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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:

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 residue theorem states that a positively oriented simple closed contour enclosing isolated poles and none on the contour has integral $2\pi i$ times the sum of their residues, provided the function is holomorphic elsewhere inside. Near a simple zero of $\psi$, $\psi (z)={\psi }^{\prime }(a)(z-a)+O((z-a{)}^2)$. HenceFor $\cot z=\cos z/\sin z$, this gives $\boxed{{\mathrm{Res}}_{z=k\pi }\cot z=1}$ for every integer $k$.

To evaluate the real integral, integrate $F(z)=\mathrm{Log}z/(z^2+a^2)$ around an upper semicircle, with a small upper semicircular indentation excluding zero. Use the branch real on the positive axis and with argument between zero and $\pi$ in the upper half-plane. The large arc tends to zero as $O(\ln R/R)$ and the small arc as $O(ϵ|\ln ϵ|)$. The only enclosed pole is $ia$, whose contribution isOn the negative real axis the upper boundary value is $\mathrm{Log}(-x)=\ln x+i\pi$. The two real segments therefore contributeComparing real parts yieldsThe integral is absolutely convergent at both zero and infinity, so no principal-value interpretation is needed.

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

Split the Fourier transform integral at zero:By Fourier inversion applied to this pair, or by the residue theorem, the other transform isFor example, inversion with $\lambda =\mu$ gives $\int e^{ikx}(k^2+{\mu }^2{)}^{-1}dk=(\pi /\mu )e^{-\mu |x|}$; evenness supplies the desired sign convention.

Take the spatial Fourier transform of the Laplace equation. For $y>0$ it gives ${\hat{\psi }}_{yy}-k^2\hat{\psi }=0$. Removing the exponentially growing solution and matching the boundary givesAt $k=0$ this formula is interpreted by continuity; a single transform value does not alter the inverse integral. The previous transform pair identifiesthe Poisson kernel for the upper half-plane. The convolution theorem therefore gives the Poisson integralThis is the bounded harmonic solution for bounded decaying data, or the corresponding solution in an appropriate Fourier class. Such a growth condition matters if vertical decay is interpreted only pointwise: the nonzero harmonic function $e^{x^2-y^2}\sin (2xy)=\mathrm{Im}{e}^{(x+iy{)}^2}$ vanishes on the real axis and tends to zero for fixed $x$ as $y\to \infty$, but grows rapidly in $x$ and has no ordinary spatial Fourier transform. It can be excluded by requiring boundedness; the stated pointwise limits alone would not establish uniqueness.

For the interval data, direct integration gives the Poisson extension of interval boundary dataAs $y\downarrow 0$ it approaches $c$ for $|x|<a$, zero for $|x|>a$, and $c/2$ at $x=\pm a$. At those jump points the prescribed isolated values cannot be attained by a continuous extension; they do not affect the integral. The solution decays vertically, with $\psi \sim 2ac/(\pi y)$ for fixed $x$ and large $y$.

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 $\beta \ge 2$.

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.

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.

Choose a branch of the complex logarithm with $-\pi /2<\arg z<3\pi /2$, placing its branch cut down the negative imaginary axis. Thus $\mathrm{Log}x=\ln x$ for $x>0$, while $\mathrm{Log}(-x)=\ln x+i\pi$ on the negative real axis for $x>0$.

Integrate $\mathrm{Log}z/(z^2+a^2)$ around an upper semicircle, with a small clockwise semicircular indentation above zero. The large arc contributes $O(\ln R/R)\to 0$, and the small arc contributes $O(\epsilon |\ln \epsilon |)\to 0$. The only enclosed pole is $ia$, withWriting $I(a)$ for the required integral, the negative real segment contributes $I(a)+i\pi {\int }_0^{\infty }dx/(x^2+a^2)$. Hence the residue theorem givesThe imaginary terms cancel, leavingThe integral converges at both endpoints because $|\ln x|$ is integrable near zero and $\ln x/x^2$ is integrable at infinity. As a real-variable check, scaling $x=au$ reduces the result to the vanishing first logarithmic moment at $a=1$, one of the logarithmic moments of the Cauchy kernel.

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.