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Given any $z\in \mathbb{C}$ that is not a nonpositive integer, choose $N$ large enough that $\mathrm{Re}(z+N)>0$ and defineThe recurrence shows that definitions from different sufficiently large $N$ agree, so this gives the unique analytic continuation from the right half-plane. It is analytic except where a denominator factor vanishes, namelyThese points are simple poles. More precisely, at $z=-n$,because one may writeand use $I(1)=1$.

Poles and residues. An isolated singularity $z_0$ is a pole of order $m$ if its Laurent series has finitely many negative powers, the lowest being a nonzero multiple of $(z-z_0{)}^{-m}$. The residue is the coefficient of $(z-z_0{)}^{-1}$.

If $h(z_0)=0$ and $h^{\prime }(z_0)\ne 0$, factor $h(z)=(z-z_0)H(z)$, where $H$ is holomorphic near $z_0$ and $H(z_0)=h^{\prime }(z_0)$. For $g(z_0)\ne 0$,The last quotient is holomorphic and nonzero at $z_0$, proving a simple pole with

The given function. For noninteger $a$, specify a branch of a multivalued function before using $z^a$. Choose $z^a=\exp (a\log z)$ with $-\pi /2<\arg z<3\pi /2$, a cut along the negative imaginary axis. This leaves all three roots of $1+z^3$ off the cut. They are $z_{+}=e^{i\pi /3}$, $z_{-}=e^{-i\pi /3}$ and $z_0=-1$, all simple. The numerator is nonzero at these points. Since the denominator has derivative $3z^2$, the residues areFor noninteger $a$, zero is a branch point, rather than an isolated pole. This remains true when $-1<a<0$ and the numerator diverges there. For nonnegative integer $a$ it is a regular point. Other logarithm branches change the phases of the residues consistently.

Integral and convergence. Near zero the integrand behaves as $x^a$, requiring $a>-1$. At infinity it behaves as $x^{a-3}$, requiring $a<2$. Therefore the exact convergence range isUse a counterclockwise sector contour with rays at angles $0$ and $2\pi /3$, inner radius $\epsilon$ and outer radius $R$. The same complex logarithm branch is valid throughout this sector. Only $z_{+}=e^{i\pi /3}$ lies inside. The small circular arc contributes $O({\epsilon }^{a+1})$ and the large one contributes $O(R^{a-2})$, so both vanish precisely in the stated range.

The outward positive-real ray contributes $I$. On the returning ray set $z=x{e}^{2\pi i/3}$: the denominator is still $1+x^3$, while $z^{a}dz=e^{2\pi i(a+1)/3}{x}^{a}dx$, with reversed limits. The residue theorem givesPut $\alpha =\pi (a+1)/3$. The right-hand phase is $-e^{i\alpha }$ and $1-e^{2i\alpha }=-2i{e}^{i\alpha }\sin \alpha$. HenceThe sine is positive in this range, consistent with a positive real integral. This is a special case of the Mellin integral of a reciprocal binomial.

Laplace transform. Use $Y(p)={\int }_0^{\infty }{e}^{-pt}y(t)dt$. The initial value is zero and the Dirac delta contributes $e^{-p}$, so the transformed equation is $pY+Y=e^{-p}$. Therefore

Bromwich inversion. The Bromwich inversion formula iswith the vertical Bromwich contour traversed upwards and its symmetric improper limit understood. For $t>1$, close to the left, where the exponential decays; the enclosed simple pole at $p=-1$ has residue $e^{-(t-1)}$. For $t<1$, close to the right, which contains no pole. Thuswhere $\Theta$ is the Heaviside step function. This agrees with the Laplace transform time-shift rule. At $t=1$, inversion returns the average $1/2$ of the two one-sided values. The value assigned at that single point does not affect the differential equation in the distributional sense.

Fourier transform. State the convention explicitly:The Fourier transform of the distributional derivative is $ik\tilde{y}(k)$, including the jump at $t=1$. The Dirac delta has transform $e^{-ik}$, giving

Inverse Fourier integral. Substitute into the inverse Fourier transform:The only pole is at $k=i$. For $t>1$, close in the upper half-plane. Jordan's lemma removes the large arc and the residue theorem gives $i{\mathrm{Res}}_{k=i}=i[e^{-(t-1)}/i]=e^{-(t-1)}$. For $t<1$, close in the lower half-plane, where there are no poles, giving zero. This reproducesIntegrating the differential equation across the impulse checks the unit jump, $y(1^{+})-y(1^{-})=1$. Away from the impulse the solution obeys $y^{\prime }=-y$, and it satisfies the stipulated zero past.

Definitions. In a punctured neighborhood of $z_0$, let the Laurent series be $f(z)={\sum }_{j=-\infty }^{\infty }{c}_j(z-z_0{)}^j$. The point is a pole of order $m\ge 1$ if $c_{-m}\ne 0$ and every coefficient with index less than $-m$ vanishes. Its residue is $\boxed{{\mathrm{Res}}_{z_0}f=c_{-1}}$. A simple pole has $m=1$.

Poles for $0<a<1$. Denote the bracket in the denominator by $D(z)$. Factoring givesPut $q=\sqrt{(1-a)/(1+a)}$, so $0<q<1$. The finite poles are $0,\pm q,\pm q^{-1}$, all simple; the numerator is nonzero at each. At zero,At a nonzero simple root $r$ of $D$, the residue formula givesFor $r^2=q^2$, $D^{\prime }(r)=8ar$; for $r^2=q^{-2}$, $D^{\prime }(r)=-8ar$. Substitution yieldsThe two zeros of a function at $z=\pm i$ do not cause additional poles: there $D=-4$ and the numerator vanishes.

The case $a=1$. Simplify before using any formulas that divide by $1-a^2$:Thus its only finite pole isAs $a\uparrow 1$, the two inner poles merge with zero while the outer two escape to infinity. If one also considers the Riemann sphere, $f$ at $a=1$ has a simple pole at infinity because $f(z)\sim z/4$; the requested unit-circle calculation only uses finite poles.

Unit-circle integral. Set $z=e^{i\theta }$. ThenAt the two angles where the tangent is undefined, the real integrand has continuous limiting value zero. Thus the substitution is valid for the improper integral, or after that continuous extension. No pole lies on $|z|=1$ for any allowed $a$.

For $0<a<1$, the sum of the enclosed residues isThe residue theorem therefore givesAt $a=1$ the enclosed residue is $1/2$, giving $I=\pi$, exactly the same formula. The direct simplification $1/(1+{\tan }^2\theta )={\cos }^2\theta$ provides an independent check of that endpoint.

Choose $\gamma >1/2$. The inverse integrand is ${\lambda }^{-s}/[2(s-1/2)]$. Because $-\log \lambda >0$, the exponential factor decays towards the left. Closing the Bromwich contour to the left and using the residue theorem picks up just the simple pole at $s=1/2$, with residue ${\lambda }^{-1/2}/2$. HenceIts defining integral checks both the answer and its convergence condition:

The denominator factors as $(z+1-i)(z+1+i)$ and neither factor cancels with the numerator. Thus there are simple poles at $z=-1\pm i$. At infinity the quotient tends to one and has a regular expansion in $1/z$, so it has a removable singularity at infinity, with extended value one.

For a simple pole, the residue is ${\mathrm{Res}}_{z_0}f={\lim }_{z\to z_0}(z-z_0)f(z)$, the coefficient $a_{-1}$ of its Laurent series. On an anticlockwise circle $z=z_0+R{e}^{it}$, integration of each Laurent term is zero except $a_{-1}/(z-z_0)$, whose integral is $2\pi i{a}_{-1}$. This proves the stated single-pole residue theorem.

For $F(z)=e^{az}/(1+e^z)$, the only pole inside the rectangle of height $2\pi$ is $z=i\pi$, with residue $e^{ai\pi }/e^{i\pi }=-e^{ai\pi }$. The residue theorem therefore gives $I(R)=-2\pi i{e}^{ai\pi }$. The upper edge is traversed from right to left and satisfies $F(x+2\pi i)=e^{2\pi ia}F(x)$, so the two horizontal contributions sum to $(1-e^{2\pi ia}){\int }_{-R}^R{e}^{ax}/(1+e^x)dx$.

On the right edge the modulus is bounded by $e^{aR}/(e^R-1)$, and on the left by $e^{-aR}/(1-e^{-R})$. Multiplying by the fixed edge length $2\pi$ shows that both contributions tend to zero because $0<a<1$. ThusUsing $1-e^{2\pi ia}=-2i{e}^{\pi ia}\sin (\pi a)$ yieldsThis rectangular contour for an exponential beta integral also gives the ordinary beta integral after the substitution $t=e^x$.

Put $d=\sqrt{a^2-b^2-c^2}>0$ and $h=b-ic$. The denominator of the complex rational function becomesThe quadratic formula gives its two zeros, hence the two poles:Their product is $(b+ic)/(b-ic)$, so $|z_{\mathrm{in}}{z}_{\mathrm{out}}|=1$. To locate them one also needs to exclude the possibility of both lying on the unit circle. Here their moduli are explicit:Both are simple poles: $P^{\prime }(z)=hz+a$ takes the nonzero values $d$ and $-d$ at them. In particular their residues are $1/d$ and $-1/d$.

For the real integral, use the change of variables $z=e^{i\theta }$. As $\theta$ runs from $-\pi$ to $\pi$, this traverses the unit circle once counterclockwise, andThus $I=(1/i){\oint }_{|z|=1}f(z)dz$. Only $z_{\mathrm{in}}$ is enclosed. The residue theorem, or Cauchy integral formula after subtracting its simple-pole part, givesThe real denominator is positive because $a>\sqrt{b^2+c^2}$. As an independent check, write $b\cos \theta +c\sin \theta =\sqrt{b^2+c^2}\cos (\theta -\varphi )$; a period shift reduces the answer to the existing reciprocal affine sine integral.

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 $\beta \ge 2$.

Iterate the Gamma function recurrence to obtain, near $z=-m$ for $m\ge 0$,The numerator is holomorphic and equals $\Gamma (1)=1$ at $z=-m$. Exactly one denominator factor, $z+m$, vanishes; the product of the remaining factors there is $(-m)(-m+1)\cdots (-1)=(-1{)}^{m}m!$. Thus there is a simple pole, with Gamma function residue at a nonpositive integerFor $m=0$ the empty product is one. The same recurrence with a sufficiently large shift defines the continuation at every point away from these integers, so no further poles occur. At the poles, the recurrence is an identity of meromorphic functions, rather than an equality of finite values.

Put $d=\sqrt{a^2-b^2}>0$. The two roots areSince $(a-d)(a+d)=b^2$, their moduli are $b/(a+d)<1$ and $(a+d)/b>1$. Thus exactly one root is inside the unit circle, and neither lies on it.

For the real integral use $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The change of variables givesThe contour is counterclockwise. At its single interior simple pole, the residue isBy the residue theorem, the reciprocal affine sine integral is thereforeThe condition $a>b>0$ also makes the real denominator strictly positive, consistent with the positive answer and the absence of contour poles on the unit circle.

The denominator vanishes at the four roots of $z^4=-1$, namely $z_k=e^{i(2k+1)\pi /4}$ for $k=0,1,2,3$. They are simple poles, since $(1+z^4{)}^{\prime }=4z^3$ is nonzero at each root and the numerator does not vanish there. The simple-pole residue formula givesExplicitly,There are no branch points or other finite singularities. At infinity the function is $O(z^{-2})$, so infinity is removable rather than a pole.

The rational factors give four simple poles at $z=\pm i,\pm 2i$; none is canceled by the numerator. The square root has branch points at $z=\pm 1$, with local behaviour proportional to $(z\mp 1{)}^{-1/2}$ in the integrand.

For this part, choose $h(z)=\sqrt{z^2-1}$ analytic off $[-1,1]$ and satisfying $h(z)\sim z$ at infinity. Its values are $h(x)=\sqrt{x^2-1}$ for $x>1$, $h(x)=-\sqrt{x^2-1}$ for $x<-1$, andThus the chosen branch cut is $[-1,1]$, and the function is analytic elsewhere apart from the four poles. Infinity is not an intrinsic branch point: $g(z)=O(z^{-2})$ on this branch.

Another legitimate convention applies the positive-axis square root of part (a) to $w=z^2-1$; it places the cuts on $(-\infty ,-1]$ and $[1,\infty )$. Its upper-half-plane values agree with the branch used here, so the contour calculation below and the positive real square root in the required integral are unchanged. The branch choice must be stated because a square root alone does not specify a single-valued complex function.

A simple zero of an analytic function has a local factorization $f(z)=(z-z_0)u(z)$, where $u$ is an analytic function and $u(z_0)=f^{\prime }(z_0)\ne 0$. Since $1/u$ is also an analytic function near $z_0$, its Taylor series givesThe nonzero coefficient of $(z-z_0{)}^{-1}$ proves that this is a simple pole, withMultiplying by the Taylor series $g(z)=g(z_0)+O(z-z_0)$ similarly givesThe assumption $g(z_0)\ne 0$ ensures that the pole is not canceled. In either formula the residue is precisely the coefficient of $(z-z_0{)}^{-1}$ in the Laurent series.

Clear the apparent negative power of $z$ to obtain the rational functionLet $\Delta =\sqrt{a^2-1}>0$. Its denominator factors as $(z-z_{-})(z-z_{+})$, withThese are distinct simple poles. Since $(a-\Delta )(a+\Delta )=1$ and $a+\Delta >1$, we have $0<a-\Delta <1$: only $z_{-}$ lies inside the unit circle. The apparent singularity at $z=0$ in the original expression is a removable singularity, with the continued value $h(0)=0$; it is not another pole.

Using the residue formula from part (a), with denominator derivative $2z-2ia$, givesThere are no other finite singularities of the continued rational function.

Put $w=z-4i$. Factor out the nonzero constant in the denominator and use the geometric series:This Taylor series converges when $|z-4i|<|3-4i|$. The simple pole at $z=3$ is the only finite obstruction to analyticity, whose distance from the expansion centre is five. Hence the radius of convergence is

For a positively oriented simple closed contour, the residue theorem states that the integral of a function holomorphic on and inside the contour except at finitely many isolated interior singularities is $2\pi i$ times the sum of its residues. There must be no singularity on the contour.

Factor the denominator as $(z^2+z+1)(z^2-z+1)$. For $R>1$, the upper semicircle encloses the two simple polesThe other two poles are in the lower half-plane. Since $D^{\prime }(z)=4z^3+2z=2z(2z^2+1)$,Writing $h=\sqrt{3}/2$ and using $1/a=1/2-ih$, $1/b=-1/2-ih$, their numerator difference simplifies toThus for every $R>1$ the closed-contour integral isTo remove the arc, on $C_R$ we have $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^4+z^2+1|\ge R^4-R^2-1$. For sufficiently large $R$, the complex line integral estimate givesThe real-line integral is absolutely convergent. Its imaginary part is zero by oddness of $\sin x/(x^4+x^2+1)$. Therefore

For $\kappa >\Omega$, put $\eta =\sqrt{{\kappa }^2-{\Omega }^2}$, so $0<\eta <\kappa$ and the poles are $i(\kappa \pm \eta )$. Closing in the upper half-plane for $t>0$ again includes both simple poles. The residue theorem givesThis overdamped response has no oscillation. Its slow tail is $e^{-(\kappa -\eta )t}/(2\eta )$, and it tends to zero because $\eta <\kappa$.

For $t>0$, close the inverse-transform contour in the upper half-plane. There $|e^{i\omega t}|\le 1$; the rational factor is $O(|\omega {|}^{-2})$, so the arc integral vanishes. For $\Omega >\kappa$, write $\beta =\sqrt{{\Omega }^2-{\kappa }^2}>0$. The two simple poles are $i\kappa \pm \beta$, and the residue theorem yieldsThis causal Green function of a damped oscillator has damped oscillations with envelope $e^{-\kappa t}$ and tends to zero as $t\to \infty$.

For the normalized equation, $x_0$ is an ordinary point when both $p$ and $q$ are analytic near $x_0$. A singular point is a regular singular point when $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically to $x_0$; equivalently $p$ has at most a simple pole and $q$ at most a double pole there.

Thus ordinary points admit ordinary power-series solutions, while the Frobenius method applies at a regular singular point. The definitions refer to the coefficients after dividing by the leading coefficient of $y^{\prime \prime }$.