What is the limit (as a distribution) of this Fourier series, similar to that of a Dirac Delta?

Question

I know that a representation of the Dirac delta function is $\sum_{n=-\infty}^\infty e^{inx}=2 \pi \delta(x)$. I am trying to figure out if the series with positive $n$ only $\sum_{n=0}^\infty e^{inx}$ has a simple limit as a distribution, but I cannot find any sources. The real part should be $\pi \delta(x)+\frac{1}{2}$, but what about the imaginary one?

Answer 1

I'll assume you are thinking in $\mathbb R/2\pi\mathbb Z$ so that your equations are correct, and there is no need for a Dirac comb.

First off, your sum converges in the sense of distributions. Indeed, taking a (periodic) test function $\phi$, the bracket amounts to summing its positive index Fourier coefficients $\phi_n = \int_0^{2\pi}\phi(x)e^{inx}dx$. By smoothness, they decay fast, i.e. $\phi_n = O(n^{-k})$ for all non positive integer $k$ since $\phi\in\mathcal C^k$, so the sum converges.

One way of doing it is to look at the limit when $r\to1^-$ (Abel summation) of: $$ f_r(x) = \sum_{n=0}^\infty r^ne^{inx} $$ You recognise the geometric series: $$ f_r(x) = \frac1{1-re^{ix}} $$ Naively setting $r=1$ you get: $$ f(x) = \frac1{1-e^{ix}} $$ The issue is that it is not a well defined distribution due to the simple poles at $x=0$. You view it as the distributional limit, which is your original sum by Abel's theorem. You can explicitly calculate the limit bracket for a (periodic) test function: $$ \langle f_r,\phi\rangle \xrightarrow{r\to1^-} 2\pi\phi(0)+\int_0^{2\pi}\frac{(\phi(x)-\phi(0))dx}{1-e^{ix}} $$ You can also relate it to the Cauchy principal value as given in lcv's answer.

Alternatively, you can relate it to the distributional derivative of: $$ g(x) = \sum_{n=1}^\infty \frac{e^{inx}}n = -\ln(1-e^{inx}) \\ f(x) = -ie^{-ix}g'(x) $$ This approach agrees with the previous ones.

Hope this helps.

Answer 2

I will just add a few details to the proof linked in the comment with no attempt on rigor. As you correctly said, the real part equals

$$ \Re{ \sum_{n=0}^\infty e^{inx} } = 1+ \sum_{n=1}^\infty \cos(nx) = \pi \sum_{n=-\infty}^\infty \delta(x-2\pi n) +\frac{1}{2}. $$

This can be proved with the formula for the full comb. For the imaginary part we need

$$ \sum_{n=1}^\infty \sin(nx) $$

Using Lagrange trigonometric identities we get

$$ \sum_{n=1}^\infty \sin(nx) = \frac{1}{2} \mathrm{PV} \cot(x/2) $$

Where the Principal Value has to be inserted to renormalize the behavior at the pole.

All in all

$$ \sum_{n=0}^\infty e^{inx} = \pi \sum_{n=-\infty}^\infty \delta(x-2\pi n) +\frac{1}{2} +i \frac{1}{2} \mathrm{PV} \cot(x/2) $$