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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 $k$.

Suppose ${\Psi }_1$ and ${\Psi }_2$ have the same boundary data, and put $u={\Psi }_1-{\Psi }_2$. Then $\Delta u=0$ in $V$ and $u=0$ on its boundary. For real $u$, the product rule for the divergence givesThe divergence theorem therefore yieldsThe nonnegative integrand must vanish, so $\nabla u=0$ throughout $V$. On each connected component $u$ is constant, and its zero boundary value makes that constant zero. Hence $\boxed{{\Psi }_1={\Psi }_2}$, proving uniqueness. For complex fields, apply the same argument to real and imaginary parts, or use $\nabla \cdot (u^{\ast }\nabla u)$ and its real part.

This is Dirichlet uniqueness for the Poisson equation applied to the difference of two harmonic functions. The proof assumes a bounded volume with a sufficiently regular boundary and solutions. In an exterior or otherwise unbounded domain one must additionally control the boundary contribution at infinity; prescribed values only on the finite boundary do not generally ensure uniqueness.

Taking traces of the given matrices gives the character on the five classes,For a complex finite-group representation, the multiplicity of an irrep $\rho$ is its character inner product with ${\chi }_D$:Every linear character has value one at both $1$ and $-1$, so its multiplicity is $(4-4)/8=0$. For the two-dimensional character,ThereforeThe dimensions check: $2\times 2=4$.

The choice of scalar field matters here, even though the displayed matrices are real. Over $\mathbb{R}$, this particular four-dimensional representation is irreducible: for any nonzero real vector $v$, the vectors $v,D(i)v,D(j)v,D(k)v$ are mutually orthogonal with the same nonzero norm. This follows from the skew-symmetry and quaternion multiplication identities of the three matrices. A nonzero invariant real subspace contains all four vectors and is therefore the whole space. Its complexification is the two-copy decomposition above.

For a one-dimensional character, write $\chi (i)=ϵ$ and $\chi (j)=\eta$. Since $\chi (-1)=1$, the relations $i^2=j^2=-1$ require $ϵ,\eta \in \{\pm 1\}$. Also $ij=k$ gives $\chi (k)=ϵ\eta$. The four sign choices therefore supply all four linear characters.

For the remaining two-dimensional character ${\chi }_2$, its value at the identity is two. Column orthogonality between the identity and $-1$ gives $4+2{\chi }_2(-1)=0$, so ${\chi }_2(-1)=-2$. Between the identity and each of the other classes, the four linear-character values sum to zero, forcing the remaining character values to vanish. Thus the character table isHere column orthogonality is the standard finite-group character theorem ${\sum }_{\rho }{\chi }_{\rho }(e{)}^{\ast }{\chi }_{\rho }(g)=0$ for $g\ne e$. The rows also have unit norm and are mutually orthogonal in the character inner product with the displayed class-size weights. This completes the character table of the quaternion group.

The elements $1,-1$ commute with the whole quaternion group, so they form singleton classes. Each of the three remaining pairs is a conjugacy class: for example conjugation by $j$ sends $i$ to $-i$, and likewise cyclically. No such pair can mix with a different pair, because the quotient by $\{\pm 1\}$ is abelian and its cosets $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$ are preserved by conjugation. HenceFor a finite group over $\mathbb{C}$, the number of inequivalent irreducible representations equals the number of conjugacy classes, so there are five. The sum of squares of irreducible representation dimensions is the group order, eight. With five positive integer dimensions, the only possibility isIndeed five one-dimensional contributions give five, and replacing one by a two-dimensional contribution adds exactly three; any further enlargement would exceed eight. These are the dimensions used in the character table of the quaternion group.

The homomorphism law and $g{g}^{-1}=g^{-1}g=e$ giveThus $D(g^{-1})$ is both a left and right inverse of $D(g)$, provingIf only multiplicativity is initially stated, $D(e{)}^2=D(e)$ together with invertibility of representation matrices first implies $D(e)=I$. This explains why a group representation takes values in the general linear group rather than arbitrary square matrices.

Work over the complex numbers, the usual scalar field for the finite-group character theory used below. A group representation on a nonzero vector space $V$ is a homomorphism $D:G\to GL(V)$, so $D(gh)=D(g)D(h)$ and $D(e)=I$.

A faithful representation is injective: distinct group elements have distinct representing maps. Equivalently, its kernel is just $\{e\}$.

An irreducible representation has no nonzero proper invariant subspace: if $W\subseteq V$ and $D(g)W\subseteq W$ for every $g\in G$, then $W=\{0\}$ or $W=V$. Irreducibility concerns invariant subspaces, whereas faithfulness concerns whether any group element is represented by the identity.