Fourier series of $\frac{1}{x}$

Question

What is the Fourier series expansion of $\frac{1}{x}$ ?

The best method I could come up with was shifting the function by 'k' (shifting the function to $\frac{1}{x - k}$), so that while calculating the coefficients you don't run into the discontinuity of 1/x.

Is there a different method to calculate the Fourier series of $\frac{1}{x}$.

On $(0,2\pi)$ it doesn't make sense but on $(-\pi,\pi)$ it does, search about "principal value" — reuns, Dec 16 '18 at 00:35
Let $h(y) = pv.(\int_{-\pi}^\pi \frac{e^{i xy}}{x}dx)$ then $h'(y) = \int_{-\pi}^\pi i e^{i xy}dx =2i \frac{\sin(\pi y)}{y}$ and since $h(0) = 0$ then $h(y) =2i \int_0^y \frac{\sin(\pi u)}{u} du$ — reuns, Dec 16 '18 at 00:38
@mathworker21 for example because $pv.(\int_{-\pi}^\pi \frac{f(x)}{x}dx ) = \int_{-\pi}^\pi \frac{f(x)-f(0)}{x}dx$ which can be taken as the definition — reuns, Dec 16 '18 at 00:45
@reuns Are you sure such one is the derivative of $h(y)=h(y,x)$ with respect to $y$ even though the function is periodic in the x axis? — Dr Potato, Mar 10 '22 at 02:06

Alex Jones · Accepted Answer · 2023-10-03T20:51:05.300

The Fourier series only exists for periodic functions which are integrable over a period. You can choose an interval and consider the periodic extension of $\frac{1}{x}$ over that interval, but if that interval contains $0$ (even as an endpoint), it will not be integrable.

At risk of cluttering what was a simple and to-the-point answer, I'd like to address some of the comments this answer has gotten over the years.

It's important to note that, when I say Fourier series, I mean the full Fourier series (not sine or cosine series), and without making use of the Cauchy principal value. For such a series, all the coefficients existing is a necessary condition for the series existing, and since $a_0 \propto \int_L f$, integrability of the function over the chosen period is a necessary condition for the series to exist.

This can be circumvented by using the above modifications. The Cauchy principal value allows odd functions with a singularity at $0$ to retain the property that integrals over a symmetric intervals vanish, so all the $a_n$ coefficients vanish as a result. The sine series, on the other hand, just assumes from the outset that the $a_n$ vanish, with the trade-off of being a series for the odd-periodic extension of $f$.

(Despite the fact that a sine series is a Fourier series, I maintain the distinction in this case because it does not arise from the standard coefficient-based definition of Fourier series without invoking the CPV. You are welcome to disagree, as this is purely a semantic distinction.)

As for square-integrability, that is a sufficient condition, but not necessary for the convergence of a Fourier series, unless you limit "convergence" to mean "convergence in the $L^2$ norm". Different types of convergence have different conditions, and square-integrability is not necessary for all of them; for example, one can construct a non-square-integrable function which has a Fourier series that is pointwise convergent almost everywhere.

I would not make such a statement. Absolute integrability of $f$ and further conditions of regularity such as local bounded variation guarantees convergence of the Fourier series at a point. But the Fourier sine series of $f(x) = 1/x$ on $[0,\pi]$ exists. Note that $b_n = \frac{2}{\pi}\int_0^\pi \frac{\sin nx}{x} , dx = \frac{2}{\pi} Si(n\pi)$ exists. — RRL, Dec 16 '18 at 06:50
Yeah as @RRL points out we would better try to give an answer before just saying there is no such an answer. Take a look to "Fourier transform of 1/x" https://math.stackexchange.com/q/1522986/622884 Never loose your hope on making sense of what is hard to think, of making the impossible possible! — Dr Potato, Dec 23 '21 at 02:05
Actually, integrability is not the criterion, but square integrability, if you want to stick to the formal definition of Fourier series. By construction, (trigonometric) Fourier series can be used to represent any function in $L^2(I)$ (which does not contain $1/x$). — DominikS, Oct 02 '23 at 09:54

Dr Potato · Answer 2 · 2022-01-31T17:25:04.483

Let $\hat{f}$ be the Fourier transform of some function $f$.

Consider the image of the integer numbers under this Fourier transform $$\{\hat{f}(n) | \ n\in\mathbb{Z}\}$$ Observe that the inverse Fourier series of this set, $$\sum\limits_{n\in\mathbb{Z}} \hat{f}(n) e^{i n}, $$ equals (a.e) some periodic function. If you calculate the Fourier Transform of $1/x$, as it is done here and here, you get the Fourier series coefficients $$c_n = \hat{f}(n).$$ In this case, if the domain of $\frac{1}{x}$ is $[-\pi, \pi)$ then $$c_n = −\frac{i}{2}sgn(n).$$

The same way it must be possible to make the same analysis for the Fourier Series.

score 0 · Answer 3 · answered Dec 16 '19 at 23:08

It is true that $f(x)=1/x$ is not locally integrable at $0$, so there is a problem.

As @reuns points out, once choice is to interpret "$1/x$" as being the Cauchy principal value integral $\lim_{\epsilon\to 0}\int_{\epsilon\le|x|\le\pi}{e^{2\pi inx}\,dx\over x}$. A similar "principal value integral" can be arranged on $[0,2\pi]$, instead, if desired.

But the non-local-integrability cannot be evaded. The "principal value integral" is not a literal integral, since the literal (improper) integral would require that the limits below $0$ and above $0$ be independent... which is not possible.

So, truly, one is probably asking about the Fourier series of a distribution, given by the principal-value-integral (as in @reuns' answer).

Fourier series of $\frac{1}{x}$

3 Answers3