Real Integrals by Residues & Jordan's Lemma

Derivation

Real Integrals by Residues & Jordan's Lemma

D-184 Home PU-205 Threads waves Depends on The Residue Theorem, Branch Points, Cuts & Riemann Surfaces

Statement

For a function \(f\) meromorphic in the upper half-plane with finitely many poles, none on the real axis, we establish two reduction techniques. First, Jordan's lemma: if \(f(z)\to 0\) uniformly in argument as \(|z|\to\infty\) for \(0\le\arg z\le\pi\), then for every \(a>0\) the integral of \(e^{iaz}f(z)\) over the upper semicircular arc \(C_R\) of radius \(R\) vanishes as \(R\to\infty\). Second, closing a real integral with such an arc (or, when a branch point sits at the origin, a keyhole) turns it into \(2\pi i\) times a sum of enclosed residues: \[\int_{-\infty}^{\infty} e^{iax}f(x)\,dx = 2\pi i \sum_{\mathrm{Im}\,z_k>0}\operatorname{Res}_{z=z_k}\!\big[e^{iaz}f(z)\big].\]

Why it matters

A vast class of definite real integrals that resist elementary antiderivatives — oscillatory Fourier integrals \(\int \cos(ax)/(x^2+b^2)\,dx\), rational integrals over the whole line, and integrals with algebraic branch points like \(\int_0^\infty x^{s-1}/(1+x)\,dx\) — collapse to a finite algebraic sum once the correct contour is chosen. The residue theorem supplies the machinery; Jordan's lemma and the keyhole supply the crucial estimates that license discarding the added arcs, which is exactly the step where naive attempts fail.

The techniques are the computational backbone of Fourier and Laplace inversion, dispersion relations, Green's functions in scattering theory, and the evaluation of loop integrals in physics. Mastery here is the difference between "the integral exists" and "the integral equals \(\pi e^{-ab}/b\)."

Assumptions

\(f\) is meromorphic in a neighbourhood of the closed upper half-plane with only finitely many poles there.With infinitely many poles the residue sum need not converge and the contour cannot be closed by a single expanding arc; one must instead use a sequence of contours chosen to avoid the poles.

No poles lie on the contour of integration (the real axis).A pole on the axis makes the integral divergent as written; it must be reinterpreted as a principal value or an \(i\epsilon\)-shifted integral, contributing an extra half-residue (\(\pi i\,\mathrm{Res}\)) from the indenting semicircle.

For Jordan's lemma, \(a>0\) and the arc closes in the upper half-plane (\(\mathrm{Im}\,z\ge 0\)); for \(a<0\) close below.Close in the wrong half-plane and \(|e^{iaz}|=e^{-a\,\mathrm{Im}\,z}\) grows exponentially on the arc, so the arc contribution diverges rather than vanishes and the method gives nonsense.

\(M(R)=\max_{\theta\in[0,\pi]}|f(Re^{i\theta})|\to 0\) as \(R\to\infty\).If \(f\) does not decay (e.g. \(f\equiv 1\)) the arc term survives and the identity fails; Jordan's lemma only needs decay, not the faster \(1/R^{1+\delta}\) that a bare arc estimate would demand.

For the keyhole, the multivalued factor (e.g. \(z^{s-1}\)) has its branch cut placed along the positive real axis and \(0<\mathrm{Re}\,s<1\) controls both endpoints.Wrong cut placement or \(\mathrm{Re}\,s\) outside \((0,1)\) lets the small circle about the origin or the large circle fail to vanish, breaking the cancellation between the two lips of the cut.

Derivation

\[ \oint_{\Gamma_R} g(z)\,dz = 2\pi i \sum_k \operatorname{Res}_{z=z_k} g(z), \qquad \Gamma_R = [-R,R]\cup C_R \]

The residue theorem applied to the closed contour formed by the real segment \([-R,R]\) and the upper arc \(C_R=\{Re^{i\theta}:0\le\theta\le\pi\}\), enclosing all upper-half-plane poles \(z_k\) once \(R\) exceeds \(\max_k|z_k|\). A

\[ \int_{-R}^{R} g(x)\,dx + \int_{C_R} g(z)\,dz = 2\pi i \sum_k \operatorname{Res}_{z=z_k} g(z) \]

Split the closed contour integral into its straight and curved parts. The residue sum is a fixed finite number once \(R\) is large; only the two integrals on the left depend on \(R\). A

\[ g(z)=e^{iaz}f(z), \qquad z=Re^{i\theta}=R\cos\theta + iR\sin\theta, \quad dz=iRe^{i\theta}\,d\theta \]

Specialise to an oscillatory integrand and parametrise the arc. This exposes the exponential \(|e^{iaz}|=e^{-aR\sin\theta}\), the object Jordan's lemma controls. A

\[ \left|\int_{C_R} e^{iaz}f(z)\,dz\right| \le \int_0^\pi e^{-aR\sin\theta}\,|f(Re^{i\theta})|\,R\,d\theta \le M(R)\,R\int_0^\pi e^{-aR\sin\theta}\,d\theta \]

Triangle inequality for integrals plus \(|dz|=R\,d\theta\), then bound \(|f|\) by its arc-maximum \(M(R)=\max_\theta|f(Re^{i\theta})|\). We have pulled the decay of \(f\) out and are left with a pure exponential integral. B

\[ \int_0^\pi e^{-aR\sin\theta}\,d\theta = 2\int_0^{\pi/2} e^{-aR\sin\theta}\,d\theta, \qquad \sin\theta \ge \frac{2\theta}{\pi}\ \ \text{on}\ \left[0,\tfrac{\pi}{2}\right] \]

Symmetry of \(\sin\theta\) about \(\theta=\pi/2\) halves the range; then apply Jordan's inequality, the concavity bound \(\sin\theta\ge 2\theta/\pi\), which is the sharp linear underestimate of \(\sin\) on the first quadrant. C

\[ 2\int_0^{\pi/2} e^{-aR\sin\theta}\,d\theta \le 2\int_0^{\pi/2} e^{-2aR\theta/\pi}\,d\theta = \frac{\pi}{aR}\left(1-e^{-aR}\right) < \frac{\pi}{aR} \]

Since \(e^{-x}\) is decreasing, the pointwise bound \(\sin\theta\ge 2\theta/\pi\) gives \(e^{-aR\sin\theta}\le e^{-2aR\theta/\pi}\); the elementary integral then evaluates in closed form. This is the crux estimate. C

\[ \left|\int_{C_R} e^{iaz}f(z)\,dz\right| \le M(R)\,R\cdot\frac{\pi}{aR} = \frac{\pi\,M(R)}{a} \xrightarrow[R\to\infty]{} 0 \]

Combine Steps 4 and 6: the two factors of \(R\) cancel, leaving \(\pi M(R)/a\), which tends to zero by the decay assumption \(M(R)\to 0\). This is Jordan's lemma — note it needs only \(M(R)\to 0\), not integrable decay. C

\[ \int_{-\infty}^{\infty} e^{iax}f(x)\,dx = \lim_{R\to\infty}\int_{-R}^{R} e^{iax}f(x)\,dx = 2\pi i \sum_{\mathrm{Im}\,z_k>0}\operatorname{Res}_{z=z_k}\!\big[e^{iaz}f(z)\big] \]

Take \(R\to\infty\) in Step 2: the arc term vanishes by Step 7 and the segment integral converges to the real-line integral, leaving exactly the enclosed residue sum. A

\[ \int_{-\infty}^\infty \cos(ax)\,f(x)\,dx = \mathrm{Re}\!\left[2\pi i \sum_k \operatorname{Res}\big(e^{iaz}f\big)\right],\quad \int_{-\infty}^\infty \sin(ax)\,f(x)\,dx = \mathrm{Im}[\,\cdots\,] \]

Take real and imaginary parts, using \(e^{iax}=\cos ax + i\sin ax\). For real-valued \(f\) this splits the single complex identity into two real integral formulas. A

Result

\[ \int_{-\infty}^{\infty} e^{iax}f(x)\,dx = 2\pi i \sum_{\mathrm{Im}\,z_k>0}\operatorname{Res}_{z=z_k}\!\big[e^{iaz}f(z)\big], \qquad a>0,\ M(R)\to 0 \]

Reading. An oscillatory integral over the entire real line equals \(2\pi i\) times the sum of residues at the poles of \(f\) lying strictly above the axis — provided the oscillation phase \(a>0\) lets us close upward, where \(e^{iaz}\) decays exponentially. Jordan's lemma is precisely the guarantee that the "imaginary" arc we appended contributes nothing, so the answer is entirely local data (residues) even though the integral is global.

Units check. Treat \(x\) as carrying dimension \([x]\); then \(a\) has dimension \([x]^{-1}\) so that \(ax\) is dimensionless inside \(\cos/\sin\). A residue \(\operatorname{Res}_{z_k}[e^{iaz}f]\) carries the dimension of \(f(z)\cdot[z]\) (one power of length is gained from the \(dz\) implicit in the residue), matching the left side \(\int f\,dx\) which has \([f]\cdot[x]\). Both sides scale as \([f][x]\); the factor \(2\pi i\) is dimensionless. Consistent.

Limiting cases

\(a\to 0^+\): \(e^{iaz}\to 1\) and Jordan's bound \(\pi M(R)/a\) blows up — the lemma degrades. One then needs genuine \(1/R^{1+\delta}\) decay of \(f\) itself; the formula still holds for such \(f\) but by the plain arc estimate, not Jordan's lemma.
\(a<0\): close in the lower half-plane; the result becomes \(-2\pi i\sum_{\mathrm{Im}\,z_k<0}\operatorname{Res}[e^{iaz}f]\) (minus sign from clockwise orientation).
Pole on the axis at \(x_0\): indent with a small semicircle; the principal value picks up \(\pi i\,\operatorname{Res}_{x_0}\) (half a residue) if indented into the upper plane.
Even rational \(f\) with no oscillation (\(a=0\)): reduces to the pure rational-function case \(\int_{-\infty}^\infty f = 2\pi i\sum\mathrm{Res}\), valid whenever \(\deg(\text{den})\ge\deg(\text{num})+2\).
Branch point at origin: the semicircle is replaced by the keyhole, and the two lips of the cut differ by the monodromy factor \(e^{2\pi i(s-1)}\), giving \(\int_0^\infty x^{s-1}f\,dx = \dfrac{2\pi i}{1-e^{2\pi i s}}\sum\mathrm{Res}\).

Breaks when

Insufficient or absent decay. If \(M(R)\not\to 0\) (e.g. \(f(z)=z/(z^2+1)\), which \(\to 0\) but only like \(1/R\), fine for Jordan but a plain arc estimate for \(a=0\) fails since \(M(R)R\not\to 0\)). For \(f\equiv 1\) the arc integral is \(O(1)\) and the identity is simply false.
Wrong half-plane for the sign of \(a\). Closing upward with \(a<0\) makes \(|e^{iaz}|=e^{-a\,\mathrm{Im}\,z}=e^{|a|\,\mathrm{Im}\,z}\to\infty\) on the arc; the arc dominates and the "residue sum" is meaningless.
Poles on the contour without regularisation. A simple pole at a real \(x_0\) makes \(\int f\,dx\) divergent; writing down \(2\pi i\sum\mathrm{Res}\) over interior poles silently drops the (finite, principal-value) real answer plus its \(\pi i\,\mathrm{Res}\) correction.
Branch cut crossing the contour. If \(f\) is multivalued and the cut is not routed outside the closed contour (or accounted for by a keyhole), the "enclosed region" is ill-defined and Cauchy's theorem does not apply.
Infinitely many poles inside. For \(f=1/\sinh z\) (poles at \(z=in\pi\)) a single expanding semicircle encloses ever more poles and the arc passes through poles; one must choose radii \(R_N\) threading between poles and check the arc bound survives.

Failure modes

"Closing up" regardless of sign. Students memorise "close in the upper half-plane" and apply it to \(\int \cos(ax)/(x^2+1)\) with \(a<0\), getting exponential garbage instead of \(\pi e^{-|a|}\).
Using \(\cos(az)\) instead of \(e^{iaz}\) on the contour. \(\cos(az)=\tfrac12(e^{iaz}+e^{-iaz})\) contains \(e^{-iaz}\), which grows in the upper half-plane. One must integrate \(e^{iaz}f\) and take the real part at the end, never \(\cos(az)f\) directly.
Forgetting the \(2\pi i\) or its \(i\). Reporting \(2\pi\sum\mathrm{Res}\) or \(\pi i\sum\mathrm{Res}\); the full closed contour gives \(2\pi i\), an axis-indentation gives \(\pi i\).
Sign of the residue at a lower pole. When forced to close downward (\(a<0\)), forgetting the overall minus from clockwise orientation.
Double-counting a pole on the axis. Treating an indented axis pole as both "half inside" and a full interior residue.
Keyhole lip sign error. Writing \(1-e^{2\pi i s}\) as \(e^{2\pi i s}-1\) and flipping the sign of the whole answer.
Applying Jordan to a non-decaying prefactor. \(\int x\sin x/(x^2+1)\) has \(M(R)\sim 1\) — Jordan still applies (needs only \(M\to0\)? here \(M\not\to0\)) — students wrongly assume it must, when in fact convergence is only conditional and needs the oscillation.

Discussion

The deep content of Jordan's lemma is that oscillation buys decay. A bare arc estimate \(\left|\int_{C_R}\right|\le \pi R\,M(R)\) requires \(M(R)=o(1/R)\), i.e. \(f\) must decay faster than \(1/R\). But the exponential \(e^{-aR\sin\theta}\) suppresses the integrand so strongly away from the endpoints \(\theta=0,\pi\) that the effective length of the arc shrinks like \(1/(aR)\), cancelling the factor \(R\) and leaving only \(M(R)\to0\). This is why \(\int_{-\infty}^\infty \sin x/x\,dx=\pi\) converges even though \(1/x\) is not absolutely integrable: the oscillation, not the decay of the amplitude, does the work.

The keyhole contour is the residue calculus meeting the theory of Riemann surfaces. When \(f\) contains \(z^{s-1}\) with non-integer \(s\), the integrand lives not on \(\mathbb{C}\) but on a branched cover; the keyhole is a loop on the base plane whose two straight lips sit on different sheets, so the integrand along them differs by the monodromy factor \(e^{2\pi i(s-1)}\). The seemingly magical cancellation \(\big(1-e^{2\pi i s}\big)\int_0^\infty\) is nothing but the difference of the two sheet values, and the whole computation is a shadow of the branch-point structure catalogued in the prerequisite result.

These contours are the analytic engine of physics. Causality in a Green's function is encoded by which half-plane you close in: the retarded propagator \(G_R(t)\propto\theta(t)\) arises because for \(t>0\) you close below the poles and for \(t<0\) you close above and enclose nothing. Kramers–Kronig relations, the optical theorem, and the analytic continuation of scattering amplitudes are all downstream of exactly the estimate proved in Steps 4–7.

A subtler point: Jordan's lemma proves the arc integral vanishes, but the resulting real integral is generally only conditionally convergent (an improper Riemann integral as a symmetric limit \(R\to\infty\)), not Lebesgue-integrable. The contour method therefore computes the principal-value/symmetric-limit object, and one must separately argue — via, say, Abel summation or an explicit tail bound — that this coincides with whatever notion of the integral the physics demands. Conflating conditional convergence with absolute convergence is the single most common gap in "rigorous" undergraduate treatments.

Common misconceptions. (i) That residues at every pole of \(f\) contribute — only the enclosed ones do, and which are enclosed depends on the closing direction. (ii) That the method requires \(f\) rational — it needs only meromorphy plus the decay hypothesis. (iii) That \(2\pi i\sum\mathrm{Res}\) is always real for a real integrand — it is the real/imaginary part that gives the physical answer, and forgetting to project produces a spurious imaginary result.

Worked examples

Example 1 — A Lorentzian Fourier integral. Evaluate \(\displaystyle I=\int_{-\infty}^{\infty}\frac{\cos(ax)}{x^2+b^2}\,dx\) for \(a>0,\ b>0\).

\[ I = \mathrm{Re}\int_{-\infty}^{\infty}\frac{e^{iax}}{x^2+b^2}\,dx, \qquad f(z)=\frac{1}{z^2+b^2}=\frac{1}{(z-ib)(z+ib)} \]

Replace \(\cos ax\) by \(\mathrm{Re}\,e^{iax}\) and factor the denominator; poles at \(z=\pm ib\), the one at \(+ib\) lies in the upper half-plane. A

\[ M(R)=\max_{\theta}\frac{1}{|R^2e^{2i\theta}+b^2|}\le \frac{1}{R^2-b^2}\to 0, \qquad a>0 \]

Check Jordan's hypothesis: \(f\to0\) uniformly on the arc, and \(a>0\) permits closing upward. Arc contribution vanishes. B

\[ \operatorname{Res}_{z=ib}\frac{e^{iaz}}{(z-ib)(z+ib)} = \frac{e^{ia(ib)}}{2ib} = \frac{e^{-ab}}{2ib} \]

Simple-pole residue: evaluate \(e^{iaz}/(z+ib)\) at \(z=ib\), giving \(e^{-ab}/(2ib)\). Symbols first: \(e^{ia\cdot ib}=e^{-ab}\). A

\[ \int_{-\infty}^\infty \frac{e^{iax}}{x^2+b^2}\,dx = 2\pi i\cdot\frac{e^{-ab}}{2ib} = \frac{\pi}{b}\,e^{-ab} \]

Multiply the residue by \(2\pi i\); the \(2i\) cancels leaving a real result, so \(\mathrm{Re}\) is the whole thing. A

\[ \int_{-\infty}^{\infty}\frac{\cos(ax)}{x^2+b^2}\,dx = \frac{\pi}{b}\,e^{-ab} \]

Reading. The Fourier transform of a Lorentzian is a decaying exponential — the width \(b\) sets both the decay rate \(e^{-ab}\) and the peak height \(\pi/b\). Numerically, with \(a=2,\ b=1\): \(I=\pi e^{-2}\approx 3.1416\times 0.1353\approx 0.4252\).

Units check. \([a]=[x]^{-1}\), \([b]=[x]\), so \(ab\) is dimensionless (good, it is an exponent) and \(\pi/b\) has dimension \([x]^{-1}\), matching \(\int (\text{dimensionless})\,dx/[x]^2 = [x]/[x]^2=[x]^{-1}\). Consistent.

Example 2 — A branch-cut integral by the keyhole. Evaluate \(\displaystyle J=\int_0^{\infty}\frac{x^{s-1}}{1+x}\,dx\) for \(0<s<1\).

\[ \oint_{\text{keyhole}}\frac{z^{s-1}}{1+z}\,dz = 2\pi i\operatorname{Res}_{z=-1}\frac{z^{s-1}}{1+z}, \quad z^{s-1}=e^{(s-1)\log z},\ \arg z\in(0,2\pi) \]

Place the branch cut on the positive real axis; the keyhole encloses the single simple pole at \(z=-1\). Fix the branch \(\arg z\in(0,2\pi)\). A

\[ \text{large circle: } |z^{s-1}/(1+z)|\cdot 2\pi R \sim R^{s-1}\to 0; \quad \text{small circle: } \sim \rho^{s}\to 0 \]

Because \(0<s<1\): the outer circle vanishes (\(R^{s-1}\to0\)) and the inner circle vanishes (\(\rho^{s}\to0\)). Both endpoint conditions are exactly the constraint on \(s\). C

\[ \underbrace{\int_0^\infty \frac{x^{s-1}}{1+x}dx}_{\text{upper lip}} - \underbrace{e^{2\pi i(s-1)}\int_0^\infty \frac{x^{s-1}}{1+x}dx}_{\text{lower lip}} = \big(1-e^{2\pi i s}\big)J \]

On the lower lip \(\arg z=2\pi\), so \(z^{s-1}=x^{s-1}e^{2\pi i(s-1)}=x^{s-1}e^{2\pi i s}\); the two lips run in opposite directions, and \(e^{2\pi i(s-1)}=e^{2\pi i s}\). This is the monodromy from the prerequisite branch-point result. C

\[ \operatorname{Res}_{z=-1}\frac{z^{s-1}}{1+z}=(-1)^{s-1}=e^{(s-1)\log(-1)}=e^{(s-1)i\pi}=e^{i\pi s}e^{-i\pi}=-e^{i\pi s} \]

Residue at the simple pole \(z=-1\); with \(\arg(-1)=\pi\), \(\log(-1)=i\pi\). Keep symbols exact. B

\[ \big(1-e^{2\pi i s}\big)J = 2\pi i\,(-e^{i\pi s}) \;\Longrightarrow\; J = \frac{-2\pi i\,e^{i\pi s}}{1-e^{2\pi i s}} = \frac{2\pi i\,e^{i\pi s}}{e^{2\pi i s}-1} \]

Solve for \(J\). Now simplify: divide num and den by \(e^{i\pi s}\). B

\[ J=\frac{2\pi i}{e^{i\pi s}-e^{-i\pi s}}=\frac{2\pi i}{2i\sin(\pi s)}=\frac{\pi}{\sin(\pi s)} \]

Use \(e^{i\pi s}-e^{-i\pi s}=2i\sin(\pi s)\). The \(2i\) cancels leaving a clean real result. A

\[ \int_0^{\infty}\frac{x^{s-1}}{1+x}\,dx = \frac{\pi}{\sin(\pi s)}, \qquad 0<s<1 \]

Reading. This is the reflection formula for the Beta/Gamma functions in disguise: \(B(s,1-s)=\Gamma(s)\Gamma(1-s)=\pi/\sin(\pi s)\). The keyhole has computed a special-function identity by pure contour manipulation. Numerically at \(s=\tfrac12\): \(J=\pi/\sin(\pi/2)=\pi\approx 3.1416\).

Units check. With \(x\) dimensionless (as it must be for \(x^{s-1}\) with non-integer \(s\)), \(J\) is a pure number; \(\pi/\sin(\pi s)\) is a pure number. Consistent, and the singularities of \(1/\sin(\pi s)\) at \(s\to0^+,1^-\) correctly flag the endpoint divergences of the original integral.

Problems

(A) Basic Lorentzian. Evaluate \(\displaystyle\int_{-\infty}^\infty\frac{dx}{x^2+4}\).
Solution
Here \(a=0\), \(f=1/(x^2+4)\) with \(\deg\)-difference \(2\), so the plain arc vanishes. Poles at \(\pm 2i\); enclose \(+2i\). \(\operatorname{Res}_{2i}=1/(z+2i)\big|_{2i}=1/(4i)\). Then \(\int=2\pi i/(4i)=\pi/2\approx 1.5708\).
(B) Oscillatory with linear numerator. Evaluate \(\displaystyle\int_{-\infty}^\infty\frac{x\sin x}{x^2+9}\,dx\).
Solution
Write as \(\mathrm{Im}\int x e^{ix}/(x^2+9)\,dx\), \(a=1>0\), close up. \(f=z/(z^2+9)\to0\), Jordan applies. Pole at \(z=3i\): \(\operatorname{Res}_{3i}\dfrac{z e^{iz}}{(z-3i)(z+3i)}=\dfrac{3i\,e^{i(3i)}}{6i}=\dfrac{e^{-3}}{2}\). So \(\int z e^{iz}/(z^2+9)=2\pi i\cdot e^{-3}/2=\pi i\,e^{-3}\); take imaginary part: \(\boxed{\pi e^{-3}}\approx 3.1416\times 0.0498\approx 0.1564\).
(C) Rational, higher-order pole. Evaluate \(\displaystyle\int_{-\infty}^\infty\frac{dx}{(x^2+1)^2}\).
Solution
\(a=0\), denominator degree 4, arc vanishes. Double pole at \(z=i\): \(\operatorname{Res}_{i}=\dfrac{d}{dz}\dfrac{1}{(z+i)^2}\Big|_{i}=\dfrac{-2}{(z+i)^3}\Big|_{i}=\dfrac{-2}{(2i)^3}=\dfrac{-2}{-8i}=\dfrac{1}{4i}\). Then \(\int=2\pi i\cdot\dfrac{1}{4i}=\dfrac{\pi}{2}\approx 1.5708\).
(D) Principal value, pole on axis. Evaluate \(\displaystyle \mathrm{P}\!\!\int_{-\infty}^\infty\frac{\sin x}{x}\,dx\).
Solution
Consider \(\oint e^{iz}/z\) over the real axis indented by a small semicircle above \(z=0\), closed by the upper arc (Jordan, \(a=1\)). No poles enclosed, so \(\mathrm{P}\!\int_{-\infty}^\infty e^{ix}/x\,dx + (\text{indent}) + (\text{arc}=0)=0\). The clockwise indent about the simple pole contributes \(-\pi i\operatorname{Res}_0(e^{iz}/z)=-\pi i\cdot 1=-\pi i\). Hence \(\mathrm{P}\!\int e^{ix}/x\,dx=\pi i\); take imaginary part: \(\int_{-\infty}^\infty \sin x/x\,dx=\pi\approx 3.1416\). (The \(\cos x/x\) real part is \(0\) by oddness/PV.)
(E) Keyhole with a parameter. Evaluate \(\displaystyle\int_0^\infty\frac{x^{1/3}}{1+x^2}\,dx\).
Solution
Keyhole with cut on \(\mathbb{R}^+\), \(z^{1/3}=e^{(1/3)\log z}\), \(\arg\in(0,2\pi)\); here \(s-1=1/3\) so \(s=4/3\), but the integrand \(x^{1/3}/(1+x^2)\) decays as \(x^{-5/3}\) and grows as \(x^{1/3}\) at 0, both integrable, so convergent. Poles at \(z=\pm i=e^{i\pi/2},e^{i3\pi/2}\). Lower-lip factor \(e^{2\pi i(1/3)}=e^{2\pi i/3}\), so \((1-e^{2\pi i/3})\,\)-wait, carefully: with the \((1+x^2)\) denominator the lip cancellation gives \((1-e^{2\pi i s})J\) with \(s=4/3\), i.e. factor \(1-e^{8\pi i/3}=1-e^{2\pi i/3}\). Residues: at \(z=e^{i\pi/2}\), \(\operatorname{Res}=\dfrac{z^{1/3}}{2z}\big|=\dfrac{e^{i\pi/6}}{2e^{i\pi/2}}=\tfrac12 e^{-i\pi/3}\); at \(z=e^{i3\pi/2}\), \(\operatorname{Res}=\dfrac{e^{i\pi/2}}{2e^{i3\pi/2}}=\tfrac12 e^{-i\pi}=-\tfrac12\). Sum \(=\tfrac12(e^{-i\pi/3}-1)\). Then \((1-e^{2\pi i/3})J=2\pi i\cdot\tfrac12(e^{-i\pi/3}-1)=\pi i(e^{-i\pi/3}-1)\). Evaluate: \(e^{-i\pi/3}-1=\tfrac12-\tfrac{\sqrt3}{2}i-1=-\tfrac12-\tfrac{\sqrt3}{2}i=e^{-i2\pi/3}\); and \(1-e^{2\pi i/3}=1-(-\tfrac12+\tfrac{\sqrt3}{2}i)=\tfrac32-\tfrac{\sqrt3}{2}i=\sqrt3\,e^{-i\pi/6}\). So \(J=\dfrac{\pi i\,e^{-i2\pi/3}}{\sqrt3\,e^{-i\pi/6}}=\dfrac{\pi}{\sqrt3}\,i\,e^{-i\pi/2}=\dfrac{\pi}{\sqrt3}\,i\,(-i)=\dfrac{\pi}{\sqrt3}\). Thus \(\boxed{J=\pi/\sqrt3}\approx 1.8138\). (Check: the standard result \(\int_0^\infty x^{\mu-1}/(1+x^2)dx=\tfrac{\pi}{2}\csc(\tfrac{\pi\mu}{2})\) with \(\mu=4/3\) gives \(\tfrac{\pi}{2}\csc(2\pi/3)=\tfrac{\pi}{2}\cdot\tfrac{2}{\sqrt3}=\pi/\sqrt3\). Agrees.)