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If a Borel summation contour meets a simple pole, an upper or lower indentation fixes the exponentially small contribution. For the transform $(1-t^2{)}^{-1}$ on the positive ray, the relevant pole is $t=1$. A lower indentation gives the Cauchy principal value minus $i\pi e^{-z}/2$. Finite geometric expansion gives an exact remainder integral, which enables optimal truncation and error-function smoothing of a Stokes multiplier.

For $z=2n+i\mu \sqrt{n}+O(1)$, the factor $t^{2n}{e}^{z(1-t)}$ has saddle scale $t=1+u/\sqrt{n}$ and becomes $e^{-u^2-i\mu u}$. If the amplitude has a simple pole there, its leading local integral contains $e^{-u^2-i\mu u}/u$. Differentiating its Cauchy principal value with respect to $\mu$ reduces it to a Gaussian integral, producing an error function. The nonsingular amplitude $(1+t{)}^{-1}$ gives $\frac{1}{2}\sqrt{\pi /n}{e}^{-{\mu }^2/4}$.

For a thin axisymmetric astrophysical disk, angular integration of the Newtonian kernel eliminates odd Legendre polynomials. Define $I_n={\int }_0^R\Sigma (s)s^{n+1}ds$ and $J_n={\int }_R^{\infty }\Sigma (s)s^{-n}ds$. The even coefficient is ${\alpha }_n=\pi [n!/(2^n((n/2)!{)}^2){]}^2$. In differentiating the paired inner and outer contributions, their moving-boundary terms cancel. The resulting galaxy rotation curve includes an inward inner-ring contribution and an outward exterior-ring contribution. For smooth surface density, the force at the disk is interpreted with a symmetric Cauchy principal value or a vanishing-thickness regularization.

The Cauchy principal value integral $\mathrm{li}(x)=\mathrm{PV}{\int }_0^{x}dt/\log t$ is the logarithmic integral. The offset version $\mathrm{Li}(x)={\int }_2^{x}dt/\log t$ differs by a constant and is often used with the prime-counting function. It has $\mathrm{Li}(x)\sim x/\log x$.

The ordinary Fourier transform does not exist. There is a real simple pole at $x=-1$, withThe two separate improper integrals diverge logarithmically. Thus no ordinary large-$k$ expansion is defined under the stated integral convention. A Cauchy principal value or a contour prescription would be additional data, and must be stated rather than silently introduced.

For completeness, the symmetric Cauchy principal value has a precise answer. The real pole contributes the principal-value Fourier transform of a real poleThe other poles are $z_{\pm }=(1\pm i\sqrt{3})/2$ with residues $r_{\pm }=1/(3z_{\pm }^2)=(-1\mp i\sqrt{3})/6$. For $k>0$, close the contour integration in the lower half-plane, with clockwise orientation. The nonreal pole contributes $-2\pi i{r}_{-}{e}^{-ik{z}_{-}}$, while the real principal-value pole supplies the half-residue above. For $k<0$, the upper-half-plane contour gives the conjugate result. With $s=\mathrm{sgn}(k)$, the exact principal-value formula for $k\ne 0$ isThese are its two nonzero asymptotic contributions: a nondecaying oscillatory real-pole term and an exponentially small complex-pole term. There is no second nonzero inverse-power term. This formula is qualified by the principal-value choice; it is not the ordinary transform requested in the statement.

Other prescriptions change the leading term. For example, replacing the real pole by its upper or lower boundary value changes the distribution by $\mp i\pi \delta (x+1)/3$, and hence changes the transform by $\mp i\pi e^{ik}/3$. This demonstrates why a pole prescription is essential. The smooth Taylor series at the origin alone would miss the decisive real-pole contribution.

For a spherical system, use the spherical shell theorem with shell mass $dm=4\pi \rho (s)s^{2}ds$. Inner shells contribute $-Gdm/r$ to the gravitational potential, while an outer shell contributes its constant interior potential $-Gdm/s$. With $\Phi (\infty )=0$, assuming the required integrals converge,On differentiation, the two terms containing $\rho (r)r$ cancel. Hence ${\Phi }^{\prime }=GM(r)/r^2$, where $M(r)=4\pi {\int }_0^r\rho (s)s^{2}ds$. Radial balance for a circular orbit gives $v_c^2=r{\Phi }^{\prime }=GM(r)/r$.

For a razor-thin axisymmetric astrophysical disk, the element of mass is $dm=\Sigma (s)sdsd{\varphi }^{\prime }$, where $\Sigma$ is the surface density. Superposing Newtonian gravitational potentials therefore givesAxisymmetry permits $\varphi =0$. Unlike the spherical case, an exterior annulus generally exerts a radial force: enclosed mass does not determine a disc rotation curve.

For the Legendre expansion of thin-disk gravity, split the radial integral at $s=R$. In the inner part the kernel expands in $s/R$, while in the outer part it expands in $R/s$. The angular integrals of odd Legendre polynomials vanish, and those of even degree $n$ equal $2{\alpha }_n$, whereIntroduce $I_n(R)={\int }_0^R\Sigma (s)s^{n+1}ds$ and $J_n(R)={\int }_R^{\infty }\Sigma (s)s^{-n}ds$. The gravitational potential becomesWhen differentiating each bracket, the moving-limit terms cancel: $R^{-(n+1)}{I}_n^{\prime }=\Sigma (R)$ and $R^n{J}_n^{\prime }=-\Sigma (R)$. Since $v_c^2=R{\Phi }^{\prime }$, this givesThe zeroth term has ${\alpha }_0=\pi$ and $M(R)=2\pi I_0$. Separating it provesThe inner correction is inward, while the exterior correction is outward. For a smooth surface density, the original potential singularity at coincident points is integrable. Its radial force is understood through a symmetric Cauchy principal value or a vanishing-thickness regularization; the paired interior and exterior terms above retain the cancellation at $s=R$. This avoids treating the two singular local force contributions separately.

For an exponential galactic disk, write $\Sigma (s)={\Sigma }_0{e}^{-s/R_d}$, so $M_{\infty }=2\pi {\Sigma }_0{R}_d^2$ and $M(R)=M_{\infty }[1-(1+R/R_d)e^{-R/R_d}]$. At large $R$, the missing mass and exterior-ring terms are exponentially small. The leading nonspherical interior term is $k=1$, with ${\alpha }_2=\pi /4$ and $I_2(\infty )=6{\Sigma }_0{R}_d^4$. Consequently the exponential-disk Keplerian asymptotic isThe positive leading correction shows that the rotation curve approaches the Keplerian limit from above. The finite-order large-radius expansion is sufficient here; an infinite moment expansion need not converge for a disk extending to infinity.

A useful special example is the Mestel disk, with surface density $\Sigma (s)=C/s$ for $s>0$, $C>0$. It has $M(R)=2\pi CR$. For every positive even $n$,All the nonspherical corrections cancel, givingThus the Mestel disk has a perfectly flat galaxy rotation curve.

This example has infinite total mass and a singular central surface density. The absolute gravitational potential cannot be set to zero at infinity, but the radial force exists as a limit of disks with increasing outer cutoff. Potential differences are $\Phi (R)-\Phi (R_0)=v_M^2\log (R/R_0)$. The earlier potential integral is therefore interpreted up to a radius-independent divergent constant for this example; the force calculation remains valid. Truncating the Mestel disk gives a more physical finite system but changes the exact flat curve near its edges.

The Hilbert-transform Fourier multiplier is bounded, and ${\xi }^r\hat{\phi }(\xi )$ is integrable for every nonnegative integer $r$. Therefore its inverse Fourier transform can be differentiated under the integral arbitrarily many times:These derivatives are continuous by dominated convergence theorem. Since $\widehat{{\phi }^{(r)}}=(i\xi {)}^r\hat{\phi }$, the Hilbert transform is smooth and commutes with differentiation:The Fourier proof avoids differentiating a singular kernel without preserving its Cauchy principal value prescription.