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I want to compute $\mathcal{F}^{-1}\{\delta(|\cdot|-1)\}(x)$, which exactly means the following computation: $$f(x) = (2\pi)^{-n/2} \int_{|\xi|=1}e^{ix\cdot\xi}\mathrm{d}\xi, \mbox{ where }~ \xi \in \mathbb{R}^n.$$ I noticed that $f \in C^{\infty}(\mathbb{R}^n)$, $f$ is radial, $f(0) = n\cdot\alpha(n) = n \pi^{n/2} / \Gamma(n/2+1)$ and when $x \neq 0$, we have $(I+\Delta)f(x) = 0$. Therefore letting $f(x) = g(|x|)$, we could get: $$g''(r) + \frac{n-1}{r}g'(r) + g(r)= 0~(r \neq 0 \text{ represents } |x|).$$ These statements above are rigorous.

But the following lacks rigour and is what I want to ask help for.

Because $f \in C^{\infty}(\mathbb{R}^n)$, so as $g$ ($g \in C^{\infty}(\mathbb{R}^1)$). Then I assume $g(r) = \sum_{m=0}^{+\infty}a_m r^m$. Substitute it into the equation above, finally I get: $$g(r) = \sum_{k=0}^{+\infty}\frac{(-1)^kn!!\cdot\alpha(n)}{(2k)!!\cdot(2k+n-2)!!} r^{2k}, ~\forall\, r \geq 0.$$ Then $$f(x) = \sum_{k=0}^{+\infty}\frac{(-1)^kn!!\cdot\alpha(n)}{(2k)!!\cdot(2k+n-2)!!} |x|^{2k}, ~\forall\, x \in \mathbb{R}^n.$$ Is this result right? If it is right, what is the kernel of $(I - \Delta)^{\alpha/2}$ for $\alpha > 0$?

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    $\begingroup$ See formula 402 (and 501) and take the distributional derivative $\endgroup$ – reuns Aug 1 '17 at 11:53
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Your derivation seems correct to me, at least formally - you assume that $g$ is real analytic instead of $C^\infty$. However, this formula looks very different from what I've seen before which leads me to be skeptical. Instead of resolving your question, I will try to point you in some other directions. Notably, this measure and its Fourier transform are intensively studied in harmonic analysis and PDEs since the delta measure on the sphere naturally yields the spherical averages by dilation which are intimately tied to the wave equation on R^n. Letting $d\sigma$ denote your delta measure above and $\widetilde{d\sigma}$ denote its Fourier transform we have $$\widetilde{d\sigma}(\xi) = 2\pi |\xi|^{1-d/2} J_{d/2-1}(2\pi|\xi|)$$ where $J_s$ is the Bessel function of order $s$. This formula is derived in several texts; for instance, in Stein--Shakarchi's "Functional Analysis" (chapter 8), Stein--Shakarchi's "Fourier analysis" (chapter 6) Stein--Weiss's "Intro to Fourier analysis on Euclidean spaces" (Theorem 3.3 of chapter IV) and Stein's "Harmonic analysis" (chapter VIII).

In short one may derive this by noting that the above Fourier transform is really about the Fourier transform of a radial function and this may be derived from that formula. The distribution $\delta(|x|-1)$ is rotationally invariant so that its Fourier tranform is too and thus $\widetilde{d\sigma}(\xi)$ only depends on $|\xi|$. Dilational symmetry of Euclidean space allows one to pull out the factor of $|\xi|^{1-d/2}$ and find an integral that defines the relevant Bessel function. Moreover, one may give an asymptotic analysis to find coefficients $c_m$ so that $$\widetilde{d\sigma}(\xi) = \sum_{m=0}^\infty c_m |\xi|^{(1-d)/2-m} e^{i|\xi|} + \bar{c_m} |\xi|^{(1-d)/2-m} e^{-i|\xi|}.$$ Here $c_m = O(1)$ and we see that $$\widetilde{d\sigma}(\xi) = O(|\xi|^{-(d-1)/2}).$$ Even that this decays at infinity is not apparent to me in your formula.

You may find formulae for the Bessel potentials $(I-\Delta)^{\alpha/2}$ in Stein's "Singular integrals..." (chapter V, section 3 and further result 6.5).

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