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A pole of order $N$ at $z_0$ has $(z-z_0{)}^{N}f(z)$ extending holomorphically to a nonzero value there. Its residue is the coefficient of $(z-z_0{)}^{-1}$ in the Laurent series. The Cauchy integral theorem gives ${\oint }_{C}fdz=0$ for a closed contour in a simply connected domain of holomorphicity. The residue theorem replaces zero by $2\pi i$ times the enclosed residues for a positively oriented simple contour avoiding finitely many poles.

For the first integral, use the upper-half-plane semicircle, with a small clockwise upper semicircle excluding the origin. Near zero,There are no poles inside the indented contour. The large arc vanishes because $a,b>0$ and the integrand there is $O(R^{-2})$; the small arc tends to $-i\pi i(a-b)=\pi (a-b)$. Therefore the real-axis complex principal value is $\pi (b-a)$. Taking real parts gives the ordinary convergent cosine-difference inverse-square integralThe real numerator cancels to second order at zero, so this final real integral itself needs no principal-value convention.

For $g(z)=e^{az}/\cosh \pi z$, the simple poles are $z_n=i(n+1/2)$, $n\in \mathbb{Z}$. Differentiating the denominator givesIntegrate around the rectangle with vertices $-R,R,R+i,-R+i$. It contains only $i/2$. On the top side $g(x+i)=-e^{ia}g(x)$, and reversed traversal therefore contributes $+e^{ia}$ times the bottom integral. The vertical sides vanish for $|a|<\pi$. The residue theorem gives $(1+e^{ia})J=2e^{ia/2}$, so the exponential moment of the hyperbolic secant is

For a pole of order $N$ at $z_0$, the residue is ${\mathrm{Res}}_{z_0}F=\frac{1}{(N-1)!}{\lim }_{z\to z_0}\frac{d^{N-1}}{d{z}^{N-1}}[(z-z_0{)}^{N}F(z)]$. For the rational integral, close the real-axis contour with an upper semicircle. The poles are double at $i$ and simple at $2i$, withThe arc integral vanishes as $O(R^{-5})$. By the residue theorem the whole real-axis integral is $\pi /9$, and evenness givesFor the second calculation, reciprocal expansion of the sine gives $1/\sin z=z^{-1}+z/6+O(z^3)$. Multiplying by $\cos z=1-z^2/2+O(z^4)$ yields $\cot z=z^{-1}-z/3+O(z^3)$. ThereforeAt every nonzero integer $n$, $\pi \cot \pi z$ has a simple pole with residue one, giving residue $1/n^2$ after division by $z^2$.

The cotangent contour summation uses the positively oriented square of half-side $R=N+1/2$. Its length is $8R$ and $|z|\ge R$ on its boundary. The supplied cotangent bound implies $|\oint Fdz|\le 8\pi C/R\to 0$. Its enclosed residues sum to $-{\pi }^2/3+2{\sum }_{n=1}^N{n}^{-2}$. The residue theorem therefore proves the Basel problem value

The residue theorem states that the contour integral of a meromorphic function around a positively oriented simple closed contour equals $2\pi i$ times the sum of its enclosed residues, provided there are no singularities on the contour. The roots of $1+z^n$ are $z_k=e^{(2k+1)i\pi /n}$, $k=0,\dots ,n-1$. They are simple poles, with residues $1/(n{z}_k^{n-1})=-z_k/n$.

Use the sector contour of angle $2\pi /n$, running outward along the positive axis and inward along the rotated ray. It encloses only $z_0=e^{i\pi /n}$. The large-arc contribution is $O(R^{1-n})$, hence vanishes. With $J={\int }_0^{\infty }dx/(1+x^n)$, the two ray contributions giveSince $1-e^{2i\alpha }=-2i{e}^{i\alpha }\sin \alpha$,For the second integral, use $z^{n/2}/(1+z^n)$ on a branch continuous across the sector away from zero, with a small inner arc of radius $ϵ$. Its large arc is $O(R^{1-n/2})$ and its small arc is $O({ϵ}^{1+n/2})$, both vanishing for $n\ge 3$. On the rotated ray, $z^{n/2}$ changes by $e^{i\pi }=-1$, so the two radial contributions add rather than subtract. The enclosed residue is $z_0^{1-n/2}/n=-i{e}^{i\pi /n}/n$. ConsequentlyBoth answers are instances of the Mellin integral of a reciprocal binomial. The direct contour argument also tracks the branch needed when $n$ is odd.

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 $t=e^x$.

Use $f(x)=(2\pi {)}^{-1}{\int }_{\mathbb{R}}\tilde{f}(k)e^{ikx}dk$. Jordan's lemma says that for $x>0$, if a function $h$ has maximum modulus $M_R\to 0$ on upper semicircles, then ${\int }_{C_R}h(k)e^{ikx}dk\to 0$; a bound is $\pi M_R/x$. For $x<0$ use lower semicircles instead. The rational factors here meet this decay condition, so the residue theorem evaluates their inverse transforms.

For $1/(a+ik)$, only the upper half-plane contains a pole, at $k=ia$. Its residue after multiplication by $e^{ikx}$ is $-i{e}^{-ax}$. The lower contour encloses no pole. ThereforeHere $H$ is the Heaviside step function, with midpoint value $H(0)=1/2$ for symmetric Fourier inversion.

For $1/(a^2+k^2)$, the upper-pole residue is $e^{-ax}/(2ia)$ when $x>0$, while the clockwise lower contour gives the matching expression for $x<0$. ThusThe final rational factor has double poles at $\pm ia$. At the upper pole,The upper inverse integral is $i$ times that residue. The lower contour for negative $x$ supplies $-x{e}^{ax}/2$, and the value at zero is zero. Consequently

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 $\beta \ge 2$.

First prove the convolution theorem with the stated negative-exponential Fourier transform. Assuming absolute integrability sufficient for Fubini's theorem, change variables $u=t-s$:

For the differential equation, integration by parts givesThe boundary terms vanish by the assumed decay of $y,y^{\prime }$. As with the convolution step, the transforms can first be justified for integrable functions and derivatives; the resulting kernel gives the desired decaying response. ThusThe convolution theorem and Fourier inversion theorem imply $y=f\ast G$.

For the inverse transform of $\tilde{G}$, useThe poles are at $ib,ia$. For $t>0$, close the inverse-transform contour above; the residues of $e^{ikt}\tilde{G}(k)$ there are $-i{e}^{-bt}/(a-b)$ and $i{e}^{-at}/(a-b)$. The inverse-transform factor $1/(2\pi )$ makes the answer $i$ times their sum. For $t<0$, close below, enclosing no poles. The large arcs vanish: the complete rational factor is $O(|k{|}^{-2})$ and the exponential is bounded on the appropriate semicircle. Therefore the causal Green function of an overdamped oscillator isIt is continuous, with $G(0)=0$ and $G^{\prime }(0^{+})=1$, so $(D_t+a)(D_t+b)G=\delta (t)$. The response can also be written $y(t)={\int }_{-\infty }^{t}G(t-s)f(s)ds$.

For the bilateral exponential forcing, split the real integral at zero:ConsequentlyFor negative time the lower half-plane contains only $k=-i$. The residue of $e^{ikt}\tilde{y}$ there is $i{e}^t/[(a+1)(b+1)]$. Clockwise closure makes the inverse transform $-i$ times that residue. For positive time the three upper residues, at $i,ia,ib$ respectively, areMultiplication of their sum by $i$ gives the bilateral exponential forcing of an overdamped oscillator:Both expressions have $y(0)=y^{\prime }(0)=1/[(a+1)(b+1)]$. This verifies the matching that prevents an extra Dirac delta function at zero. Direct substitution gives the forcing on either side, and all terms decay at their corresponding infinities. A causal Green function does not imply zero response before $t=0$ when the forcing has already been acting at negative times. The assumptions $a,b\ne 1$ avoid repeated poles; their limiting cases would contain polynomial times exponential terms.

For $\epsilon \ne 0$, use $z=e^{i\theta }$. The interval traverses $|z|=1$ once counterclockwise, $\cos \theta =(z+z^{-1})/2$, and $d\theta =dz/(iz)$. ThusPut $d=\sqrt{1-{\epsilon }^2}>0$. The two simple poles areTheir moduli are $|\epsilon |/(1+d)<1$ and $(1+d)/|\epsilon |>1$, so exactly $z_{+}$ is enclosed, for either sign of $\epsilon$. Its residue isThe residue theorem yieldsThis also follows from the reciprocal affine sine integral after an angular shift. At $\epsilon =0$, the original integral is directly $2\pi$, agreeing with the boxed formula. The displayed two-pole formulas require $\epsilon \ne 0$: at zero the contour integrand reduces to $1/(iz)$ with a single pole. Thus the source's claim of two simple poles has this removable special-case exception.

The integrand has simple poles at $z=ia$ and $z=-ia$. Only the first is inside the upper contour, and its residue isThe residue theorem therefore gives, for every $R>a$,On the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^2+a^2|\ge R^2-a^2$. The estimation lemma bounds the arc contribution by $\pi R/(R^2-a^2)\to 0$. Hence the real-axis integral has the same limit, and taking its real part givesThe imaginary part vanishes independently because $\sin x/(a^2+x^2)$ is odd. The real integral converges absolutely, since the denominator grows quadratically.

The only possible singularity of $f(z)/(z-z_0)$ inside the contour is at $z_0$, with residue $f(z_0)$. It is at most a simple pole; when $f(z_0)=0$, it is removable. Thus the residue theorem gives the Cauchy integral formula,for a counterclockwise simple contour containing $z_0$ once, with $f$ analytic on a neighbourhood of the contour and its interior. More generally the answer is $2\pi i\mathrm{Ind}(C,z_0)f(z_0)$, where the winding number accounts for multiple turns or reversed orientation. A clockwise simple contour has the opposite sign.

Use the principal branch of the complex logarithm for $\mathrm{Log}(z+i)$. Its branch cut is the ray $z=-i-s$, $s\ge 0$, lying below the real axis. Thus it is analytic in the upper half-plane, where the only pole of the integrand is $z=i$.

Close the real interval by a large upper semicircle. The estimation lemma bounds the arc integral by $O(\ln R/R)$, so it vanishes. The simple-pole residue isThe residue theorem then givesFor real $x$, $\mathrm{Re}\mathrm{Log}(x+i)=\frac{1}{2}\ln (x^2+1)$. Taking real parts and then using evenness of the real integrand yieldsBoth endpoints converge: the integrand is $O(x^2)$ near zero and $O(\ln x/x^2)$ at infinity. This also agrees with the logarithmic parameter integral with Cauchy weight at parameter one.

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.

Close the interval $[-R,R]$ with an upper semicircle, traversed counterclockwise. On its large arc,so the estimation lemma bounds the arc integral by $\pi R^3/(R^4-1)\to 0$. The enclosed residues are those at $(1+i)/\sqrt{2}$ and $(-1+i)/\sqrt{2}$, whose sum is $-i/(2\sqrt{2})$. The residue theorem therefore givesThe positive sign agrees with the nonnegative real integrand.

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.

Let $f$ be meromorphic on and inside a positively oriented simple closed contour $C$, with finitely many isolated poles $z_k$ inside and none on $C$. The residue theorem statesThe positive orientation is counterclockwise. A clockwise contour reverses the sign, and the residue is the coefficient of $(z-z_k{)}^{-1}$ in the local Laurent series.

Substitution givesThe inverse Fourier transform is $(2\pi {)}^{-1}{\int }_{\mathbb{R}}\tilde{y}(\omega )e^{i\omega t}d\omega$. For $t<0$, close clockwise in the lower half-plane, where there are no poles; hence $y(t)=0$.

For $t>0$, close counterclockwise in the upper half-plane. The three simple-pole residues of the full inversion integrand areMultiplying their sum by $i$ and combining the conjugate exponential contributions yieldsThe large arcs vanish because the rational factor is $O({\omega }^{-3})$ and the exponential is bounded on the chosen half-plane. At zero, the right-hand value and first derivative are both zero, matching the negative-time branch, so no unwanted delta or delta derivative appears in the second-order equation.

For an independent dynamical check, this is the exponential forcing of an underdamped oscillator with damping $\gamma =1$, natural frequency $\sqrt{5}$, damped frequency $2$, exponent $\beta =2$ and force amplitude $-1$. For positive time its substitution gives $y^{\prime \prime }+2y^{\prime }+5y=-e^{-2t}$.

Use the branch from part (i) and close the real-axis path with a large upper semicircle, indenting clockwise around $-1$ and $1$. On the segment between them, take the upper-bank value of the branch cut. This is an instance of square-root endpoint integrals by an upper-half-plane contour.

The contributions can be controlled separately. Since $g(z)=O(z^{-2})$, the large arc contributes $O(R^{-1})\to 0$. Near either branch point, $g=O({\epsilon }^{-1/2})$, so an arc of length $O(\epsilon )$ contributes $O(\sqrt{\epsilon })\to 0$. On the upper bank of $(-1,1)$,which is odd in $x$, so its integral is zero. On the exterior real intervals $h(-x)=-h(x)$, while the rational numerator is odd. Hence $g(-x)=g(x)$ and the two exterior contributions sum to $2J$.

The enclosed residues use $h(i)=i\sqrt{2}$ and $h(2i)=i\sqrt{5}$:The residue theorem gives $2J=2\pi i[-i/(3\sqrt{2})-i/(6\sqrt{5})]$. ThereforeAs an independent check, $u=\sqrt{x^2-1}$ converts the integral towhich yields the same value. This check also confirms the branch signs and the factor of two in the contour argument.

The inverse Fourier transform isFor $t>0$, close in the upper half-plane, where $e^{i\omega t}$ decays. The single enclosed pole has residue $-i{e}^{-t}$, so counterclockwise integration gives $f=e^{-t}$. For $t<0$, close clockwise in the lower half-plane; there is no enclosed pole and the result is zero. ThusHere $H$ is the Heaviside step function. At the jump, symmetric Fourier inversion gives $f(0)=1/2$, the average of the one-sided limits. The transform itself cannot fix a function’s value at a single point, so this is the conventional representative rather than extra frequency information.

A direct check is ${\int }_0^{\infty }{e}^{-(1+i\omega )t}dt=(1+i\omega {)}^{-1}=-i/(\omega -i)$. Distributionally, $f^{\prime }=\delta (t)-e^{-t}H(t)$.

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.