Test question regarding convergence of Fourier series

Question

I'm preparing for a test and I have no clue how I should solve the following question.

Let $f:\Bbb{R}\to\Bbb{R}$ be $2\pi \text{-periodic}$ function such that $f(0)=1$ and $$\forall x\in[-\pi,\pi]\setminus\{0\}, \qquad f(x)=1+\sin\left(\frac{\pi^2}{x}\right)$$ Is Fourier series of $f$ converging at $x=0$? If so, what is the sum at $x=0$?

So, I tried to evaluate the term for Fourier series coefficients: $$\hat{f(n)}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left[1+\sin\left(\frac{\pi^2}{x}\right)\right]e^{-inx}dx$$

Now, how can I continue with this integral? what should I do with $x=0$?

Please help, thank you!

Answer 1

The answer is NO. Furthermore, since the function has no limit as x converges to 0 then Fourier series doesn't exist at all.

Hope it Helps

Answer 2

The function is integrable in [-$\pi$, $\pi$] so it does have a Fourier series. Use the convergence criterion for the Fourier series at $0$:

A necessary and sufficient condition for the Fourier series $T(x)$ of $f$ to converge pointwise to $c$ at $x$ is that there exists a fixed $\delta$ such that $ 0 < \delta<\pi$ and $\int_0^\delta {{g}(u)\frac{{\sin \left( {nu} \right)}}{u}du \to 0} $ pointwise as n tends to infinity.

Here ${g_c}(u) = \frac{1}{2}\left( {f(x + u) + f(x - u) - 2c} \right)$.

Take $x =0$, $c =1$, then ${g_c}(u) =0$. Thus trivially your Fourier series converges to $1$ at $0$.

You can also use Dini's test.

For details on convergence criterion see

Convergence of Fourier Series

https://037598a680dc5e00a4d1feafd699642badaa7a8c.googledrive.com/host/0B4HffVs7117IbmZ2OTdKSVBZLVk/Fourier%20Series/Convergence%20of%20Fourier%20Series.pdf