Find $\sum_{k \geq 1} e^{itk}$ in the sense of distribution - $\delta(x-a)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{i(x-a)t}dt$ I have to solve $Z(t)=\sum_{k \geq 1} e^{itk}$ in the sense of distribution (generalized function), i.e., $<\sum_{k \geq 1} e^{itk}, \varphi>$, where $\varphi$ is a test function. So far, by the Poisson Summation Formula, we know that if $$f(x)=e^{\frac{itx}{2 \pi}} \implies \sum_{k \in \mathbb{Z}} f(2 \pi k)=\sum_{k \in \mathbb{Z}} e^{itk} = \sum_{k \in \mathbb{Z}} \hat{f}(k),$$ with $\hat{f}(k)=\int_{-\infty}^{\infty} f(x) e^{-2i \pi kx} = \int_{-\infty}^{\infty} e^{ix(\frac{t}{2 \pi}-2 \pi k)} dx$. I think I could find $Z(t)$ in function of the Dirac-$\delta$-function, because $\delta(x-a)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{i(x-a)t}dt$. However, I don't know how to conclude this question. Is there anyone could help me to finish this problem?
 A: I'm going to post my answer, even though I think it constitutes only about 80% of a proof.
Let
$$
Z(t) = Z_c(t) + i Z_s(t)\, ,
$$
where
$$
Z_c(t) = \sum_{k = 1}^{\infty}\cos(k t)
$$
and
$$
Z_s(t) = \sum_{k = 1}^{\infty}\sin(k t)
$$
Now, using the complex exponent version of $\cos(kt)$, $Z_c(t)$ can be written as
\begin{align}
Z_c(t) &= \frac{1}{2}\left[\sum_{k = -\infty}^{\infty}\exp(i k t) - 1\right]\\
&= \frac{1}{2}\left[2\pi \sum_{k = -\infty}^{\infty}\delta(t - 2\pi k) - 1\right]\, ,
\end{align}
where the latter identity can be proven using Poisson's summation formula.
\begin{align}
Z_s(t) &= \lim_{M\rightarrow\infty} \sum_{k = 1}^{M}\sin(k t)\\
&= \frac{1}{2\tan(t/2)} - \lim_{M\rightarrow\infty}\frac{\cos(t(M+1/2))}{2 \sin(t/2)}\, .
\end{align}
All of the above can be proven by representing $\sin(kt)$ as a sum of complex exponentials, and using the sum of a geometric series along with some trig identities.
Now, here comes the non-rigorous bit that perhaps a real mathematician can improve upon. If we imagine integrating $Z_s(t)$ with respect to some smooth test function $\varphi(t)$, then the last term above (the limit term with the $M$ in it) will oscillate arbitrarily fast as $M$ becomes large, and will "average" to zero. The principle value of the point at $t=0$ will also be zero.
Thus we are left with:
$$
Z(t) = -\frac{1}{2} + \pi\sum_{k=-\infty}^{\infty}\delta(t - 2\pi k) + \frac{i}{2\tan(t/2)}\, .
$$
