$\int_{\mathbb{R}}x^2e^{-2\pi i wx}dx$ exists for no value of $w$.

While reading Fourier transform theory, I have a remark that says:

The Fourier transform does not always exist. For example, the function $f (x) = x^2$ does not admit a Fourier transform because the integral $\int_{\mathbb{R}}x^2e^{-2\pi i wx}dx$ exists for no value of $w$.

But I found a difficulty in proving it. Any idea please

Solution 1:

This is the definition for the Fourier Transform of a function $f(x)$: $$\hat{f}(\xi) = \int_{-\infty}^\infty f(x)\ e^{- 2\pi i x \xi}\,dx,$$ for any real number $ξ$.

We assume $f(x)$ is an integrable function, Lebesgue-measurable on the real line, and satisfy: $$\int_{-\infty}^\infty |f(x)| \, dx < +\infty.$$

If you check for the Lebesgue-measurability of $f(x) = x^2$ you will find it doesn't belong to that class of functions, indeed:

$$\int_{\mathbb{R}} |x^2 e^{-2\pi i \omega x}|\ \text{d}x = \int_{\mathbb{R}} |x^2| \text{d}x = +\infty$$