Contour integration

Contour integration integrates complex functions along oriented curves and evaluates many real integrals through residues.

A pair of oppositely directed contours on a branch cut produces the difference of the two boundary values integrated over their actual endpoint interval. Replacing finite endpoints by zero and infinity requires a limiting argument and convergence of the combined improper integral. Individual side integrals may diverge even when their difference converges, so cancellation must precede separation of endpoint limits.

For real $a>b>0$, put $d=\sqrt{a^2-b^2}$. The change of variables $z=e^{i\theta }$ changes the integral to ${\oint }_{|z|=1}(2/b)dz/[z^2+2i(a/b)z-1]$. The roots are $-i(a-d)/b$ and $-i(a+d)/b$, with only the first inside the circle because $a-d=b^2/(a+d)$. The residue there is $1/(id)$, and the residue theorem proves the displayed result. A shift of the angular variable replaces sine by cosine. More generally for real $a>|b|$, the same answer follows by shifting by $\pi$ when $b<0$, and the case $b=0$ follows directly.

Let $a>0$ and let $R$ be a rational function with $R(z^2)$ finite for real $z$ and $R(u)=O(u^{-1})$ at infinity. Choose the branch $h(z)=\sqrt{z^2-a^2}$ analytic off $[-a,a]$, with $h(z)\sim z$ at infinity. Put $g(z)=zR(z^2)/h(z)$. On the real exterior, $g(-x)=g(x)$; on the upper bank of the cut, $h(x+i0)=i\sqrt{a^2-x^2}$, so $g(x+i0)$ is odd in $x$.

Close a real-axis contour with a large upper semicircle, making clockwise upper indentations at the two branch points. The large arc vanishes because $g=O(z^{-2})$. Each small indentation is $O(\sqrt{\epsilon })$ and vanishes. The integral along the upper bank of $[-a,a]$ is zero by oddness; the two exterior pieces sum to twice the required integral. The residue theorem therefore givesResidues must use the chosen branch, notably $h(iy)=i\sqrt{y^2+a^2}$ for $y>0$. The substitution $u=\sqrt{x^2-a^2}$ provides an independent real-variable check, converting the integral to ${\int }_0^{\infty }R(u^2+a^2)du$.

Take $a>0$ and define $z^s$ using $-\pi /2<\arg z<3\pi /2$, so the branch cut lies on the negative imaginary axis. An indented upper semicircle encloses the simple pole $ia$. Its two straight segments contribute $(1+e^{i\pi s})$ times the desired real integral, while the residue theorem gives $\pi a^{s-1}{e}^{i\pi s/2}$. The large arc vanishes as its radius to the power $s-1$, and the small arc as its radius to the power $s+1$, giving the stated convergence range and result. This contour avoids placing a branch cut on either straight integration segment.

This contour integral defines an analytic function away from the oriented integration curve, under the relevant integrability assumptions. Its limiting values across the curve are related by the Sokhotski–Plemelj theorem. On an oriented real axis, with $Hf(\rho )={\pi }^{-1}\mathrm{PV}\int f(s)/(\rho -s)ds$, the signed Cauchy boundary operators are $C^{\pm }f=\pm f/2+iHf/2$.

For continuous boundary data on the counterclockwise unit circle and its Cauchy transform $\Phi$, $\Phi (r\omega )-\Phi (r^{-1}\omega )=P\varphi (r\omega )$ for $0<r<1$. This follows by subtracting the two Cauchy kernels and obtaining the Poisson kernel on the circle. Consequently the paired interior-minus-exterior difference tends to $\varphi (\omega )$ as $r\uparrow 1$. For $r\downarrow 1$ from above, the same written difference has the opposite sign. This paired limit does not require each individual boundary value of the transform to exist.

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A commutator contour around two branch points cancels the total monodromy of its integrand. For the contour circling $1,0$ clockwise and then $1,0$ anticlockwise, with initial positive-real branches, the Beta function obeysThe contour integral is entire in $z$, giving a meromorphic continuation of the Beta function.

A complex integration contour is an oriented piecewise continuously differentiable curve in the complex plane. For a parametrization $\gamma (s)$, its contour integral is ${\int }_{\Gamma }f(z)dz=\int f(\gamma (s)){\gamma }^{\prime }(s)ds$. Reversing orientation changes the sign. Unbounded contours require a convergent improper or explicitly regularized limit.

An oriented piecewise differentiable path in the complex plane, used as the domain of a contour integral. Its orientation controls the signs of residues in the residue theorem.

A piecewise smooth curve has a continuous function as a parametrization on a compact interval, with a finite partition into subintervals on each of which the parametrization is continuously differentiable. One-sided derivatives may differ at joining points, so polygonal contours are allowed. A contour integral is the sum of the integrals over those smooth pieces.

The orientation of a contour specifies the direction in which its parametrization is traversed. Reversing that direction negates its contour integral. For a simple closed curve in the complex plane, positive orientation of a contour is the direction keeping its bounded interior on the left.

An integration ray can be rotated through an analytic sector when endpoint and connecting-arc contributions vanish. Complex powers gain a phase from the rotated differential and power. This relates conditionally convergent oscillatory gamma integrals to decaying Laplace integrals.

A contour encircling a branch cut travels on its two sides and joins them by small and large circles. It converts the phase change of a complex power into a real integral through the residue theorem. The two circular contributions must be bounded before taking their limiting radii.

For a piecewise differentiable parametrized contour $z(t)$, the contour integral is $\int f(z(t))z^{\prime }(t)dt$. It depends on the orientation of the contour. For a holomorphic function, the Cauchy integral theorem allows a contour deformation through regions without singularities.

For a simple closed curve in the standard oriented complex plane, positive orientation keeps the bounded interior on the left while traversing the curve. A circle is therefore traversed counterclockwise; reversing direction changes the sign of its contour integral.

If $|f(z)|\le M$ along a contour of length $L$, then

If continuous functions $f_n$ converge uniformly to $f$ on the image of a rectifiable curve $\gamma$, thenIndeed, the absolute difference is at most the length of $\gamma$ times ${\sup }_{\gamma }|f_n-f|$.

Contour shifting deforms a complex integration path through a region where the integrand is meromorphic. The residue theorem says that the change equals $2\pi i$ times the sum of residues of the poles crossed.

A holomorphic function has an antiderivative on a domain only if its integral around every closed curve vanishes. Thus a single nonzero period, such as ${\int }_{|z|=1}dz/z=2\pi i$, prevents a global antiderivative.

Jordan's lemma controls exponential contour integrals on large semicircles and makes their arc contributions vanish.

For real $a\ne 0$ and $|x|<\pi$, integrate $z\sin (xz)/((a^2+z^2)\sin \pi z)$ around a thin positively oriented strip containing the integers. Its integer residues give the series. Closing the two strip sides outward instead gives minus the sum of the residues at $z=\pm ia$, each $\sinh (ax)/(2\sinh (a\pi ))$. The factor $|x|<\pi$ ensures exponential decay on those closures. The bilateral series is interpreted by symmetric partial sums; at $a=0$ its continuous extension is $-x$ with the zero-index summand defined to be zero.

Integrating along a straight segment bounds a complex integral by segment length times the supremum of the integrand.

A Hankel contour runs along both banks of a branch cut and circles its branch point, converting the jump of a complex power into a sine factor.

Subtracting the local Taylor expansion at the encircled branch point extends a Hankel integral successively across left half-planes; prefactors often remove the introduced poles.

When the complex power becomes an integer power, a collapsed Hankel contour extracts the coefficient of $t^{-1}$ in the local Laurent series.

Keep the circular portion of a Hankel contour at fixed nonzero radius. Its numerator $I(z)={\int }_H{t}^{z-1}{e}^{t}dt$ is an entire function of $z$, because the contour avoids zero and its tails decay exponentially, uniformly on compact parameter sets. At a positive integer $n$, the integrand is single-valued and entire in $t$, so the bank integrals cancel and the circular integral is zero. Thus $I(n)=0$, canceling the simple zero of $2i\sin (\pi z)$ in the Hankel representation of the gamma function. At a nonpositive integer $-n$, the circular integral instead gives $2\pi i/n!$, yielding residue $(-1{)}^n/n!$.