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Factor the cubic denominator as $(\omega -i)({\omega }^2+4)$. Substitution givesThere is a double pole at $i$ and simple poles at $2i,-2i$. For $t>0$, close the inverse-transform contour counterclockwise in the upper half-plane. The residue at the double pole isand the residue at $2i$ is $-i{e}^{-2t}/2$. Multiplication by $i$ from the residue theorem therefore gives $y(t)=(t/3-5/9)e^{-t}+e^{-2t}/2$.

For $t<0$, the clockwise lower closure encloses only $-2i$, with residue $-i{e}^{2t}/18$. The orientation supplies $-i$, giving $y(t)=-e^{2t}/18$. Hence the continuous solution isThe large inverse-transform arcs vanish: the rational factor is $O({\omega }^{-3})$, while the exponential is bounded on the appropriate closure.

The graph tends to zero from below as $t\to -\infty$, passes through $y(0)=-1/18$ with slope $-1/9$, reaches a small negative minimum after zero, then crosses the axis, reaches a positive maximum and tends to zero from above. The positive-time stationary points solve $8/9-t/3=e^{-t}$, and the crossing solves $t/3-5/9+e^{-t}/2=0$; these equations determine the sketch without guessing signs.

For a distributional check, $y$ and $y^{\prime }$ are continuous, while $y^{\prime \prime }(0^{+})-y^{\prime \prime }(0^{-})=1$. The jump conditions for a third-order Green function therefore produce the delta in $f^{\prime }=\delta -e^{-t}H(t)$, and each smooth branch satisfies the remaining equation. The nonzero solution for $t<0$ is consistent with the stipulated decay at both temporal infinities. Those are two-sided conditions, not a causality assumption: this is the setting of a decaying two-sided Green function of a constant-coefficient ordinary differential equation and causality versus stability of a Green function.

First use $\sin \theta /(a-\sin \theta )=a/(a-\sin \theta )-1$. If $J={\int }_{-\pi }^{\pi }(a-\sin \theta {)}^{-1}d\theta$, the required integral is $aJ-2\pi$.

Set $z=e^{i\theta }$. The unit circle is traversed counterclockwise once, $d\theta =dz/(iz)$, and $\sin \theta =(z-z^{-1})/(2i)$. ThusThe function $h(z)/z=-2i/(z^2-2iaz-1)$ is regular at zero. Its only enclosed pole is $z_{-}$ from part (b), and its residue isBy the residue theorem, $J=(1/i)2\pi i/\Delta =2\pi /\Delta$. ThereforeAs a sign and scale check, this is positive and behaves as $\pi /a^2$ for large $a$, in agreement with expanding the integrand and integrating its first nonzero, ${\sin }^2\theta /a^2$, term.

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 simple-contour form of the residue theorem, let $C$ be a positively oriented, simple, closed, piecewise smooth contour. Assume $f$ is holomorphic on a neighbourhood of $C$ and its interior except at finitely many isolated singularities $z_1,\dots ,z_s$, all strictly inside $C$, with none on $C$. ThenThe more general version weights each residue by the contour's winding number, with the usual requirement that the contour be homologous to zero in the domain of analyticity containing it and the nonsingular region.

For the requested application, assume $f$ is holomorphic on a neighbourhood of the contour and its closed interior, and $z_0$ is strictly inside it. Its Taylor series showsOnly the second term contributes a residue. The residue theorem therefore proves the derivative form of the Cauchy integral formula:

Use $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The unit circle is traversed anticlockwise, and the integral becomesOnly the simple pole $z=i/3$ lies inside the contour. Its residue isHence the residue theorem gives

Close the real segment $[-R,R]$ by an anticlockwise semicircle in the upper half-plane. For $F(z)=1/[(z^2+1)(z^2+4)]$, the contour integral over the arc is $O(R^{-3})$: its length is $\pi R$, whereas $|z^2+1|\ge R^2-1$ and $|z^2+4|\ge R^2-4$. The enclosed simple poles are $i$ and $2i$, with residuesLetting $R\to \infty$ in the residue theorem yields

If $f$ is analytic on and inside a positively oriented simple closed contour except for finitely many isolated poles $z_j$ strictly inside it, then the residue theorem statesThere must be no singularity on the contour. More generally each residue is multiplied by the contour's winding number about its point.

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 $-1<s<1$.

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.

For this real integral choose $0<\arg z<2\pi$, so the upper bank of the positive real axis has $z^{1/4}=x^{1/4}>0$. Use a counterclockwise keyhole contour with inner radius $\epsilon$ and outer radius $R$, enclosing $\pm i$. The upper-bank contribution tends to $J$. On the lower bank, $z^{-1/4}=e^{-i\pi /2}{x}^{-1/4}=-i{x}^{-1/4}$, and the reversed orientation makes its contribution $+iJ$.

The large arc is $O(R^{-5/4})$ because $g=O(R^{-9/4})$, while the small arc is $O({\epsilon }^{3/4})$. Both vanish, givingThe residues on this branch areThe residue theorem yieldsThereforeThe branch cut in (a) was arbitrary; here placing it on the integration axis makes the bank contributions directly proportional to the desired positive-real integral.

For a double pole at $z_0\in \{i,-i\}$, differentiate the remaining holomorphic factor:Since $z_0^2=-1$ and $1+z_0+z_0^2=z_0$, this simplifies toOn the selected branch $\sqrt{i}=e^{i\pi /4}$ and $\sqrt{-i}=e^{3i\pi /4}$, givingThe residue theorem therefore givesThe equality $C=2I$ uses the limiting contour integrals established in (c).

A simple pole at $z_0$ means that $f$ is holomorphic in a punctured neighbourhood and $(z-z_0)f(z)$ extends holomorphically to a nonzero value at $z_0$. Equivalently its Laurent series begins with $a_{-1}/(z-z_0)$ and has no higher negative powers. The residue isIn particular, if $f=g/h$ with $g(z_0)\ne 0$, $h(z_0)=0$ and $h^{\prime }(z_0)\ne 0$, this becomes $\mathrm{Res}(f,z_0)=g(z_0)/h^{\prime }(z_0)$. The associated residue theorem isfor a positively oriented simple contour with no poles on it.

Use the counterclockwise rectangular contour with vertices $-R,R,R+i\pi ,-R+i\pi$. It encloses just the simple pole $i\pi /2$, whose residue is $e^{k\pi /2}/i$. Let $I_R={\int }_{-R}^R{e}^{-ikx}/\cosh xdx$. The identitymakes the top edge, traversed right to left, contribute $e^{k\pi }{I}_R$. On either vertical edge,Thus the sum of the vertical integrals tends to zero, with absolute value at most $2\pi e^{|k|\pi }/\sinh R$. The residue theorem and $R\to \infty$ now giveThe argument works for every fixed real $k$, including zero and negative values, and gives the Fourier transform of the hyperbolic secant with the convention $\tilde{f}(k)=\int e^{-ikx}f(x)dx$.

Let $B=|b|>0$. For $a>0$, integrate $e^{iaz}/(z^2+B^2)$ over the upper semicircle. Jordan's lemma removes the arc, and the residue theorem at $z=iB$ givesTaking the real part and using evenness in both $x$ and $a$ yieldsThe case $a=0$ follows directly from an inverse tangent primitive. The complex exponential is essential on the contour: the cosine alone has a growing exponential component.

For $a>0$, differentiation under the integral sign gives $-I_1^{\prime }(a)={\int }_0^{\infty }x\sin (ax)/(x^2+B^2)dx$. This derivative is a conditionally convergent improper integral; the Dirichlet test makes its tails uniform on compact subintervals of $a>0$, justifying the differentiation. Setting $a=1$ givesFinally let $J(a)={\int }_0^{\infty }\sin (ax)/[x(x^2+B^2)]dx$. Its differentiated integrand is absolutely integrable, so $J^{\prime }(a)=I_1(a)$ and $J(0)=0$. HenceIf $b=0$, $I_1$ and $I_3$ diverge at the origin, whereas $I_2={\int }_0^{\infty }\sin x/xdx=\pi /2$ is the corresponding Dirichlet integral.

For fixed $z\notin \mathbb{Z}$, integratecounterclockwise around the specified square of half-side $R_N=N+1/2$, choosing $N>|z|$. The integer poles inside it are $-N,\dots ,N$, and their residues are $1/(n^2-z^2)$ because $\pi \cot (\pi w)$ has residue one at each integer.

The two other possible poles are $w=z$ and $w=-z$. Their residues are each $\pi \cot (\pi z)/(2z)$, hence sum to $\pi \cot (\pi z)/z$. This formula still applies when the numerator vanishes and a possible pole is removable. The residue theorem givesOn the square, $|w|\ge R_N$, its perimeter is $8R_N$, and the supplied cotangent bound givesThe series converges absolutely by part (b), so its symmetric partial sums have the same limit as the full sum. ThereforeAt half-integers the cotangent form makes the value zero explicit; the reciprocal-tangent notation is understood by its continuous extension there. The cotangent contour summation rests on the contour estimate, not only on formally matching poles.

The residue $\mathrm{res}(f;a)$ is the coefficient $c_{-1}$ of $(z-a{)}^{-1}$ in the Laurent series at the isolated singularity. Equivalently it is $(2\pi i{)}^{-1}$ times the integral around a sufficiently small counterclockwise circle enclosing $a$.

The residue theorem states that if $f$ is holomorphic on a neighborhood of a positively oriented simple closed contour and its interior except for finitely many isolated singularities $a_j$ inside, thenNo singularity may lie on the integration contour. A more general contour weights each residue by its winding number.

For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence

Use the principal logarithm $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z<\pi$, whose branch cut is the negative real axis. For this choice,has three simple poles away from the cut: $z=1$, $z=ia$, and $z=-ia$. The pole at $1$, caused by $\mathrm{Log}1=0$, must be included.

Choose $0<\epsilon <\min \{1,a\}$ and $R>\max \{1,a\}$, and use a positively oriented keyhole contour around the negative real axis: the large circle is traversed counterclockwise, the upper bank from $-R$ to $-\epsilon$, the small circle clockwise, and the lower bank from $-\epsilon$ to $-R$. Formally, use radial segments at arguments $\pm (\pi -\eta )$, then let $\eta \downarrow 0$ with $R,\epsilon$ fixed. The function has continuous limiting values on each compact bank, which justifies this limit.

Put $L=\log a$. The residues areThe first follows from $\mathrm{Log}z=(z-1)+O((z-1{)}^2)$; the other two follow from the simple zeros of $z^2+a^2$. Their sum isThese expressions are valid also for $a=1$, since $L\pm i\pi /2$ never vanish for real $L$.

On the outer circle, $|\mathrm{Log}z|\ge \log R$ for $R>1$ and $|z^2+a^2|\ge R^2-a^2$. On the inner circle, $|\mathrm{Log}z|\ge |\log \epsilon |$ and $|z^2+a^2|\ge a^2-{\epsilon }^2$. Thus the estimation lemma bounds the two arc integrals byrespectively, as $R\to \infty$ and $\epsilon \downarrow 0$.

On the upper bank, $z=-x+i0$ gives $\mathrm{Log}z=\log x+i\pi$; on the lower bank it gives $\log x-i\pi$. Accounting for the opposite orientations, their combined integral isThe real improper integral converges: its integrand is bounded near zero by $1/(a^2{\pi }^2)$ and at infinity by $1/({\pi }^2{x}^2)$. Taking the arc limits in the residue theorem gives $-2\pi iI=2\pi i\Sigma$, henceThe final subtraction is exactly the contribution of the pole at $1$.