I am trying to deduce the dispersive estimate for the free wave equation in $\mathbb{R}^{d+1}$$\mathbb{R}^{d+1}\equiv\{(x,t) : x\in\Bbb R^d \wedge t\in\Bbb R\}$ $$u_ {tt}-\Delta_{\mathbb{R}^d}u=0$$$$u_ {tt}-\Delta_xu=0$$ The fundamental solutions of this equation are given by $K_t^0(x)=cos(t|\nabla|)\delta(x)$ and $K_t^1(x)=\frac{sin(t|\nabla|)}{|\nabla|}\delta(x)$. According$$ \begin{align} K_t^0(x) &=\cos(t|\nabla|)\delta(x), \\ \\ K_t^1(x)&=\frac{\sin(t|\nabla|)}{|\nabla|}\delta(x). \end{align}$$ According to Tao's dispersive pde textbook, given a Scharwtz function $\phi(x)$, we have the following time decay $$|K^0_t*\phi_\lambda(x)|\lesssim_{\phi,d}\lambda^d<\lambda t>^{-\frac{d-1}{2}}$$,$$|K^0_t*\phi_\lambda(x)|\lesssim_{\phi,d}\lambda^d\langle\lambda t\rangle^{-\frac{d-1}{2}}\;,$$ where $<y>=(1+|y|^2)^\frac{1}{2}$$\langle y\rangle=(1+|y|^2)^\frac{1}{2}$. I am trying to deduce the above estimate using stationary phase method. Rewrite the LHS with Fourier transform, we get $$LHS=\int_{\mathbb{R}^d}e^{it|\xi|+ix\cdot\xi}\hat\phi(\lambda^ {-1}\xi) d\xi$$$$\text{LHS}=\int_{\mathbb{R}^d}e^{it|\xi|+ix\cdot\xi}\hat\phi(\lambda^ {-1}\xi) d\xi$$ Since $D^2|\xi|$ degenerates in radial direction, in order to apply stationary phase I rewrote it in polar coordinates, where $\xi=rw, w\in S^{d-1}$ $$\int_0^\infty e^{itr}r^{d-1}\hat\phi(\lambda^{-1}r)\int_{S^ {d-1}}e^{irw\cdot x}dw dr$$ But I am not sure how to proceed from here.
