Fourier transform of an L1 function - Stein and Shakarchi, 2.25

Question

This is from Stein and Shakarchi's Real Analysis (2.25). I'm really not so sure what to do with the hint. Can someone please walk me through the steps? I'm a little lost... thank you!

Answer 1

Let $f$ be given by

$$f(x)=\int_0^\infty e^{-\pi (t+|x|^2/t)}t^{\varepsilon-1-d/2}\,dt$$

where $x\in \mathbb{R}^d$ and $\varepsilon>0$. Applying Fubini's Theorem, we see that

$$\begin{align} \int_{\mathbb{R}^d}|f(x)|\,dx&=\int_0^\infty \underbrace{ \left(\int_{\mathbb{R}^d}e^{-\pi |x|^2/t}\,dx\right)}_{=O(t^{d/2})} t^{\varepsilon-1-d/2}e^{-\pi t}\,dt\\\\ &=O(1)\int_0^\infty e^{-\pi t}t^{\varepsilon-1}\,dt \end{align}$$

Inasmuch as $\varepsilon>0$, $f\in \mathbb{L}^1(\mathbb{R}^d)$ as was to be shown.

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The Fourier Transform, $\hat f$, of $f$ is given by

$$\begin{align} \hat f(\xi)&=\int_{\mathbb{R}^d} f(x) e^{-i2\pi \xi x}\,dx\\\\ &=\int_{\mathbb{R}^d} \left(\int_0^\infty e^{-\pi (t+|x|^2/t)}t^{\varepsilon-1-d/2}\,dt\right) e^{-2\pi i\xi x}\,dx\\\\ &=\int_0^\infty t^{\varepsilon-1-d/2}e^{-\pi t}\underbrace{\left(\int_{\mathbb{R}^d} e^{-\pi |x|^2/t}e^{-i2\pi \xi x}\,dx\right)}_{=t^{d/2}e^{-t\pi |\xi|^2}}\,dt\\\\ &=\int_0^\infty e^{-\pi |\xi|^2 t} t^{\varepsilon-1}e^{-\pi t} \,dt \end{align}$$

as was to be shown!

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Finally, enforcing the substitution $x\mapsto x/(\pi(1+|\xi|^2))$, we find that

$$\begin{align} \hat f(\xi)&=\int_0^\infty e^{-\pi x (1+|\xi|^2)}x^{\varepsilon-1}\,dx\\\\ &=\int_0^\infty e^{-x}\left(\frac{x}{\pi(1+|\xi|^2)}\right)^{\varepsilon-1}\,\left(\frac1{\pi(1+|\xi|^2)}\right)\,dx\\\\ &=\pi^{-\varepsilon}\Gamma(\varepsilon)\frac1{(1+|\xi|^2)^\varepsilon} \end{align}$$

as expected again!