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For $0<r<R$, suppose $f_n,f$ are holomorphic on a neighborhood of the closed radius-$R$ disc. Subtract their Cauchy integral formula expressions on that circle. Its length is $2\pi R$, and the distance from an interior point with modulus at most $r$ to the circle is at least $R-r$. The estimation lemma gives the displayed bound. Uniform convergence on that circle therefore implies uniform convergence of each fixed-order derivative on the smaller disc. The positive separation from the contour is essential for a uniform bound.

For an entire function $f$, use the complex logarithm $L$ with argument in $(0,2\pi )$. The unit-circle integral in the title uses the branch's one-sided boundary values at its crossing with the cut. On a keyhole contour, the outward upper bank has $L=\log x$, and the inward lower bank has $L=\log x+2\pi i$. Their combined integral of $f(L-i\pi )$ is $-2\pi i{\int }_{\epsilon }^{1}f(x)dx$. The inner circle contributes $O(\epsilon |\log \epsilon |)$, which vanishes. The Cauchy integral theorem on the slit plane therefore gives the identity.

On $z=e^{i\theta }$, $|L-i\pi |=|\theta -\pi |\le \pi$. The estimation lemma consequently givesSubtracting the middle branch value $i\pi$ centres the weight and produces this bound. The contour crosses the cut only in the limiting boundary-value interpretation, so it is not a closed contour inside the logarithm's domain.

Let $a=N+\frac{1}{2}$. The useful identity on the square contour isOn the vertical edges the first term equals one; on the horizontal edges $|y|=a$ and $\sinh \pi a>1$. Thus $|1/\sin \pi z|\le 1$ on every edge. Also $|z|\ge a$ and the contour has length $8a$, so the estimation lemma givesAt every nonzero integer $k$, the residue is $(-1{)}^k/(\pi k^2)$. At zero the Laurent series isso the residue there is $\pi /6$. The positive orientation and the residue theorem therefore giveLetting $N\to \infty$ yields the alternating reciprocal-square sum

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.

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.

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.

Choose $0<\arg z<2\pi$, so $\sqrt{z}=\sqrt{R}{e}^{i\theta /2}$ on the arc. Its clockwise orientation givesBecause $f$ is an analytic function at zero, $f(R{e}^{i\theta })=f(0)+O(R)$ uniformly on the small arc. Integration then givesThereforeThe estimation lemma also gives $|I(R)|\le \pi \sqrt{R}{\max }_C|f|$. Hence a sufficient large-arc condition isFor example $f(z)=O(|z{|}^{-1/2-\eta })$ uniformly on the arc, with $\eta >0$, suffices. Under the stronger literal assumption that $f$ is analytic at infinity, its Laurent expansion is $f(z)=f(\infty )+O(1/z)$; the necessary and sufficient condition is then $f(\infty )=0$. A nonzero constant term would instead give the growing contribution $2\sqrt{2R}f(\infty )$.

For the usual piecewise smooth complex path $\gamma$, let $\ell (\gamma )$ denote its finite length. The estimation lemma givesBy uniform convergence, the supremum tends to zero, soThe uniform limit is continuous on the path, so its integral exists. This is uniform convergence and contour integration; the path length is fixed while the functions vary.

Put $D(z)={\sinh }^{2}z+{\cosh }^{2}a$ and $S=\sinh (2a)>0$. Traverse the rectangle $-R\to R\to R+\pi i\to -R+\pi i\to -R$ with positive orientation of a contour, taking $R>a$. Part (i) shows that the only enclosed poles of $f(z)=z/D(z)$ areThey are simple poles becauseThe residue formula for a quotient at a simple denominator zero givesTheir imaginary terms cancel and their residues sum to $-2a/S$. Thus Cauchy's residue theorem gives

The choice of numerator $z$ makes the horizontal sides useful. The hyperbolic sine satisfies $\sinh (z+\pi i)=-\sinh z$, so $D(x+\pi i)=D(x)$. Consequently the lower and upper contour integrals, with their actual orientations, add toIt remains to justify discarding both vertical sides. On either side $z=\pm R+iy$, $0\le y\le \pi$, the reverse triangle inequality and part (b)(ii) implyThis lower bound is positive for sufficiently large $R$. Also $|z|\le R+\pi$ and each vertical side has length $\pi$. The estimation lemma bounds the absolute value of the sum $V_R$ of the vertical contributions byThe real improper integral converges absolutely, since its integrand is continuous on the real axis and decays like $4e^{-2|x|}$ at infinity. Combining all four sides and taking $R\to \infty$ givesThereforeAn independent real-variable check uses the hyperbolic tangent substitution $u=\tanh x$: the integral becomes ${\int }_{-1}^{1}du/({\cosh }^{2}a-u^2{\sinh }^{2}a)$. Its antiderivative uses the inverse hyperbolic tangent, giving $2\mathrm{arctanh}(\tanh a)/(\sinh a\cosh a)=4a/\sinh (2a)$. As $a\downarrow 0$, this tends to $2$, agreeing with ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^{2}xdx=2$ and checking the factor of two.

For $\varphi (t)=r{e}^{it}$, ${\varphi }^{\prime }(t)=ir{e}^{it}$ and $|{\varphi }^{\prime }(t)|=r$. The triangle inequality for an integral applied to the parametrized contour integral givesThis is the circular-contour instance of the estimation lemma.

Fix $0<r<1$ and $k\ge 1$. On $|z|=1$ and $|z_0|<r$, the reverse triangle inequality gives $|z-z_0|\ge 1-r$. Subtract the two Cauchy integral formula expressions and apply the estimation lemma, using circle length $2\pi$:Thus every fixed-order derivative converges uniformly on each smaller disc. The right-hand factor tends to zero by the assumed uniform convergence on $D$. Keeping a positive gap $1-r$ from the integration contour is the essential feature of this Cauchy derivative convergence bound. If $r\le 0$, the specified open disc is empty and the claim is vacuous.

Morera's theorem states that a continuous function on a domain is holomorphic if its contour integral around every triangle whose filled-in interior lies in the domain is zero.

The uniform limit $f$ is a continuous function, by the uniform limit theorem. For any such triangle $T$, the estimation lemma givesEach integral of $f_n$ vanishes by the Cauchy integral theorem. Hence the integral of $f$ vanishes too, and Morera's theorem proves $f$ is holomorphic on $D$. This is the basic locally uniform convergence of holomorphic functions argument; the given uniform convergence on all of $D$ is more than needed.

For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence

Use the principal logarithm $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z<\pi$, whose branch cut is the negative real axis. For this choice,has three simple poles away from the cut: $z=1$, $z=ia$, and $z=-ia$. The pole at $1$, caused by $\mathrm{Log}1=0$, must be included.

Choose $0<\epsilon <\min \{1,a\}$ and $R>\max \{1,a\}$, and use a positively oriented keyhole contour around the negative real axis: the large circle is traversed counterclockwise, the upper bank from $-R$ to $-\epsilon$, the small circle clockwise, and the lower bank from $-\epsilon$ to $-R$. Formally, use radial segments at arguments $\pm (\pi -\eta )$, then let $\eta \downarrow 0$ with $R,\epsilon$ fixed. The function has continuous limiting values on each compact bank, which justifies this limit.

Put $L=\log a$. The residues areThe first follows from $\mathrm{Log}z=(z-1)+O((z-1{)}^2)$; the other two follow from the simple zeros of $z^2+a^2$. Their sum isThese expressions are valid also for $a=1$, since $L\pm i\pi /2$ never vanish for real $L$.

On the outer circle, $|\mathrm{Log}z|\ge \log R$ for $R>1$ and $|z^2+a^2|\ge R^2-a^2$. On the inner circle, $|\mathrm{Log}z|\ge |\log \epsilon |$ and $|z^2+a^2|\ge a^2-{\epsilon }^2$. Thus the estimation lemma bounds the two arc integrals byrespectively, as $R\to \infty$ and $\epsilon \downarrow 0$.

On the upper bank, $z=-x+i0$ gives $\mathrm{Log}z=\log x+i\pi$; on the lower bank it gives $\log x-i\pi$. Accounting for the opposite orientations, their combined integral isThe real improper integral converges: its integrand is bounded near zero by $1/(a^2{\pi }^2)$ and at infinity by $1/({\pi }^2{x}^2)$. Taking the arc limits in the residue theorem gives $-2\pi iI=2\pi i\Sigma$, henceThe final subtraction is exactly the contribution of the pole at $1$.

The improper integral converges: its integrand is bounded near zero and at most $x^{-n}$ for $x\ge 1$, which is integrable because $n\ge 2$.

Set $\theta =\pi /n$, $\omega =e^{2i\theta }$ and $\zeta =e^{i\theta }$. Apply the residue theorem to $F(z)=1/(1+z^n)$ on the positively oriented sector contour consisting of the segment $0\to R$, the arc $R{e}^{it}$ for $0\le t\le 2\theta$, and the segment $R\omega \to 0$, with $R>1$. No pole lies on the boundary. Exactly one root of $z^n=-1$ is inside this sector, namely $\zeta$; it is a simple pole. Part (a) givesOn the returning ray, put $z=\omega x$ with $x$ decreasing from $R$ to zero. Since ${\omega }^n=1$, the two ray contributions areOn the arc, $|1+z^n|\ge R^n-1$, so the estimation lemma bounds its contour integral byThe residue theorem and the convergence already established now giveUsing $1-e^{2i\theta }=-2i{e}^{i\theta }\sin \theta$, we concludeThe sector also works for $n=2$, when it is the upper semicircle. Its angle is chosen so that $z^n$ has the same value on the two rays.

Factor the denominator as $(z^2+1)(z^2+4)$ and integrateover $[-R,R]$ followed by the upper semicircle, for $R>2$. The enclosed simple poles are $i$ and $2i$. The local quotient formula from part (a) givesOn this arc, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^2+1|\ge R^2-1$, $|z^2+4|\ge R^2-4$. Thus the estimation lemma givesOn the real line, $|F(x)|\le 1/[(x^2+1)(x^2+4)]$, so the improper integral is absolutely convergent. Hence the residue theorem yieldsTaking real parts is justified by absolute convergence. ThereforeThe imaginary part vanishes as a further check, since the sine integrand is odd.

On the upper semicircular contour, $|z|=R$ and the reverse triangle inequality gives $|z^4+1|\ge R^4-1>0$. Its arc length is $\pi R$, so the estimation lemma givesThe arc is open, so an ordinary integral sign is appropriate even though the source uses a closed-integral symbol there. Subtracting the vanishing arc contribution from the closed contour integral gives the real-line improper integral $\pi /\sqrt{2}$. Its integrand is an even function, and its tail is $O(x^{-2})$, ensuring convergence. Thus

On the small clockwise contour $z=ϵe^{i\theta }$, $\theta$ decreases from $\pi$ to $-\pi$. Uniformly on that arc, $e^{z-\sqrt{z}}=1+O(\sqrt{ϵ})$, and henceThe large arcs also vanish. To make this precise for the displayed geometry, take them on the left semicircle $z=\gamma +R{e}^{i\theta }$, $\pi /2\le \theta \le 3\pi /2$, split at the cut. On the principal holomorphic branch, $\mathrm{Re}\sqrt{z}\ge 0$, $|z|\ge R-\gamma$, and $\mathrm{Re}z=\gamma +R\cos \theta$. The estimation lemma givesFor the last bound, symmetry and $\sin u\ge 2u/\pi$ on $0\le u\le \pi /2$ imply that the angular integral is at most $\pi /R$. The exact choice of large arcs does not affect the limiting result. The vertical improper integral converges absolutely because $e^{-\mathrm{Re}\sqrt{\gamma +iy}}/|\gamma +iy|$ decays as $O(e^{-c\sqrt{|y|}}/|y|)$.

Taking limits in the closed contour integral now yields $2\pi iJ+2iK-2\pi i=0$, where $K={\int }_0^{\infty }{e}^{-x}\sin (\sqrt{x})dx/x$. ThusAs an optional explicit evaluation, put $x=t^2$ and define $K(a)=2{\int }_0^{\infty }{e}^{-t^2}\sin (at)dt/t$. Differentiation under the integral sign and the Gaussian integral give $K^{\prime }(a)=\sqrt{\pi }{e}^{-a^2/4}$ and $K(0)=0$. Therefore $K=K(1)=\pi \mathrm{erf}(1/2)$ and $J=\mathrm{erfc}(1/2)\simeq 0.479500$ in terms of the error function and complementary error function.

Let $\Gamma$ be a piecewise smooth contour forming a simple closed curve, positively oriented: travel counterclockwise, with the interior on the left. Suppose $f$ is holomorphic on an open set containing $\Gamma$ and its interior, except at finitely many isolated singularities $a_1,\dots ,a_m$ strictly inside. ThenNo isolated singularity may lie on $\Gamma$: the ordinary contour integral must be defined there. Reversing the direction of the contour reverses the sign of the answer. The theorem applies to essential singularities as well as poles. ^[ref]

Return to $f(z)=1/(z^2+1)$. A counterclockwise circle $|z|=2$ encloses both simple poles, soYet $f$ is nonzero everywhere it is defined. A zero contour integral does not imply a zero function, or even absence of poles: the enclosed residues can cancel. By contrast, the counterclockwise circle $|z-i|=1/2$ encloses only $i$, givingThe picture compares these two selections. Changing the shape of a contour while keeping the same enclosed poles leaves the contour integral unchanged, provided the contour deformation stays where $f$ is holomorphic.

This is the practical appeal of the residue theorem: local Laurent coefficients determine a global contour integral. For a rational function, a few simple pole computations can replace a long parametrized calculation. It also makes real improper integrals accessible by treating the real line as part of a closed contour in the complex plane.

To compute $I={\int }_{-\infty }^{\infty }dx/(1+x^2)$, take $R>1$ and travel along the real line from $-R$ to $R$, then return along the upper semicircle $A_R$, parametrized by $z=R{e}^{it}$ for $0\le t\le \pi$. This contour is positively oriented and encloses the simple pole $i$ only; $-i$ is outside. HenceThe added arc must be controlled before taking a limit. On $A_R$, the reverse triangle inequality gives $|z^2+1|\ge R^2-1$, and the arc length is $\pi R$. The estimation lemma therefore yieldsThe denominator grows quadratically while the arc length grows only linearly. This quantitative comparison, rather than a sketch alone, explains why closing the contour does not alter the limiting real improper integral. It even supplies an error bound: the difference between $\pi$ and the finite real integral is at most $\pi R/(R^2-1)$.

For a numerical check, $R=10$ gives the finite real integral $2\arctan (10)\approx 2.94226$, leaving about $0.19934$ to reach $\pi$. The arc bound is $10\pi /99\approx 0.31733$, comfortably larger than that error. The graph shows both quantities approaching zero. The exact tail here can be checked with the antiderivative $\arctan x$; the residue theorem method remains useful for improper integrals where a convenient real antiderivative is unavailable.

Finally, both one-sided improper integrals converge absolutely: for $|x|\ge 1$, $0<1/(1+x^2)\le 1/x^2$, whose tails are integrable. Thus the symmetric limit computes the ordinary improper integral, not merely a Cauchy principal value, andClosing below instead would enclose the simple pole $-i$. With the same left-to-right traversal of the real line, that contour is clockwise, so its contour integral is $-2\pi i\mathrm{Res}(f,-i)=\pi$ again. This provides a useful sign check: the opposite residue is accompanied by the opposite direction around the enclosed region.

The same strategy extends to many other rational functions: choose a contour, identify its enclosed isolated singularities, compute their residues, and bound the added pieces. The choice of contour and the bound are as much part of the calculation as the residues. ^[ref]