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 $\alpha > 0$?