I am working on Exercise 8 of Chapter 6 of Fourier Analysis by Stein. The essence of the problem is to prove the identity $$e^{-2\pi|x|} = \int_0^\infty\frac{e^{-u}}{\sqrt{\pi u}}e^{-\pi^2|x|^2/u}\,\mathrm{d}u,$$ where $x\in \mathbb{R}^d$, and use this to show $$P_y^{(d)}(x) = \int_{\mathbb{R}^d}e^{2\pi \mathrm{i}x\cdot\xi}e^{-2\pi|\xi|y}\,\mathrm{d}\xi = \frac{\Gamma((d+1)/2)}{\pi^{(d+1)/2}}\frac{y}{(|x|^2+y^2)^{(d+1)/2}}.$$

Now I am stuck on showing the first identity. The hint was to take the Fourier transform of both sides - I managed to do it for the right hand side, but I had no idea how to obtain the Fourier transform of the left hand side.

My thoughts & attempts:

Since $e^{-2\pi|x|}$ is radial, so is its Fourier transform. Thus, $\hat{f}(\xi) = \hat{f}(|\xi|,0,\dots,0) = \int_{\mathbb{R}^d}e^{-2\pi|x|}e^{-2\pi\mathrm{i}|\xi|x_1}\,\mathrm{d}x.$ Let $u = x_1$, and transform $x_2,\dots,x_d$ using spherical coordinates, and I got $$\hat{f}(\xi) = \int_{\mathbb{R}}\int_0^\infty e^{-2\pi\sqrt{u^2 + r^2}}e^{-2\pi\mathrm{i}|\xi|u}r^{d-2}A_{d-2}\,\mathrm{d}r\mathrm{d}u,$$ where $A_{d-2} = \frac{2\pi^{d/2 - 1}}{\Gamma(d/2 - 1)}$ is the area of the unit hypersphere $S^{d-2}$.

Otherwise, transform $x_1,\dots, x_d$ using spherical coordinates, I got $$\hat{f}(\xi) = \int_{\mathbb{R}}\int_0^\pi e^{-2\pi r}e^{-2\pi\mathrm{i}\xi r\cos\theta}r^{d-1}A_{d-2}\sin^{d-2}(\theta)\,\mathrm{d}\theta\mathrm{d}r.$$

I have no idea how to find either integral.

I found a similar post on stackexchange where $d = 3$. However, $d = 3$ is much easier since $r\mathrm{d}r = \frac{1}{2}\mathrm{d}r^2$ for method 1 and $\sin(\theta)\mathrm{d}\theta = -\mathrm{d}\cos(\theta)$ for method 2. There is also a post that proves the identity by change of variables, but the answer did not use Fourier transform. I'd appreciate any hints on how to evaluate the Foureir transform of LHS for general $d$.


1 Answer 1


I think the Exercise 2.2.10 and Exercise 2.2.11 in in Grafakos' 'Classical Fourier Analysis' can help you.

And I post my proof of 2.2.11 as below (2.2.11(b) is exactly what you want):

(a) Take $f(x)=e^{-tx^2}$ in Exercise 2.2.10, we have \begin{align*} e^{-2t}&=\cfrac{\int_{-\infty}^\infty e^{-tx^2-t/x^2}dx}{\int_{-\infty}^\infty e^{-tx^2}dx}\\ &=\cfrac{1}{\sqrt \pi}\int_{-\infty}^\infty \sqrt{t}e^{-tx^2-t/x^2}dx\\ &=2\cfrac{1}{\sqrt \pi}\int_{0}^\infty \sqrt{t}e^{-tx^2-t/x^2}dx\\ (y=tx^2) ~~~~~&=\frac{1}{\sqrt{\pi}}\int_0^\infty e^{-y-t^2/y}\cfrac{dy}{\sqrt{y}}. \end{align*}

(b) Set $t=\pi |x|$ and integrate with respect to $e^{-2\pi i\xi \cdot x} dx$ in (a) and then we see that \begin{align*} (e^{-2\pi |x|})^{\wedge}(\xi)&=\frac{1}{\sqrt{\pi}}\int_0^\infty e^{-y-\pi^2|x|^2/y}\int_{x\in\Bbb R^n}\cfrac{1}{\sqrt{y}}e^{-2\pi i \xi\cdot x}dxdy\\ &=\cfrac{1}{\sqrt{\pi}}\int_0^\infty e^{-y}y^{-1/2} (e^{-\pi{(|\sqrt{\pi}x/\sqrt{y}|)}^2})^\wedge(\xi)dy\\ &=\pi^{-\frac{n+1}{2}}\int_0^\infty e^{-y}y^{\frac{n-1}{2}}e^{-y|\xi|^2}dy\\ &=\cfrac{1}{\pi^{\frac{n+1}{2}}}\cfrac{\int_0^\infty e^{-t}t^{\frac{n-1}{2}}dt}{(1+|\xi|^2)^{\frac{n+1}{2}}}\\ &=\cfrac{\Gamma(\frac{n+1}{2})}{\pi^{\frac{n+1}{2}}}\cfrac{1}{(1+|\xi|^2)^{\frac{n+1}{2}}}. \end{align*}


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