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Let $T_q\phi (x)={\sum }_{k=0}^q{\phi }^{(k)}(0)x^k/k!$ denote the Taylor polynomial at zero. On $0<|x|<1$, the integrand defining the Hadamard finite-part reciprocal-power distribution isThe first term is odd, so it cancels under symmetric truncation. The Taylor remainder in the second term has absolute value at most $|x{|}^m\Vert {\phi }^{(m)}{\Vert }_{\infty }/m!$. The integral near zero therefore has a finite limit. For $|x|\ge 1$, the original terms are integrable: $x^{-m}\phi$ has compact support, and each subtracted power has exponent $k-m\le -2$.

This proves existence and linearity. It also gives the global estimateThusIn particular the order of a distribution is bounded by one integer on all compact sets, not merely separately on each of them.

For the recurrence, write the truncated Hadamard finite-part integral aswhere parity evaluates all subtraction integrals explicitly:Set $C_1=0$ and ${\Lambda }_1=\mathrm{pv}(1/x)$. On the two truncated intervals, integration by parts givesThere are no boundary contributions at infinity. The coefficients of the subtraction terms satisfyThis is exactly the negative of the boundary term's Taylor polynomial through degree $m$; the degree-$m$ term itself vanishes by parity. HenceTaking the limit proves the full distributional identityThe same calculation includes $m=1$. Starting with the distributional derivative of the logarithmic modulus from part (i), induction yieldsThis identity holds on all of $\mathbb{R}$ as a distributional identity. It is stronger than matching the ordinary derivatives away from zero, which would leave possible differentiated Dirac delta distributions at zero undetermined.