Set $z=e^{i\theta }$. Then $d\theta =dz/(iz)$ and $\tan \theta =(z^2-1)/[i(z^2+1)]$, so the requested integral isThe denominator factors asFor $0<a<1$, its two inner roots are $z=\pm r$, $r=\sqrt{(1-a)/(1+a)}<1$, and the other two roots have modulus $1/r>1$. There is also a simple pole at zero. Their residues areThe latter values follow by differentiating the quadratic product, or from the supplied hint. The sum of the enclosed residues isThe residue theorem therefore givesAt the angles where the tangent is undefined, the original reciprocal has a removable zero limit. Its continuous extension gives the same integral; the transformed rational integrand has no pole on the unit circle.

Put $s=1/n$, so $0<s<1$. For the complex exponentiation $z^s$, choose $-\pi /2<\arg z<3\pi /2$: the branch cut is the negative imaginary axis. Integrate $z^s/(a^2+z^2)$ around the upper half-plane semicircle, with a small semicircular indentation excluding the origin. The only enclosed simple pole is $z=ia$, with residueThe large arc is $O(R^{s-1})$ and the small arc is $O({ϵ}^{s+1})$, so both vanish. On the negative real segment, $z^s=e^{i\pi s}|z{|}^s$; accounting for the direction of traversal, the two real segments therefore sum to $(1+e^{i\pi s})I$, where $I$ is the desired positive-axis integral. The residue theorem givesSince $1+e^{i\pi s}=2e^{i\pi s/2}\cos (\pi s/2)$,The endpoint estimates also prove convergence. More generally the same calculation is the fractional-power integral from an upper-half-plane contour, valid for $-1<s<1$.

Let $f=u+iv$ be analytic on a neighbourhood of a bounded planar region $\Omega$ and its piecewise smooth boundary $C$, oriented with $\Omega$ on the left. The Cauchy integral theorem statesFor a region with holes, $C$ includes the clockwise inner boundaries as well as the counterclockwise outer one. In particular, the integral around a closed contour bounding a region throughout which $f$ is analytic is zero.

To prove this classical form, expand the line integral and apply Green's theorem to its real and imaginary parts:Both integrands vanish by the Cauchy-Riemann equations, $u_x=v_y$ and $u_y=-v_x$. Analyticity supplies the continuous first partial derivatives needed by this Green's theorem proof. The requirement that the enclosed region contain no singularities is essential.

Reflect the source across the plane, defining ${\mathbf{r}}^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$ for an interior point $z^{\prime }>0$. Superpose the direct Green function from part (b) and an image of opposite sign:The direct term supplies the delta source in the upper half-space; the image singularity lies outside that domain and contributes no source there. On $z=0$ the two distances are equal, so the terms cancel, giving homogeneous Dirichlet boundary conditions. Both decay at infinity, which is the additional condition selecting this half-space solution. This is the method of images realization of the Dirichlet image kernel for the three-dimensional modified Helmholtz equation. It includes the unscreened case $k=0$.

Write $C=A\sqrt{\pi /(2k)}$, so the radial field is $\Phi =C{e}^{-kr}/r$. Integrate its equation over a ball of radius $ϵ$ centred at the source. The divergence theorem givesThe second term is $4\pi k^{2}C{\int }_0^{ϵ}r{e}^{-kr}dr=O({ϵ}^2)$ and vanishes in the limit. MeanwhileThusThe minus sign follows from the convention $(\Delta -k^2)\Phi ={\delta }^{(3)}$, opposite to the positive static Yukawa Green function for $-\Delta +k^2$. For arbitrary real $k$, replace it in the exponential by $|k|$ to select decay at infinity. The $k=0$ limit is $-1/(4\pi r)$.

For the real decaying Bessel form used here, take $k>0$. The original PDF has $K_m(r)$ as the last factor in its auxiliary differential equation, although both differentiated factors have argument $kr$; that last argument must also be $kr$ for the stated Modified Bessel differential equation.

Put $B(r)=K_m(kr)$. The corrected identity is $B^{\prime \prime }=-B^{\prime }/r+(k^2+m^2/r^2)B$. Since the field is radial, the radial Laplacian givesAway from the point source this must vanish. For $k>0$, a constant ratio of $r{B}^{\prime }$ to $B$ would make $B$ a pure power, incompatible with the $k^2{r}^2$ term of its equation. Therefore both displayed coefficients must vanish:The two orders give the same Modified Bessel function of the second kind: $K_{-1/2}(z)=K_{1/2}(z)=\sqrt{\pi /(2z)}{e}^{-z}$.^[ref]

The source describes $k$ merely as real, but the following displayed real square-root formula implicitly needs $k>0$. For the operator with arbitrary real $k$, its decaying fundamental solution instead uses $\kappa =|k|$; $k=0$ is obtained by a limit. The unmodified real $A{K}_m(kr)$ ansatz is not a real Bessel formula for all negative $k$.