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The Laplace-transform formula for a semigroup resolvent is locally uniformly convergent in the half-plane $\mathrm{Re}z>\omega$, so it may be differentiated under the Bochner integral:The resolvent identity gives $R^{\prime }(z)=-R(z{)}^2$ and, by mathematical induction,ThereforeFor real $\lambda >\omega$, the integral triangle inequality and the Gamma integral giveConsequentlyThe exponent $-n+1$ printed in the question is a typographical error: already at $n=0$ it contradicts the first-resolvent bound.

Because $0<\alpha <n$, $|x{|}^{-\alpha }$ is locally integrable at the origin and has only polynomial growth at infinity, so it defines a tempered distribution. It is a homogeneous distribution of degree $-\alpha$ and is radial. Its Fourier transform is therefore radial and homogeneous of degree $\alpha -n$, so it must have the form $c_{\alpha }|\lambda {|}^{\alpha -n}$. In particular, $\boxed{\beta =n-\alpha }$.

To determine the constant, use the stated Gamma integral representation, Fubini's theorem, and the Fourier transform of a Gaussian:The change of variables formula $s=|\lambda {|}^2/(4\tau )$ then givesThis is precisely the Fourier transform of the Riesz kernel.

The Gamma integral $u^{-1/2}={\pi }^{-1/2}{\int }_0^{\infty }{t}^{-1/2}{e}^{-ut}dt$ givesFor each $t>0$, the last factor is a Gaussian kernel, and the remaining weight is nonnegative. The diagonal integral equals ${\alpha }^{-1/2}<\infty$, so part ii shows that the displayed function is a positive-semidefinite kernel.

The intended oscillatory factor is $e^{-ik\tau }$. Give the early-time endpoint the usual i-epsilon prescription and set $x=-k\tau$. For an integer $p\ge 0$,It is therefore purely imaginary. Equivalently, rotating the contour to the negative imaginary $\tau$ axis turns the remaining integral into a real Gamma integral and leaves one overall factor of $i$.