# Explicit example of a Fourier transform of $f \in L^2(\mathbb R) \setminus L^1(\mathbb R)$

The Fourier transform shall be defined by

$$\hat f(\xi) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i \cdot \xi x} dx$$

The Fourier transform is well-defined for $$f \in L^1(\mathbb R)$$, that is, $$f$$ absolutely integrable, as can easily be checked.

However, we also know that the Fourier transform is well-defined for $$f \in L^2(\mathbb R)$$. In textbooks, this is usually shown by defined the Fourier transform over the Schwartz space and then taking the norm closure.

I am wondering whether there is an explicit and computable example of a function $$f \in L^2(\mathbb R) \setminus L^1(\mathbb R)$$ whose Fourier transform can be computed explicitly. Ideally, the example should be suitable for an undergraduate class, i.e., as simple as possible.

• Standard example: $\frac {\sin x }x$. Commented Dec 1, 2023 at 7:54
• Can you give a reference where that example is fully worked out? It seems to depend on cancellation properties. Commented Dec 1, 2023 at 8:09
• Compute the FT of the characteristic function of the interval $(-\frac 1{2\pi},\frac 1{2\pi})$. Commented Dec 1, 2023 at 8:16
• I know. What is the best way of showing it is not absolutely integrable but has a well-defined Fourier transform? Commented Dec 1, 2023 at 8:26
• @shuhalo Please let me know how I can improve my answer. I really want to give you the best answer I can Commented Apr 30 at 13:15

Let's look at the example $$f(x)=\frac{\sin(x)}{x}$$ as refernced in the comments. Clearly, $$f\in L^2$$ since

\begin{align} \int_{-\infty}^\infty\left|f(x)\right|^2\,dx&=\int_{|x|\le 1}\left|\frac{\sin(x)}{x}\right|^2\,dx+\int_{|x|\ge 1}\left|\frac{\sin(x)}{x}\right|^2\,dx\\\\ &\le \int_{-1}^1(1)^2\,dx+2\int_1^\infty \frac1{x^2}\,dx\\\\ &=4 \end{align}

Next, we show that $$f\notin L^1$$. Proceeding, let $$f^+$$ and $$f^-$$ denote the positive and negative parts of $$f$$, respectively. Then, we can write

\begin{align} \int_{-(2N+2)\pi}^{(2N+2)\pi}\left|f(x)\right|\,dx&=2\int_{0}^{(2N+2)\pi} |f(x)|\,dx\\\\ &=2\int_{0}^{(2N+2)\pi} (f^+(x)-f^-(x))\\\\ &=2\sum_{n=0}^N \left(\int_{2n\pi}^{(2n+1)\pi}\frac{\sin(x)}{x}\,dx\right)-2\sum_{n=0}^N \left(\int_{(2n+1)\pi}^{2(n+1)\pi}\frac{\sin(x)}{x}\,dx\right)\\\\ &\ge \sum_{n=0}^N \frac{4}{(2n+1)\pi}+\sum_{n=0}^N \frac{2}{n+1}\\\\ \end{align}

which clearly diverges as $$N\to \infty$$.

However, we have

\begin{align} \mathscr{F}\{f\}(\xi)&=\int_{-\infty}^\infty \frac{\sin(x)}{x}e^{-i2\pi \xi x}\,dx\\\\ &=\int_{-\infty}^\infty \frac{e^{i(1-2\pi \xi) x}-e^{i(1+2\pi \xi) x}}{i2x}\,dx\\\\ &=\int_{-\infty}^\infty \frac{\sin((1-2\pi \xi) x-\sin((1+2\pi \xi) x)}{2x}\,dx\\\\ &=\frac{\pi}{2}\left(\text{sgn}(1-2\pi \xi)-\text{sgn}(1+2\pi \xi)\right) \end{align}

• @shuhalo Please let me know how I can improve my answer. I really want to give you the best answer I can Commented Apr 30 at 13:15