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An analytic continuation extends a holomorphic function beyond its original domain without changing its existing values. Formally, if $U\subset V$ are domains and $f$ is holomorphic on $U$, a holomorphic function $F$ on $V$ with $F{|}_U=f$ is an analytic continuation. More locally, holomorphic functions on overlapping open sets continue one another when they agree on the overlap, allowing a chain of patches along a path. The identity theorem for holomorphic functions forces agreement throughout a connected overlap once they agree on a nonempty open set in it. It also makes an extension to a fixed connected domain unique. ^[ref]

The geometric series illustrates why a formula can reach farther than the power series that first represents it. For $z\ne 1$, its partial sum isWhen $|z|<1$, $z^{N+1}\to 0$, so the series converges to $f(z)=1/(1-z)$. The rational function $F(z)=1/(1-z)$ is holomorphic on $\mathbb{C}\setminus \{1\}$ and agrees there with the sum on the unit disc. It therefore continues $f$ to every point except its simple pole at $1$. In particular,Thus analytic continuation gives a value at $2$ without making the original geometric series converge there. Its radius of convergence remains $1$: on $|z|\ge 1$ the terms $z^n$ do not even tend to zero.

The pole at $1$ limits the original Taylor series, although most points of its convergence circle are regular for $F$. Recentring at $-1$ gives another genuinely convergent power series,At $z=-3/2$ this yields $\frac{1}{2}{\sum }_{n\ge 0}(-1/4{)}^n=2/5$, even though the original powers of $-3/2$ grow. This is the useful idea: overlapping local descriptions can represent one holomorphic function on a larger region.

Uniqueness on a connected overlap does not imply that every path leads to the same terminal germ of a holomorphic function. Continuing a complex logarithm around $0$ changes its value by $2\pi i$, giving monodromy; the return patch can carry a different branch of a multivalued function. If two patches have a disconnected overlap, agreement on one component imposes no agreement on the others. ^[ref] Continuation using the rational function $1/(1-z)$ is single-valued: any path avoiding $1$ gives the same terminal germ of a holomorphic function, even if it winds around the pole.

Removing factorial growth from the coefficients of a formal inverse-power series produces its Borel transform. When continuation and exponential growth bounds allow it, ${\int }_{\mathcal{C}}{e}^{-zt}\mathcal{B}(t)dt$ reconstructs the series by Laplace transform. A singularity on the integration direction requires a lateral contour prescription. A simple pole contributes a residue when that prescription is changed, explaining how Stokes phenomenon arises from otherwise identical algebraic expansions.

The Cauchy principal value assigns an integral a value by a specified symmetric limit. Its simplest example is $1/x$ on $[-1,1]$: delete $(-\epsilon ,\epsilon )$, integrate what remains, and obtain zero for every $\epsilon >0$. The positive signed area under a graph cancels the negative one, although each grows without bound as the gap closes. This is a useful cancellation rule, but it does not make the ordinary improper integral exist.

Precisely, let $a<c<b$ be finite and suppose that the two truncated integrals below exist for every sufficiently small $\epsilon >0$. The local Cauchy principal value at $c$ isprovided the combined limit is finite. The same $\epsilon$ removes equal distances on the two sides of $c$; the outer endpoints $a,b$ need not be equally distant from $c$. By contrast, the ordinary improper integral requires both one-sided limits to be finite separately before adding them. Whenever those separate limits exist, the Cauchy principal value agrees with their sum.^[ref]

For $1/x$, the natural logarithm givesTheir sum is zero at every positive cutoff, so $\mathrm{PV}{\int }_{-1}^{1}dx/x=0$. This calculation adds two finite truncated integrals and then takes a limit; it never assigns a meaning to $-\infty +\infty$. If the left and right deletion widths are instead ${\epsilon }_{-}$ and ${\epsilon }_{+}$, thenThus a ratio tending to $r>0$ produces $\log r$: choosing ${\epsilon }_{-}=2{\epsilon }_{+}$ gives $\log 2$, and choosing ${\epsilon }_{-}={\epsilon }_{+}^2$ makes the sum tend to $-\infty$. Both widths can tend to zero without selecting a unique answer. The local Cauchy principal value selects the ratio $1$.

Unequal outer endpoints leave a finite contribution even with symmetric local deletion. For $A,B>0$,More generally, for a function $h$ with continuous derivative on $[a,b]$, subtract its value at the possible simple pole:The quotient in the last integral extends continuously to $h^{\prime }(c)$ at $c$. This formula explains both the cancellation and a practical way to compute it: integrate a continuous remainder instead of subtracting two large, nearly equal divergent contributions.

A finite Cauchy principal value is weaker than absolute convergence. Indeed,Taking the absolute value destroys the cancellation. Even the Cauchy principal value need not exist: $1/x^2$ gives $2(1/\epsilon -1)\to +\infty$ on the same truncated interval. Absolute integrability makes the cutoff prescription unnecessary; an ordinary improper integral can also converge without absolute convergence, as the Dirichlet integral illustrates. The three notions should therefore be checked separately.

There is an independent Cauchy principal value at infinity. For a function integrable on every finite interval, definewhen the limit is finite. The ordinary improper integral instead requires convergence of the two tails separately.^[ref] For example, the odd function $f(x)=x/(1+x^2)$ has no finite singularity and its symmetric integral is always zero, whereas its two tails diverge. Unequal truncations giveLetting $B/A\to r>0$ as $A,B\to \infty$ yields $\log r$. Symmetry at infinity is therefore a separate choice from symmetry near a finite pole. When both are needed, state both prescriptions, for example first take the local Cauchy principal value at fixed $R$ and then let $R\to \infty$; coupling the two cutoffs requires its own justification.

The local cancellation also defines the principal-value reciprocal distribution. Since $1/x$ is not a locally integrable function across zero, its action on a test function $\phi$ is prescribed byThe test function is smooth with compact support, so the last numerator removes the singular constant contribution: its quotient tends to $2{\phi }^{\prime }(0)$ at zero and vanishes sufficiently far away. This defines a distribution, not a finite pointwise value of $1/x$ at zero. It extends to Schwartz functions as a tempered distribution, and is also the distributional derivative of the logarithmic modulus.^[ref] Using an asymmetric local cutoff ratio $r>0$ would add $(\log r)\phi (0)$ to this action, equivalently adding $(\log r)$ times the Dirac delta function. Thus the cutoff choice matters even after passing to distributions.

Why keep this prescription? A Fourier transform naturally encounters such a kernel. With $\hat{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$, pairing $x$ and $-x$ gives, for $k\ne 0$ and $0<\epsilon <R$,The cosine contribution is an odd function and cancels locally; the sine contribution has a removable singularity and its tail converges by the Dirichlet integral. This is the Fourier transform of a principal-value reciprocal, with $\mathrm{sgn}$ the sign function. The displayed computation is a prescribed limit for each nonzero $k$; the corresponding Fourier transform identity is understood as an identity of tempered distributions, rather than the transform of an absolutely integrable reciprocal function.

The Hilbert transform uses exactly this principal-value reciprocal distribution as a convolution kernel. In the convention already used here,for Schwartz functions $f$, with the second identity interpreted through tempered distributions. Hence the singular integral has a simple description as a Fourier multiplier.^[ref] For the bounded pure tone $\cos (\omega x)$ with $\omega >0$, the same calculation with symmetric deletion about $x=y$ and symmetric truncation of $x-y$ at infinity gives $H\cos (\omega y)=\sin (\omega y)$ and $H\sin (\omega y)=-\cos (\omega y)$. These tones lie outside the Schwartz space; their prescribed Cauchy principal values explain the quarter-cycle shift in the figure. Both figures are calculated mathematical illustrations. Reversing the Hilbert denominator reverses these signs; NIST uses that opposite convention.^[ref]

Complex analysis studies functions of a complex number, especially holomorphic functions. Its distinctive feature is that a single complex derivative controls variation in every direction of the complex plane. The map $f(z)=z^2$ gives a first picture: straight coordinate lines become curved, yet small perpendicular arrows remain perpendicular wherever $f^{\prime }(z)\ne 0$.

The bottom panels show displacements rather than absolute positions. Multiplication by $f^{\prime }(1+i)=2+2i$ predicts the image of each small arrow; the exact image differs by $h^2$. The origin is exceptional because $f^{\prime }(0)=0$. This distinction between local conformal maps and a critical point is central: complex differentiability imposes strong geometric structure, without making every holomorphic function globally one-to-one.

The remarkable next step is that holomorphic functions are determined inside a circle by their values on its boundary. Precisely, let $R>0$, let $D=\{z:|z-a|<R\}$, and suppose $f$ is a holomorphic function on an open set containing the closed disc $\overline{D}$. Orient its boundary counterclockwise. The Cauchy integral formula states that, for every $w\in D$,The contour integral combines boundary values with a kernel depending on the requested interior point. It is a reconstruction formula, not merely a trick for evaluating an integral. The hypothesis about the closed disc ensures that all boundary values and the contour integral are well defined. ^[ref]

At the centre, substituting $z=a+R{e}^{i\theta }$ cancels the denominator against $dz=iR{e}^{i\theta }d\theta$:Thus the centre value is the angular average of the boundary values. For the polynomial $f(z)=z^2$, $a=1+i$ and $R=1/2$, expand the boundary data asBoth oscillating terms have zero average, leaving $f(a)=2i$. Away from the centre the weights change: on the unit circle,With $f(z)=z^2$ and $w=1/2$, expanding the denominator as a geometric series leaves only its second-power term after integration, giving $f(1/2)=1/4$. This also explains how the Cauchy integral formula recovers an entire Taylor series from boundary data.

When a holomorphic function has isolated singularities, a different boundary calculation detects their residues. Suppose $\Gamma$ is a counterclockwise simple closed contour and $f$ is holomorphic on a neighbourhood of it and its interior except at finitely many interior isolated singularities $a_j$. The residue theorem givesA residue is the coefficient of $(z-a_j{)}^{-1}$ in the Laurent series. A short proof idea is to deform the contour to small circles around the isolated singularities using the Cauchy integral theorem. Integrating the Laurent series on each circle kills every power except $-1$. ^[ref]

Here is how a complex pole determines a real improper integral. Set $g(z)=1/(1+z^2)$ and use the contour consisting of the segment from $-R$ to $R$, followed by the upper semicircle $A_R$, with $R>1$. Only the simple pole $i$ lies inside, andThe arc must actually disappear. On $A_R$, the reverse triangle inequality gives $|1+z^2|\ge R^2-1$, and its length is $\pi R$, soThe real integrand is bounded by $x^{-2}$ for $|x|\ge 1$, so both tails converge separately. Taking the limit therefore proves the ordinary improper integralFor example, at $R=10$ the real segment contributes $2\arctan (10)\approx 2.942255$, and the arc contributes $0.199337$, below the bound $10\pi /99\approx 0.317333$. The usefulness of the residue theorem is that the same strategy works for many rational functions whose real antiderivatives are cumbersome: find the enclosed poles, compute their residues, and control the connecting arc. ^[ref]

Historically, Cauchy's 1825 memoir on integrals with imaginary limits develops complex contour integration and its use in transforming and evaluating definite integrals. The opening sections define complex path integrals, discuss changing the path, and examine what happens near singular points. These are recognizable ancestors of the modern link between holomorphic functions, contours and residues. ^[ref]

The ratio of modular discriminants has weight zero and level ${\Gamma }_0(2)$. It has one simple pole at infinity and one simple zero at the zero modular cusp, with no zeros or poles in the half-plane. It is a spherical coordinate by the single-pole criterion for a spherical coordinate. The Klein j-invariant is $(t+256{)}^3/t^2$, with $t(e^{2\pi i/3})=-256$.

For real $|a|<\pi$, integrate $g(z)=e^{az}/\cosh \pi z$ around a rectangle of height $i$ and width $2R$. Only the simple pole $i/2$ lies inside, with residue $e^{ia/2}/(\pi i)$. Since $g(x+i)=-e^{ia}g(x)$, the reversed top edge adds $e^{ia}$ times the bottom integral. The two vertical edges vanish as $R\to \infty$, using $|a|<\pi$. The residue theorem gives $(1+e^{ia})J=2e^{ia/2}$, proving the formula. The condition is also the exact convergence range: the two tails decay with exponents $\pi -a$ and $\pi +a$.

If $f$ has a pole of positive order $m$ at $a$, then $h=e^f$ has an essential singularity there. The function $h$ is holomorphic and nonzero on a sufficiently small punctured disk.

Suppose instead that $h$ had a removable singularity or a pole. A nonzero meromorphic function then has the local form $h(z)=(z-a{)}^{k}v(z)$ with integer $k$ and $v$ holomorphic and nonzero at $a$. Its logarithmic derivative iswhich has at most a simple pole. But $h\prime /h=f\prime$, and differentiating the leading term of the pole of $f$ gives a pole of order $m+1\ge 2$. This is a contradiction. The classification of isolated singularities leaves only the essential case.

For example, $e^{1/z}=1+z^{-1}+z^{-2}/2!+\cdots$ has infinitely many terms in its principal part. The logarithmic-derivative argument also covers a pole with several principal coefficients and avoids assuming that an informal exponential expansion cannot cancel negative powers.

A first-order low-pass filter with positive constant gain $K$ and time constant $\tau >0$ obeys the ordinary differential equationThe derivative term limits how quickly the plant output follows a changing plant input. A larger time constant means slower adjustment. This standard form, with $K=1$, is developed in Derek Rowell, MIT 2.004, lecture 20, section 1.1: first-order systems; the following calculation keeps the gain $K$ explicit.

Write $y_0=y(0)$. The Laplace transform of the derivative is $sY(s)-y_0$, soSetting the initial condition $y_0=0$ gives the transfer function $G(s)=K/(\tau s+1)$. Its single simple pole is $s=-1/\tau$, which has negative real part. The unforced response decays, and the input/output map has BIBO stability. The eventual gain for a constant plant input is $G(0)=K$.

For a unit step, represented by the Heaviside step function $u(t)=H(t)$, the Laplace transform is $U(s)=1/s$. From zero initial condition,Take $K=2$ and $\tau =3\mathrm{s}$, with dimensionless plant input and plant output. Then $y(t)=2(1-e^{-t/(3\mathrm{s})})$ rises toward $2$. After one time constant, at $t=3\mathrm{s}$, it reaches $2(1-e^{-1})\approx 1.2642$, or $63.2\%$ of its final value. At $t=9\mathrm{s}=3\tau$, it reaches $2(1-e^{-3})\approx 1.9004$, or $95.0\%$. These percentages describe the rise from zero, rather than an instantaneous jump to the target.

With a nonzero initial condition, the same step instead givesFor $K=2$ and $\tau =3\mathrm{s}$, compare $y_0=0$ with $y_0=1$. Their plant outputs differ by $e^{-t/(3\mathrm{s})}$ despite having the same transfer function and the same plant input. The extra exponential decay term must be supplied from the initial condition; $G$ and $u$ alone cannot determine it.

For real $k$, integrate $e^{-ikz}/\cosh z$ around a rectangle of height $\pi$. Its top-edge value is $-e^{\pi k}$ times the bottom-edge value, and the only enclosed simple pole is $i\pi /2$, with residue $e^{\pi k/2}/i$. The vertical-edge integrals vanish as their real parts tend to positive and negative infinity. The residue theorem gives $(1+e^{\pi k})\tilde{f}(k)=2\pi e^{\pi k/2}$, proving the stated Fourier transform identity. This is an example of a function reproducing its shape under a Fourier transform, up to rescaling.

In a fixed frame, a first-order linear differential equation is Fuchsian at a finite point when its coefficient matrix has at most a simple pole there. Its residue is $R$. The global object is the matrix-valued one-form $A(z)dz$: in the coordinate $t=1/z$ its coefficient is $-t^{-2}A(1/t)$. A Fuchsian system has intrinsic regular singular points, but an intrinsically regular singular system may have higher-order coefficient poles in a different meromorphic frame.

The fundamental strip of a Mellin transform is the interior of the set of real parts $\sigma$ for whichViewed in the complex plane, it is a vertical strip $\alpha <\mathrm{Re}s<\beta$; either bound may be infinite, and the interior may be empty. Both endpoints of the improper integral must be checked. Convergence on a boundary line, whether absolute convergence or conditional convergence, requires a separate test. The Mellin transform is holomorphic throughout a nonempty fundamental strip.^[ref]

A practical sufficient test uses Big O notation: if $f(x)=O(x^p)$ near zero and $f(x)=O(x^{-q})$ near infinity, with real $p,q$, then comparison with elementary improper integrals givesThus $-p<\sigma <q$ guarantees absolute convergence. These upper bounds on $|f|$ need not locate the maximal Mellin fundamental strip; faster actual decay can widen it. The strip property itself follows by bounding the weight at any intermediate $\sigma$ by the sum of two weights already known to be integrable. On a closed narrower strip, neighboring integrable weights also dominate the natural logarithm factors introduced by differentiation under the integral sign, making the transform a holomorphic function.

An exact two-endpoint example isHere both bounds are sharp: at $\sigma =-2$ the zero-end absolute convergence test contains ${\int }_0^{1}dx/x$, and at $\sigma =3$ the infinity-end test contains ${\int }_1^{\infty }dx/x$. The displayed rational function continues meromorphically outside the Mellin fundamental strip and has simple poles at $-2$ and $3$. Its other finite continued values outside the strip are values of a meromorphic continuation, not of a convergent defining integral. More general endpoint information leads to meromorphic continuation of a Mellin transform from an asymptotic expansion.

The Gamma function extends factorials to a continuous argument. Its defining Gamma integral iswhere $t^{z-1}=\exp ((z-1)\log t)$ uses the real natural logarithm for $t>0$. The Gamma function recurrence gives its meromorphic continuation to the complex plane, with simple poles at $0,-1,-2,\dots$ and no zeros. See Gamma function residue at a nonpositive integer and Gamma function has no zeros for proof outlines.^[ref]

For nonnegative integers $n$, the factorial identity has a shift of one:Thus $\Gamma (n)=(n-1)!$ for positive integers $n$. Half-integer arguments supply a different familiar constant:The first value follows from the Gaussian integral; the second follows from the Gamma function recurrence.^[ref]

The plot shows only positive real arguments; the gamma function also accepts complex numbers away from its poles. Matching the factorials alone does not select a unique interpolation, even if the Gamma function recurrence is imposed: $f(x)=\Gamma (x)\exp (\sin (2\pi x))$ has the same positive-integer values and the same recurrence relation. The Bohr-Mollerup theorem supplies the extra condition that selects the gamma function.

The useful idea is to replace a discrete product of numbers by an integral whose exponent varies continuously. The Gamma integral then evaluates whole families of integrals at once, provides the normalizing constant for the Gamma distribution in probability theory, and appears in physics, for example in the Bose integral. Its large-positive-argument growth reconnects this continuous description to factorials through Stirling's formula.^[ref]

Historically, Leonhard Euler studied the interpolation of factorials in his paper E19, written in 1729 and published in 1738. His integral over $(0,1)$ becomes the Gamma integral after a change of variables, with the same one-step shift in the argument.^[ref]

The Gamma function obeys the recurrence relationInitially, for $\mathrm{Re}z>0$, integration by parts in the Gamma integral gives the short proofThe boundary term vanishes at zero because $|t^z|=t^{\mathrm{Re}z}\to 0$, and at infinity because of exponential decay. Together with $\Gamma (1)={\int }_0^{\infty }{e}^{-t}dt=1$, the Gamma function recurrence immediately gives $\Gamma (n+1)=n!$ for every nonnegative integer $n$.^[ref]

Read backwards, the Gamma function recurrence defineswhere $m$ is a positive integer. These overlapping formulas agree by the Gamma function recurrence, giving the meromorphic continuation. Since $\Gamma (1)=1$, they give a simple pole at zero and then at every negative integer; Gamma function residue at a nonpositive integer records the residues. The recurrence relation therefore holds throughout the continuation as an identity of meromorphic functions.

The Gamma function recurrence and $\Gamma (z)\sim 1/z$ at zero imply a simple pole at every $z=-n$, $n\ge 0$, with residue $(-1{)}^n/n!$.

Suppose a meromorphic function is holomorphic on two concentric annuli about $a$, with only finitely many poles between chosen inner and outer coefficient circles. Let $a_n,b_n$ denote the corresponding Laurent coefficients. Subtract their contour integral formulas and apply the residue theorem to the intervening annular region. It givesThe sum is over the poles crossed, including their multiplicities through the residue. This identity applies to every integer index. Thus changing the annulus can change even the coefficients of nonnegative powers; it is not merely a change in the principal part.

For example, $f(z)=1/\sin z$ has simple poles at $\pm \pi$, each with residue $-1$. Moving a coefficient circle centered at zero from radius below $\pi$ to radius between $\pi$ and $2\pi$ changes its $n$th coefficient by $-{\pi }^{-n-1}-(-\pi {)}^{-n-1}$. Even indices remain unchanged, while odd indices acquire a nonzero correction.

For real $\beta >1$, choose a branch of the complex logarithm whose argument interval contains $[0,2\pi /\beta ]$ strictly, and integrate $\mathrm{Log}z/(1+z^{\beta })$ over an annular sector of angle $\alpha =2\pi /\beta$. The one enclosed simple pole is $e^{i\pi /\beta }$, with residue $-i\pi e^{i\pi /\beta }/{\beta }^2$. The circular arcs tend to zero, with bounds $O(ϵ|\log ϵ|)$ and $O(R^{1-\beta }\log R)$.

Writing $I$ for the title integral and $J={\int }_0^{\infty }(1+x^{\beta }{)}^{-1}dx$, the two radial segments and the residue theorem giveDivide by $e^{i\pi /\beta }$ and equate real and imaginary parts. This gives $J=\pi /[\beta \sin (\pi /\beta )]$ and the title formula. When $1<\beta <2$, the sector extends below the negative real axis; the principal complex logarithm therefore cannot be used throughout that sector. A cut along a ray just below the positive axis avoids this issue for every $\beta >1$.

The Mellin transform is an integral transform suited to changes of scale: replacing $f(x)$ by $f(ax)$ multiplies its transform by $a^{-s}$ for $a>0$. For a locally integrable function $f:(0,\infty )\to \mathbb{C}$, definewhere the improper integral converges. The complex number $x^{s-1}=\exp ((s-1)\log x)$ uses the real natural logarithm because $x>0$. Its absolute value is $x^{\sigma -1}$, so absolute convergence depends on the real part $\sigma$ rather than the imaginary part $\tau$. The Mellin fundamental strip records the open range of such real parts.^[ref]

A familiar example is the exponential function $f(x)=e^{-x}$:This is exactly the Gamma integral. Near zero $e^{-x}$ tends to one, so absolute convergence requires $\sigma >0$; at infinity $x^r{e}^{-x}\to 0$ for every real $r$, so there is no upper boundary. The Gamma function has a meromorphic continuation with simple poles at $0,-1,-2,\dots$, but a finite continued value elsewhere in $\sigma \le 0$ does not make the defining improper integral converge.^[ref]

The Mellin transform of an interval indicator gives an elementary example with absolute convergence for every complex number $s$. The Mellin transform in logarithmic coordinates explains why changes of scale become translations, connecting the scaling property of the Mellin transform to the Fourier transform.

Optimal truncation chooses how many terms to retain for a useful finite approximation. In a divergent asymptotic expansion, increasing the truncation order eventually makes the terms larger rather than smaller. Stopping near a smallest term often gives the best direct approximation, but the actual remainder must be checked. A small last term alone is not a universal error bound.^[ref]

For a divergent factorial series can approximate a convergent integral, $T_k=k!/z^k$ decreases while $k+1<z$ and increases while $k+1>z$. Thus the useful cutoff is near $k=z$, rather than at arbitrarily large order. At $z=10$, $T_9=T_{10}=0.00036288$ are the two smallest terms. Retaining $N=10$ terms givesThe actual error is about $0.0001770$. Retaining twenty terms instead gives $S_{20}(10)\approx 0.9076137084$, with error about $0.008020$: more work produces a worse answer.

For this example the bound $N!/z^N$ is explicit, so a requested tolerance can be tested before using a truncation. For a different integral, a nearby saddle point or simple pole can change the relevant remainder scale. This is why the smallest-term rule is a useful guide, rather than a replacement for remainder analysis.

The cutoff for factorial terms $n!z^{-n}$ has $n=|z|+O(1)$. Better analysis can expose contributions smaller than every retained inverse power, linking optimal truncation to the Stokes phenomenon. Such contributions can be important even when many asymptotic coefficients agree. Directly summing ever more terms does not automatically recover them.

At any $z\ne 0$, take sufficiently small $h$ so $z+h\ne 0$. The complex difference quotient isHence the complex derivative exists at every point of the punctured plane, andThe excluded point is essential to the domain statement: the reciprocal has a simple pole there.

Use the branch from part (b) and close the real interval $[-R,R]$ with the upper semicircle, oriented counterclockwise. For $\mathrm{Im}z\ge 0$, the number $z+i\alpha$ has positive imaginary part, so the numerator is a holomorphic logarithm throughout a neighbourhood of the contour and its interior. The branch point $-i\alpha$ lies strictly below the contour. The only enclosed pole ofis the simple pole at $i\beta$, with residueTo justify discarding the arc, take $R>\max \{\beta ,\alpha +1\}$. On it, $|z+i\alpha |$ lies between $R-\alpha$ and $R+\alpha$, and the argument has bounded absolute value. HenceThe real-axis integral is absolutely convergent, since its integrand is $O(\log |x|/x^2)$ at infinity and has no real singularities. The residue theorem therefore givesFinally, $\mathrm{Re}\mathrm{Log}(x+i\alpha )=\frac{1}{2}\log (x^2+{\alpha }^2)$. Taking twice the real part yieldsThe branch choice affects the first integral's imaginary part; the second integral is real and independent of that choice.