Is the Fourier Transform in $L^1$

Question

Let $f$ be defined $$f = \begin{cases} \frac{1}{(t+1)^3}, & t > 0 \\ 0, & t \leq 0 \end{cases}$$

Is the Fourier transform of $f$ in $L^1(\mathbb{R})$?

I know that the decay of the Fourier transform can be induced from smoothness of the original function, but I don't see how this is used here, as $f$ is not continuous at $0$.

I think this may have something to do with the inverse Fourier transform: if the Fourier transform is in $L^1$, then it itself has a Fourier transform which has to be continuous everywhere.

Answer 1

We know that $f \in L^1(\mathbb R)$, then $\hat{f}$ is bounded and continuous. If $\hat{f} \in L^1(\mathbb{R})$ then almost everywhere $$f(x)=\dfrac{1}{2\pi}\int_{\mathbb R}e^{i\xi x}\hat{f}(\xi)\ \mathrm{d}\xi$$ This means that $f$ is equal almost everywhere to a continuous function $g$. We find a contradiction around $x=0$ because we know that there exists $\delta > 0$ such that for $x \in [-\delta, \delta], \ g(x) \in [g(0)-0.1, g(0)+0.1]$. Therefore for almost every $x \in [-\delta, \delta], \ f(x) \in [g(0)-0.1, g(0)+0.1]$, which is not the case.

Answer 2

Let

$$\hat{f}(\omega) = \int_{-\infty}^{\infty} f(t)e^{-i\omega t} \, dt = \int_{0}^{\infty} \frac{1}{(1+t)^3}e^{-i\omega t} \, dt. $$

We make the following manipulation, which is better suited for analyzing asymptotic behavior.

\begin{align*} \hat{f}(\omega) &= \int_{0}^{\infty} \left(\frac{1}{2}\int_{0}^{\infty} x^2 e^{-(t+1)x} \, dx \right)e^{-i\omega t} \, dt \\ &= \frac{1}{2}\int_{0}^{\infty} x^2 e^{-x}\left(\int_{0}^{\infty} e^{-(x+i\omega)t} \, dt \right) \, dx \\ &= \frac{1}{2}\int_{0}^{\infty} \frac{x^2 e^{-x}}{x+i\omega} \, dx \\ &= \frac{\omega^2}{2}\int_{0}^{\infty} \frac{u^2 e^{-|\omega|u}}{u+i\operatorname{sign}(\omega)} \, du, \end{align*}

where in the last line we further assumed that $\omega \neq 0$ and applied the substitution $x = |\omega|u$. Now we claim that

$$ \int_{0}^{\infty} \frac{u^2 e^{-|\omega|u}}{u+i\operatorname{sign}(\omega)} \, du = \frac{2}{i\omega^3} + \mathcal{O}(\omega^{-4}) \qquad\text{as} \quad |\omega|\to\infty. $$

Indeed, notice that $\frac{2}{i\omega^3} = \int_{0}^{\infty} \frac{u^2 e^{-|\omega|u}}{i\operatorname{sign}(\omega)} \, du$ and hence

$$ \left| \int_{0}^{\infty} \frac{u^2 e^{-|\omega|u}}{u+i\operatorname{sign}(\omega)} \, du - \frac{2}{i\omega^3} \right| \leq \int_{0}^{\infty} u^3 e^{-|\omega|u} \, du = \frac{6}{\omega^4}, $$

proving the claim. Therefore we have

$$ \hat{f}(\omega) = \frac{1}{i\omega} + \mathcal{O}(\omega^{-2}) \qquad \text{as} \quad |\omega|\to\infty. $$