,dx \end{align}$$

And we are done!