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At $z=1$, numerator and denominator both have simple zeroes. Their derivatives give the removable valueAt each positive integer $n\ge 2$, the numerator is nonzero and the denominator has a simple zero, so these are simple poles. At zero the expansion on the principal sheet iswhich contains a fractional power. Thus zero is a branch point with pole-like leading growth, not an ordinary isolated pole. The negative real axis is the chosen branch cut. On local continuations across its nonzero points, negative integers also give simple poles because the continued square root is $\pm i\sqrt{|n|}$, never one. They lie outside the open principal branch domain; noninteger negative points are merely cut points, not intrinsic singularities.

Suppress the common time factor $e^{i\omega t}$, take $\omega >0$, and fix the spatial Fourier transform conventionA plus transform is supported on $x>0$ and analytic above $\Gamma$; a minus transform is supported on $x<0$ and analytic below it. For the outgoing radiation condition with this time convention, initially take $k_0=\omega /c_0-i\epsilon$, $\epsilon >0$, and pass to the limit at the end. Thus $+k_0$ lies below the contour and $-k_0$ above it.

Choose $\gamma =(k^2-k_0^2{)}^{1/2}$ to have positive real part on the real transform line. Its branch cuts run from $+k_0$ into the lower half-plane and from $-k_0$ into the upper half-plane, without crossing $\Gamma$; downward and upward vertical rays are suitable. In the zero-absorption limit, $\gamma >0$ for real $|k|>k_0$ and $\gamma =+i\sqrt{k_0^2-k^2}$ between the branch points. This ensures that $e^{-\gamma |y|}$ represents decay or outgoing radiation, rather than an incoming exterior field.

Evenness in $y$ and the Helmholtz equation give the transformed fieldsAt $y=b$, the one-sided normal derivatives of the scattered field agree for $x<0$, because that part of the interface is open. For $x>0$ they both vanish by rigidity. Their common trace is therefore a minus function, denoted $V^{-}(k)={\partial }_y{\Phi }^{-}{|}_{y=b}$. ConsequentlyThe jump in the total scattered transform isFor $x<0$, continuity of the total mass density requires the scattered jump to cancel the incident jump, so $\varphi (b^{+})-\varphi (b^{-})=e^{i{k}_{0}x}$. Its minus transform is $-i/(k+k_0)$. The unknown plate-side jump is the plus function $J^{+}=[{\Phi }^{+}{]}_{b^{-}}^{b^{+}}$. Thus $J=J^{+}-i/(k+k_0)$, and the Wiener-Hopf equation follows:

Use the Wiener-Hopf factorization $L=L^{+}{L}^{-}$, with factors analytic and nonzero in their designated half-planes. Their analytic continuations allocate outgoing modal zeros to $L^{+}$ below the contour, and the opposite zeros to $L^{-}$ above it. Set $C=L^{+}(-k_0)$. Multiplication by $L^{+}$ and pole subtraction giveThe pole in the upper expression is removable by its numerator. The two expressions analytically continue to the common entire function, which is zero under the stipulated edge/growth assumption. Henceand the transformed fields areA constant reciprocal rescaling of the factors does not alter $C{L}^{-}$ or the physical field.

Inside the guide, $\cosh (\gamma y)$ and $\gamma \sinh (\gamma b)$ are even entire functions of $\gamma$, while $L^{-}$ is analytic in the lower half-plane. Thus continuation across the lower branch cut changes none of the interior transform: that cut is removable. The exterior expression retains the cut, corresponding to radiation into the open exterior.

For $x>0$, the inverse-transform contour closes downwards, clockwise. Away from modal cutoffs, its enclosed singularities are the outgoing simple poleswith positive real part for propagating modes and negative imaginary part for decaying modes; $k_0$ is the $n=0$ root. The opposite roots lie above the contour or cancel against zeros of $L^{-}$. Symmetry excludes odd transverse modes. At $n\ge 1$, $\gamma =in\pi /b$, and differentiation of $D(k)=\gamma \sinh (\gamma b)$ givesEach inverse-transform contribution is $-i$ times its residue. Therefore the outgoing modes of an open rigid acoustic waveguide areThe original time factor multiplies this expression. Thus every cut-on mode propagates in the positive $x$ direction; cut-off modes are evanescent waves decaying into the guide, not additional backward waves. Strictly, a complete mode sum includes these evanescent modes as well as propagating ones. The displayed simple-pole amplitudes apply away from exact cutoffs; cutoff values use the outgoing limiting-absorption continuation before taking the limit. The reflected plane-wave amplitude relative to the unit incident wave is $A_0$ above. It has the expected negative sign in the long-wavelength leading kernel approximation $L\simeq b(k^2-k_0^2)$.

On the parabolic domain with the nonpositive real axis removed, first apply the principal square root, then scale by $\pi /2$, and then apply the map of part (b). This gives the stated composite $C(z)$ and maps bijectively onto the slit unit disc.

There is no genuine branch cut in the composite. The function ${\tan }^2(\pi s/4)$ is even in $s$, so changing the local square root has no effect. Near zero its Taylor series starts asgiving a holomorphic extension with $C(0)=0$. There are no poles within the parabolic domain: the possible squared locations are $z=(2+4k{)}^2$, all on the excluded real portion $x\ge 4$.

For $z=-r^2$ with $r\ge 0$, the continuation givesAs $r$ ranges from zero to infinity, these values run once through $(-1,0]$. They fill precisely the slit omitted by the previous mapping, and no point away from the negative real axis has an image in that slit. Therefore the extended map is a bijection onto the whole unit disc:It is holomorphic throughout, so this bijection is the desired conformal map from a parabolic domain to the unit disc; in particular $C^{\prime }(0)={\pi }^2/16\ne 0$.

Use a genus-zero Polyakov path integral, target signature $(-,+,\dots ,+)$ and all four momenta incoming. Define the Mandelstam variables by $s=-(p_1+p_2{)}^2$, $t=-(p_1+p_3{)}^2$, $u=-(p_1+p_4{)}^2$. The string mass-shell condition is $p_i^2=4/{\alpha }^{\prime }$ and momentum conservation is ${\sum }_i{p}_i=0$, soThe reduced amplitude omits the overall momentum-conservation Dirac delta distribution and any conventional overall scattering-matrix phase. Interpret each tachyon vertex operator as normal ordered; self-contractions must not be included.

The free-boson worldsheet propagator is $⟨X^a(z)X^b(w)⟩=-({\alpha }^{\prime }/2){\eta }^{ab}\log |z-w{|}^2$. Wick theorem then evaluates the normal-ordered exponential correlator as the Koba-Nielsen factorThe worldsheet zero mode supplies momentum conservation. Four vertices give $g_s^4$ and the sphere contributes $g_s^{-2}$ through the string genus expansion, leaving $g_s^2$.

The sphere's residual Möbius transformations fix three insertion points. In sphere gauge fixing for four string vertices, take $(z_1,z_2,z_3,z_4)=(z,0,1,\infty )$. The worldsheet ghost fields, equivalently the Faddeev-Popov determinant for this residual group, supply $|z_{23}{z}_{24}{z}_{34}{|}^2$. At finite $z_4$, this determinant grows as $|z_4{|}^4$, while the matter correlator decays as $|z_4{|}^{-4}$ by momentum conservation and the tachyon mass shell; their product has a finite limit. Equivalently the weight-$(1,1)$ matter operator at infinity is normalized with $|z_4{|}^4$, while the ghost pair uses the inverse factor. After stripping the fixed-position normalization, onlyremains. Here choose the standard complex-coordinate measure $d^{2}z=idz\wedge d\bar{z}=2dxdy$. With $dxdy$ instead, the reduced normalization must acquire a factor two to give the same requested amplitude. Absolute vertex/sphere normalization is a convention; this choice fixes it consistently with the displayed $2\pi$ prefactor.

Set $a=-1-{\alpha }^{\prime }s/4$, $b=-1-{\alpha }^{\prime }t/4$, $c=-1-{\alpha }^{\prime }u/4$. Then $a+b+c=1$ and the integrand is $|z{|}^{2a-2}|1-z{|}^{2b-2}$. To evaluate the complex beta integral first work where $\mathrm{Re}a,\mathrm{Re}b,\mathrm{Re}c>0$. This ensures convergence near $0$, $1$ and infinity. Set $v=1-a$ and $w=1-b$. Schwinger parameterization givesCompleting the square, the $dxdy$ Gaussian integral is $\pi (\lambda +\rho {)}^{-1}\exp [-\lambda \rho /(\lambda +\rho )]$. Change variables to $q=\lambda +\rho$ and $x=\lambda /q$; their Jacobian determinant is $q$. Integrating $q$ gives $\Gamma (v+w-1)[x(1-x){]}^{-(v+w-1)}$. The remaining integral is the ordinary beta function $B(1-w,1-v)=B(b,a)$. ThusRestoring $d^{2}z=2dxdy$ and substituting the Mandelstam variables proves the Virasoro–Shapiro amplitude:For physical scattering the original position integral generally fails to converge. The formula defines the amplitude by analytic continuation from the convergence domain, with the desired scattering boundary value at real poles; the convergent integral should not be claimed valid for every physical momentum.

The gamma function is a meromorphic function, with simple poles at nonpositive integers and no zeros, while its reciprocal $1/\Gamma$ is an entire function. Consequently it is a meromorphic function of the independent invariants and a crossing-symmetric scattering amplitude: permuting $s,t,u$ leaves it unchanged. Its generic channel poles areand the same tower in $t$ and $u$. They represent the exchanged bosonic string mass spectrum: the tachyon, massless states including the graviton, and an infinite sequence of massive closed-string levels. There are no threshold branch cuts at this tree order.

A precise check of factorization is the Virasoro–Shapiro amplitude pole residue. Near $a=-n$, set $c=1+n-b$ in the nonsingular factor. The gamma function recurrence givesCombining this with $\Gamma (a)\sim (-1{)}^n/[n!(a+n)]$ showsThe residue is a degree-$2n$ polynomial in $t$, consistent with exchange up to spin $2n$ and scattering-amplitude factorization. At exceptional kinematics reciprocal Gamma zeros can remove apparent poles. In particular, when both $a$ and $b$ approach nonpositive integers, the zero from $1/\Gamma (a+b)$ cancels the putative double pole, leaving channel simple-pole terms. Thus one must not count products of numerator poles without using $a+b+c=1$.

Choose the square root with $-\pi /2<\arg z<3\pi /2$, so the branch cut is the negative imaginary axis. The change of variable $u=e^{-i\pi /4}\sqrt{z}$ maps this sheet onto $\mathrm{Re}u>0$ and gives $z=i{u}^2$. Writing the exponential function as $e^{k\varphi (z)}$, its phase becomes exactly quadratic:The saddle point is therefore $u=1$, or $z_s=i$. On the descending line $u=1+iv$, $v\in \mathbb{R}$, the phase is $-1-v^2$. Its image is the parabolaThe contour runs from left to right, corresponding to $v$ decreasing from $+\infty$ to $-\infty$. A contour deformation in the right half of the $u$-plane moves the original indented contour to this line. The connecting tails vanish in the descending sectors, and no branch point or pole is crossed. Since $dz=-2(1+iv)dv$, the Gaussian integral and the odd function $v{e}^{-k{v}^2}$ giveHere the method of steepest descent actually gives an exact answer for $k>0$, because the quadratic phase and linear transformed amplitude have no further even correction.

For a fixed pole away from the saddle point, the smooth amplitude there is $(i-z_0{)}^{-1}$. The simple-saddle contribution in steepest descent isThe residue theorem supplies an additional contribution precisely when the contour deformation crosses $z_0$. The region swept above the original contour has positive orientation: the original contour from left to right followed by the reversed saddle contour encloses it counterclockwise. Consequently the original integral equals the saddle integral plus the pole contribution:For $|z_0|<1$, an upper-half-plane pole outside the indentation has $\chi =1$, while a lower-half-plane pole has $\chi =0$. The entire upper unit half-disk lies below the saddle parabola, so there is no further case inside that half-disk. With a fixed indentation radius $\delta$, a pole inside the upper semicircle, $|z_0|<\delta$, is also below the original contour and has $\chi =0$; real points in the indentation gap are excluded as well. Taking $\delta <|z_0|$ for a fixed nonzero pole gives the usual upper/lower classification. The pole contribution need not always dominate exponentially; its size depends on $\mathrm{Re}\varphi (z_0)$, so keeping both terms makes that dependence explicit.

A pole on the original contour requires a stated Cauchy principal value or an indentation prescription. A pole parameter on the negative imaginary branch cut still defines the integral: the contour stays on its fixed sheet, and only the denominator uses $z_0$. Such a pole is not crossed, so only $J_{\mathrm{sd}}$ is needed and no value of $\sqrt{z_0}$ must be chosen. The residue exponential above is evaluated only when $\chi =1$. At $z_0=0$ the original indentation excludes the singular point, so no additional residue is crossed. These conventions matter before taking any limiting pole position.

Because $\mathrm{Re}a>0$, the endpoint factor $(1+t^2{)}^a$ vanishes at the branch points $t=\pm i$. A finite contour may therefore run from $-i$ to $i$ on either side of a chosen branch cut; the diagram shows contours bowing through the left and right half-planes.

For $a=1$, $f(t)=1$ is entire, so path independence makes either contour give, up to orientation,Hence a nonzero constant multiple gives the solution

A branch point of a multivalued analytic function is a point around which analytic continuation along a closed loop can return a different value. A branch cut is a curve or union of curves removed from the domain so that loops causing this monodromy are excluded and one single-valued branch can be chosen.

On taking $z$ once anticlockwise around zero, $\theta$ increases by $2\pi$ and $w$ changes by $2\pi i$; hence zero is a branch point. Under the coordinate $\zeta =1/z$, a loop around $\zeta =0$ corresponds to a loop around infinity and produces the opposite increment, so infinity is also a branch point. Removing any curve joining the two branch points prevents such loops. With $0<\arg z<2\pi$, one possible branch cut is therefore

On the slit plane $\mathbb{C}\setminus [-a,a]$, chooseso that $F(z)\sim z$ as $z\to \infty$. This branch cut gives, for $-a<x<a$, the boundary values

Integrate $F$ over the boundary of a large disk with a thin slit removed. The upper bank is traversed from $-a$ to $a$ and the lower bank from $a$ to $-a$, so together they contributeAt infinity the Laurent series isand hence the counterclockwise large-circle integral tends to $-\pi i{a}^2$. By the Cauchy integral theorem, the large-circle and slit-boundary contributions sum to zero. Thereforewhich yields

Putand take the branch cut $[-1,1]$. On $\mathbb{C}\setminus [-1,1]$, choose the analytic branch of a square root whose upper boundary value isIts lower boundary value is $s(x-i0)=-\sqrt{1-x^2}$. This fixes the integrand to be $+1$ at $0^{+}$ as required.

For $0\le \arg z\le \pi$, start at $0^{+}$ and use any path to $z$ lying above the cut; path deformations within that region do not alter the integral by the Cauchy integral theorem. For $\pi <\arg z<2\pi$, start in the same way, continue counterclockwise around the branch point $-1$ from the upper bank to the lower bank, and then continue to $z$ through the lower half-plane. This second prescription is the continuation obtained as $argz$ increases through $\pi$.

An analytic branch of a square root determined by the branch $\ell$ isIf ${\psi }_1$ and ${\psi }_2$ are two such branches, their ratio $h={\psi }_1/{\psi }_2$ is analytic and satisfies $h^2=1$. Since a domain is connected and $h$ takes values in the discrete set $\{1,-1\}$, $h$ is constant. Thus ${\psi }_1={\psi }_2$ throughout $D$ or ${\psi }_1=-{\psi }_2$ throughout $D$.

For $z=r{e}^{i\theta }$, the principal square root isOn $D_2$, one may instead takeThe respective removed half-axes are their branch cuts.

On $\mathbb{C}\setminus [-1,1]$, defineIt is analytic there, its square is $1-z^2$, andso it is the required branch. Substitution gives, for $0<|z|<1$,The binomial series yieldshence the first three terms of the Laurent series are

Sinceits residue at zero is $-i$. Under the change of variable $z=1/\zeta$, the two orientation reversals cancel, and the residue theorem gives

Taking the Laplace transform in $t$ and using the zero initial condition givesThe solution bounded as $x\to \infty$ isThe transformed boundary condition isso this is the sinusoidally forced heat equation on a half-line. Integrating in $x$ gives

Use the principal square root, with a branch cut along the negative real axis. The Bromwich contour is a vertical line to the right of $p=\pm i\omega$ and the branch point $p=0$. For $t>0$, close it in the left half-plane, indent around the cut, and let the large semicircle recede. The simple poles at $p=\pm i\omega$ contributeHence

On the upper and lower sides of the cut, put $p=-s$; the two values of $\sqrt{p}$ are $i\sqrt{s}$ and $-i\sqrt{s}$. Their jump and the opposite contour orientations combine to giveThis is precisely Bromwich inversion with a square-root branch cut: the residues produce the permanent periodic response, while the cut integral is the transient that decays with time.

Harmonic parts. Write the holomorphic function as $f=u+iv$. The Cauchy-Riemann equations say $u_x=v_y$ and $u_y=-v_x$. Equality of mixed partial derivatives therefore givesThus both the real and imaginary parts are harmonic functions and satisfy two-dimensional Laplace's equation.

Single-valued logarithm. A convenient choice isusing the principal logarithm for each term. Its branch cuts in the $z$ plane are $(-\infty ,-1]$ and $[1,\infty )$. The branch points are $-1$ and $1$: going once around the first adds $2\pi i$, while going once around the second subtracts $2\pi i$. Removing the two exterior rays makes this a consistent branch of a multivalued function and in particular leaves the whole open unit disk available.

Potential inside the cylinder. Put $z=r{e}^{i\theta }$ with $r<1$. Multiplication by $1-\overline{z}$ givesThe denominator is positive and the real part of the numerator is positive. Thus its argument lies strictly between $-\pi /2$ and $\pi /2$, andAs $r\uparrow 1$, this tends to $\pi /2$ on the upper open semicircle and $-\pi /2$ on the lower open semicircle. Scale the imaginary part to obtainIt is a harmonic function, has the required boundary values, and gives $\Phi (0)=0$. The two joining points of the boundary arcs have discontinuous boundary data, so no common continuous boundary value is demanded there. Along the real diameter, including radial approaches to either joining point, the displayed solution is zero. This is the disk potential with two semicircular boundary values.

For a second check, the Taylor series of the logarithms gives $f(z)=2{\sum }_{j\ge 0}{z}^{2j+1}/(2j+1)$. Taking the imaginary part yields the Fourier serieswhose boundary limit is exactly the required alternating pair of constant semicircle values.

Branches and cuts. For the square root of z squared minus one, a branch point is a point around which analytic continuation along an arbitrarily small closed loop can change the value of a chosen branch of a multivalued function. Near $z=1$, the square root behaves as $\sqrt{2}(z-1{)}^{1/2}$; near $z=-1$ it is a nonzero analytic factor times $(z+1{)}^{1/2}$. One circuit around either point reverses its sign. Thus the finite branch points are $\boxed{z=\pm 1}$. Infinity is not a branch point: in $w=1/z$ the expression is $w^{-1}\sqrt{1-w^2}$, which has two single-valued meromorphic choices near zero.

A single branch cut can join $-1$ to $1$; its shortest choice is the real segment $[-1,1]$. Any surviving loop then surrounds either both branch points or neither, so it produces no sign change. Alternatively, use two branch cuts, $(-\infty ,-1]$ and $[1,\infty )$, each joining a branch point to infinity. On the resulting slit region one can choose a consistent branch of a multivalued function.

For the finite cut, fix $f(z)\sim z$ at infinity, equivalently $f(x)>0$ for $x>1$. To give its actual values on both sides, putFor $y\ne 0$, the selected square root isHere $\mathrm{sgn}(0)=0$; when $x=0$, $U=0$. Squaring verifies the formula because $UV=|xy|$. Its real part has the sign of $x$, its imaginary part has the sign of $y$, and it matches the chosen value at infinity. In particular, the two boundary values on the branch cut areOutside the cut, $f(x)=\mathrm{sgn}(x)\sqrt{x^2-1}$. Reversing the global branch of a multivalued function reverses every displayed value.

Differential equation. Write $f(\zeta )=\sqrt{{\zeta }^2-1}$ with this finite cut, and orient $C$ counterclockwise with one circuit around it. Since $C$ is fixed and avoids the branch points, differentiation under the integral sign givesOn the contour $f^{\prime }=\zeta /f$. ThereforeThe last contour integral vanishes because the expression in brackets returns to its original value after circling both branch points. This proves $\boxed{z{u}^{\prime \prime }+u^{\prime }-zu=0}$, including at $z=0$; division by $z$ is unnecessary.

Value at zero. Near either endpoint ${\zeta }_0=\pm 1$, the integrand at zero is $O(|\zeta -{\zeta }_0{|}^{-1/2})$. A small arc of radius $\epsilon$ has length $O(\epsilon )$ and hence contributes $O(\sqrt{\epsilon })\to 0$. The upper lip runs from $1$ to $-1$, and the lower lip from $-1$ to $1$. ThusUsing $x=\sin t$ evaluates the final integral as $\pi$, and soThe same deformation gives $u(z)=2i{\int }_{-1}^1{e}^{zx}(1-x^2{)}^{-1/2}dx$. The sign of $u(0)$ reverses if either the global square-root choice or the contour orientation is reversed; the differential equation is unaffected.

Choose the square root of one minus z squared with $f(0)=1$: explicitly $f(z)=\sqrt{1-z}\sqrt{1+z}$ with principal square roots. It is single valued and holomorphic away from the two specified exterior real branch cuts. If $q(x)=\sqrt{x^2-1}>0$ for $|x|>1$, its boundary values areFor example, above the right cut $1-z$ approaches the negative real axis from below, selecting $-i\sqrt{x-1}$; above the left cut $1+z$ approaches it from above, selecting $+i\sqrt{-x-1}$. Continuing once around either branch point changes the sign, which explains the need for cuts.

Apply the residue theorem to $F(z)=1/[zf(z)]$ on a large disk slit along both cuts, with small arcs excluding $\pm 1$. The only enclosed pole is at zero, and its residue is $1/f(0)=1$. On the positively oriented boundary of the slit domain, the upper right lip runs from one to $R$, and the lower right lip returns from $R$ to one. The jump in $F$ gives their combined integral as $2i{I}_R$, where $I_R={\int }_1^{R}dx/[x\sqrt{x^2-1}]$. The left lips similarly give $2i{I}_R$ after the substitution $x=-t$.

The small arcs contribute $O(\sqrt{\epsilon })$ because the branch singularities have order $-1/2$. The large arc contributes $O(R^{-1})$ because $F(z)=O(|z{|}^{-2})$. Taking limits therefore yields $4iI=2\pi i$, and henceThe real substitution $x=\sec \theta$ independently reduces the integral to ${\int }_0^{\pi /2}d\theta$, confirming its value and sign.

The square-root factor gives an algebraic branch point at $z=0$. The logarithm is locally analytic there but has value an odd multiple of $i\pi$, which is nonzero, so it cannot cancel that branching. The zero and pole of $(z-1)/(z+1)$ give logarithmic branch points at $z=1$ and $z=-1$, respectively. Thus the finite singularities areInfinity is also a branch point. On the logarithm branch tending to zero there, $\log [(z-1)/(z+1)]=-2/z+O(z^{-3})$, so the product behaves as $-2z^{-1/2}$. Other logarithm branches add $2\pi ik\sqrt{z}$ and retain branching. A chosen branch cut joins or separates these branch points; its ordinary interior points are not additional branch points of the underlying multivalued function.

For a positively oriented simple closed contour $C$, with no pole on $C$ and with $f$ holomorphic inside except at finitely many poles, the residue theorem statesFor a more general closed contour, each residue is multiplied by its winding number about the pole.

The powers must be defined using the same branch of the complex logarithm as the numerator. A branch suitable for both requested calculations isThe branch cut is a ray just below the positive real axis, and $z^{\beta }:=e^{\beta \mathrm{Log}z}$. Since $\beta >1$, the displayed choice of $\delta$ is possible. It includes the required upper half-plane and the sector used below. The origin remains an excluded branch point.

In $0\le \arg z<\pi$, all zeros of $1+z^{\beta }$ areEach is a simple pole, since $\beta {\zeta }_j^{\beta -1}=-\beta /{\zeta }_j\ne 0$. Its residue isThe strict inequality matters if a root lies on the negative real axis, which is outside the specified region.

To evaluate the two improper integrals, write them as $I$ for the logarithmic numerator and $J$ for numerator one, and set $\theta =\pi /\beta$, $\alpha =2\theta$. Use a counterclockwise annular sector with $ϵ<|z|<R$ and $0<\arg z<\alpha$. It encloses precisely ${\zeta }_0=e^{i\theta }$. The outward positive-axis segment contributes $I$ in the limit. On the inward segment $z=r{e}^{i\alpha }$, the logarithm is $\log r+i\alpha$ and $z^{\beta }=r^{\beta }$, so that segment contributes $-e^{i\alpha }(I+i\alpha J)$.

The small arc is $O(ϵ|\log ϵ|)$, and the large arc is $O(R^{1-\beta }\log R)$; both vanish. These estimates follow from the arc lengths and $|1+z^{\beta }|\ge ||z{|}^{\beta }-1|$. The real integrals also converge absolutely at both endpoints for $\beta >1$. The residue theorem now givesAfter division by $e^{i\theta }$, the real and imaginary parts areSince $0<\theta <\pi$, $\sin \theta >0$. Hence the logarithmic beta integral from a sector contour yieldsFor $1<\beta <2$ the sector angle exceeds $\pi$. This is why choosing the principal complex logarithm on the whole sector would be invalid; the displaced cut above handles that range as well as $\beta \ge 2$.