The Residue Theorem and the Evaluation of Real Integrals

Let $U:=\{z\in \mathbb{C}\setminus {\mathrm{\Gamma }}^{\ast }:n(\mathrm{\Gamma },z)\mathrm{\ne }0\}\mathrm{.}$ By the local constancy part of [L1], $U$ is open in $\mathbb{C}\setminus {\mathrm{\Gamma }}^{\ast }$. The same local constancy also shows that if $z\mathrm{\notin }{\mathrm{\Gamma }}^{\ast }\cup U$, then a whole neighbourhood of $z$ lies outside $U$, so every limit point of $U$ outside the trace already lies in $U$. Therefore $\bar{U}\subseteq U\cup {\mathrm{\Gamma }}^{\ast }$.

By the far-from-the-trace clause of [L1], the set $U$ is bounded. Since the trace ${\mathrm{\Gamma }}^{\ast }$ is compact, step 1.1 makes $\bar{U}$ a bounded closed subset of $\mathbb{C}$, hence compact.

Because $\mathrm{\Gamma }$ is admissible, its trace avoids the pole set. Thus [given, step 2.1, L2] ∎ $S\cap \bar{U}=S\cap U=\{a\in S:n(\mathrm{\Gamma },a)\mathrm{\ne }0\}\mathrm{.}$ The right-hand set is a closed discrete subset of the compact set $\bar{U}$ by, and every closed discrete subset of a compact metric space is finite. So only finitely many poles have nonzero index.

By [L1], the set $A:=\{a\in S:n(\mathrm{\Gamma },a)\mathrm{\ne }0\}$ is finite. For each $a\in A$ choose $r_a>0$ so small that the closed discs $\overline{D(a,r_a)}$ are pairwise disjoint, lie in $\mathrm{\Omega }$, meet no pole other than $a$, and are disjoint from ${\mathrm{\Gamma }}^{\ast }$. Let $C_a$ be the positively oriented circle $\mathrm{\mid }\zeta -a\mathrm{\mid }=r_a$.

Put [step 1.1, L4] $\mathrm{\Delta }:={\sum }_{a\in A}n(\mathrm{\Gamma },a)C_a\mathrm{.}$ For every $p\in \mathbb{C}\setminus (\mathrm{\Omega }\setminus A)$ one has $n(\mathrm{\Gamma },p)=n(\mathrm{\Delta },p)$. Indeed, if $p\mathrm{\notin }\mathrm{\Omega }$ then admissibility makes $n(\mathrm{\Gamma },p)=0$, and every $C_a$ lies in $\mathrm{\Omega }$, so gives $n(\mathrm{\Delta },p)=0$ as well. If $p=a\in A$, then [L4] gives $n(C_a,a)=1$ and $n(C_b,a)=0$ for $b\mathrm{\ne }a$, so $n(\mathrm{\Delta },a)=n(\mathrm{\Gamma },a)$. Therefore $\mathrm{\Gamma }$ and $\mathrm{\Delta }$ are homologous in $\mathrm{\Omega }\setminus A$.

The function $f$ is holomorphic on $\mathrm{\Omega }\setminus A$, so [L3] applied to step 2.1 yields ${\int }_{\mathrm{\Gamma }}f(z)dz={\int }_{\mathrm{\Delta }}f(z)dz={\sum }_{a\in A}n(\mathrm{\Gamma },a){\int }_{C_a}f(z)dz\mathrm{.}$

Each $C_a$ encloses only the pole $a$, so [L2] gives ${\int }_{C_a}f(z)dz=2\pi i\mathrm{Res}(f,a)\mathrm{.}$ Substituting this into step 3.1 proves the displayed formula. Since the set $A$ is finite, the residue sum has only finitely many nonzero terms.

The circle $C(a,r)$ is null-homologous in any open set containing the closed [given, L1] disc it bounds, so applies to it.

By [L2], every pole $b$ with $\mathrm{\mid }b-a\mathrm{\mid }<r$ contributes the factor [step 1.1, L2] ∎ $n(C(a,r),b)=1$, while every pole outside the circle contributes the factor $0$. Substituting those indices into gives the formula.

Along ${\gamma }_T^{+}$ one has $\mathrm{\mid }{\gamma }_T^{+}(t)\mathrm{\mid }=T$, so $\mathrm{\mid }R({\gamma }_T^{+}(t))\mathrm{\mid }\le \frac{1}{T}{\sup }_{0\le s\le \pi }\mathrm{\mid }{\gamma }_T^{+}(s)R({\gamma }_T^{+}(s))\mathrm{\mid }\mathrm{.}$

The arc length of ${\gamma }_T^{+}$ is $\pi T$. Therefore the ML estimate gives $\mid {\int }_{{\gamma }_T^{+}}R(z)dz\mid \le \pi T\cdot \frac{1}{T}{\sup }_{0\le s\le \pi }\mathrm{\mid }{\gamma }_T^{+}(s)R({\gamma }_T^{+}(s))\mathrm{\mid }=\pi {\sup }_{0\le s\le \pi }\mathrm{\mid }{\gamma }_T^{+}(s)R({\gamma }_T^{+}(s))\mathrm{\mid }\mathrm{.}$

The right-hand side tends to $0$ by hypothesis, so the arc integral tends to $0$ as claimed.

For $z=T{e}^{it}$ on the upper semicircle, $\mathrm{\mid }{e}^{i\lambda z}\mathrm{\mid }=\mathrm{\mid }{e}^{i\lambda T(\cos t+i\sin t)}\mathrm{\mid }=e^{-\lambda T\sin t}\mathrm{.}$

On $[0,\pi \mathrm{/}2]$ the function $g(t)=-\sin t$ has [L1, algebra] $g^{\mathrm{\prime }\mathrm{\prime }}(t)=\sin t\ge 0$, so makes $g$ convex there. A convex graph lies below the chord joining its endpoint values, hence $-\sin t\le -2t\mathrm{/}\pi$ and therefore $\sin t\ge \frac{2t}{\pi }(0\le t\le \frac{\pi }{2})\mathrm{.}$ By symmetry, also $\sin t\ge \frac{2(\pi -t)}{\pi }(\frac{\pi }{2}\le t\le \pi )\mathrm{.}$

Put $M_T:={\sup }_{0\le t\le \pi }\mathrm{\mid }{\gamma }_T^{+}(t)R({\gamma }_T^{+}(t))\mathrm{\mid }\mathrm{.}$ Then along the arc $\mathrm{\mid }{e}^{i\lambda z}R(z)dz\mathrm{\mid }\le e^{-\lambda T\sin t}\frac{M_T}{T}Tdt=M_T{e}^{-\lambda T\sin t}dt\mathrm{.}$ So $\mid {\int }_{{\gamma }_T^{+}}{e}^{i\lambda z}R(z)dz\mid \le M_T{\int }_0^{\pi }{e}^{-\lambda T\sin t}dt\mathrm{.}$

Splitting the integral at $\pi \mathrm{/}2$ and using step 1.2 gives ${\int }_0^{\pi }{e}^{-\lambda T\sin t}dt\le 2{\int }_0^{\pi \mathrm{/}2}{e}^{-2\lambda Tt\mathrm{/}\pi }dt=\frac{\pi }{\lambda T}(1-e^{-\lambda T})\le \frac{\pi }{\lambda T}\mathrm{.}$ Hence $\mid {\int }_{{\gamma }_T^{+}}{e}^{i\lambda z}R(z)dz\mid \le \frac{\pi M_T}{\lambda T}\mathrm{.}$

Since $M_T\to 0$, the bound in step 3.1 tends to $0$. Therefore the arc integral tends to $0$.

Write $f(z)=c_{-1}(z-a{)}^{-1}+h(z)$ as in [L1]. Along the arc, put $\theta (t)=(1-t)\alpha +t\beta$. Then $z-a=\epsilon e^{i\theta (t)}$ and $dz=i(\beta -\alpha )\epsilon e^{i\theta (t)}dt$, so ${\int }_{{\gamma }_{\epsilon }}\frac{c_{-1}}{z-a}dz=i(\beta -\alpha )c_{-1}=i(\beta -\alpha )\mathrm{Res}(f,a)\mathrm{.}$

The holomorphic function $h$ is bounded on a small closed disc around $a$, say by $M$. Hence $\mid {\int }_{{\gamma }_{\epsilon }}h(z)dz\mid \le M\epsilon \mathrm{\mid }\beta -\alpha \mathrm{\mid },$ which tends to $0$ with $\epsilon$.

Adding the two parts from steps 1.1 and 2.1 proves the limit formula. The two indentation special cases are the choices $(\alpha ,\beta )=(\pi ,0)$ and $(\alpha ,\beta )=(\pi ,2\pi )$.

By [L3], the real improper integral converges. The same degree gap implies $zR(z)\to 0$ as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$, because the numerator degree is at least two less than the denominator degree.

Let ${\mathrm{\Gamma }}_T$ be the contour formed by $[-T,T]$ and the upper semicircle [L1] ${\gamma }_T^{+}$. For all sufficiently large $T$, the contour avoids every pole of $R$ and encloses exactly the poles of $R$ with positive imaginary part. So gives ${\int }_{-T}^{T}R(x)dx+{\int }_{{\gamma }_T^{+}}R(z)dz=2\pi i{\sum }_{\mathrm{\Im }a>0}\mathrm{Res}(R,a)\mathrm{.}$

By step 1.1 and [L2], the arc integral tends to $0$ as $T\to \mathrm{\infty }$. The straight-piece integral tends to ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}R(x)dx$ by step 1.1. Passing to the limit in step 2.1 yields the residue formula.

For large $T$ and small pairwise disjoint indentation radii ${\epsilon }_j$, form the contour specified in the statement. It is an admissible positively oriented cycle, avoids every real pole, and encloses precisely the nonreal poles in the upper half-plane, so [L2] applies.

The contour pieces have their asserted limits as $T\to \mathrm{\infty }$ and every ${\epsilon }_j\downarrow 0$.

The residue theorem and step 2.1 give $\mathrm{PV}\text{\hspace{0.17em}⁣}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}R(x)dx-i\pi {\sum }_j\mathrm{Res}(R,a_j)=2\pi i{\sum }_{\mathrm{\Im }a>0}\mathrm{Res}(R,a)\mathrm{.}$ Hence the joint limit exists, and moving the indentation term to the right proves the formula.

Assume first that $\lambda >0$ and that $R$ has no real pole. Apply [L2] to [assume-case positive, L1, L2] the contour formed by $[-T,T]$ and the upper semicircle. The arc term tends to $0$ by, so the real-line integral is the sum of the residues of $e^{i\lambda z}R(z)$ in the upper half-plane.

If $\lambda <0$ and $R$ has no real pole, close instead by the lower semicircle. The same computation gives a minus sign because the positively oriented contour now traverses the real segment from $T$ back to $-T$, so the real integral equals $-2\pi i$ times the sum of the residues in the lower half-plane.

If $R$ has simple real poles and $\lambda >0$, indent them above the axis. Each indentation excludes its pole and contributes $-i\pi$ times its residue by [L3]; the residue theorem therefore gives the first displayed principal-value formula. If $\lambda <0$, use lower indentations and the lower semicircle. Each indentation contributes $+i\pi$ times its residue, while the clockwise outer contour contributes $-2\pi i$ times the lower-half-plane residue sum, giving the second formula.

Steps 1.1, 1.2, and 1.3 prove all cases listed in the statement.

Put $z=e^{i\theta }$. By [L1], $\cos \theta =\frac{z+z^{-1}}{2},\sin \theta =\frac{z-z^{-1}}{2i}\mathrm{.}$ Differentiating $z=e^{i\theta }$ gives $dz=iz{e}^{0}d\theta =izd\theta$, so $d\theta =dz\mathrm{/}(iz)$.

As $\theta$ runs from $0$ to $2\pi$, the variable $z$ traverses the unit circle once in the positive direction. Substituting the identities of step 1.1 into the real integral gives exactly the contour integral of $G(z)$. The hypothesis that $G$ has no pole on $\mathrm{\mid }z\mathrm{\mid }=1$ is what makes the contour integral well defined.

On the upper lip of the slit one has $\mathrm{Arg}(x+i0)=0$, so $\mathrm{Log}(x+i0)=\log x$ and therefore ${\lim }_{y\downarrow 0}(x+iy{)}^{\alpha -1}=\exp ((\alpha -1)\log x)=x^{\alpha -1}\mathrm{.}$

On the lower lip one has $\mathrm{Arg}(x-i0)=2\pi$, so $\mathrm{Log}(x-i0)=\log x+2\pi i\mathrm{.}$ Hence ${\lim }_{y\downarrow 0}(x-iy{)}^{\alpha -1}=\exp ((\alpha -1)(\log x+2\pi i))=e^{2\pi i(\alpha -1)}{x}^{\alpha -1}=e^{2\pi i\alpha }{x}^{\alpha -1}\mathrm{.}$

Let ${\mathrm{\Gamma }}_{\epsilon ,\rho }$ be the keyhole contour of inner radius $\epsilon$ and outer radius $\rho$. Applying [L2] to $f(z)=z^{\alpha -1}R(z)$ on the slit annulus gives ${\int }_{{\mathrm{\Gamma }}_{\epsilon ,\rho }}f(z)dz=2\pi i\sum \mathrm{Res}(f,a),$ where the sum is over the enclosed poles away from the positive axis.

The contour integral splits into outer circle, upper lip, inner circle, and lower lip. By hypothesis the two circular contributions vanish. The assumed convergence of the improper integral and the upper boundary value in [L1] make the upper lip tend to ${\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}R(x)dx$. The lower lip has the reverse orientation and boundary value $e^{2\pi i\alpha }{x}^{\alpha -1}R(x)$, so it tends to $-e^{2\pi i\alpha }{\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}R(x)dx$.

Taking $\epsilon \downarrow 0$ and $\rho \to \mathrm{\infty }$ in step 1.1 and substituting the boundary terms from step 1.2 yields $(1-e^{2\pi i\alpha }){\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}R(x)dx=2\pi i\sum \mathrm{Res}(f,a)\mathrm{.}$ The finite set of poles of the rational factor is eventually enclosed, so this is the sum over all poles of $f$ away from the positive axis and hence the claimed keyhole identity.

Let $F(z)=\pi \cot (\pi z)f(z)$. By [L1], $\sin (\pi z)$ has a simple zero at [L1, L2, algebra] each integer $n$, so applied to $\pi \cos (\pi z)f(z)\mathrm{/}\sin (\pi z)$ gives $\mathrm{Res}(F,n)=\frac{\pi \cos (\pi n)f(n)}{\pi \cos (\pi n)}=f(n)\mathrm{.}$

Integrate $F$ around the rectangle with vertices $N+\frac{1}{2}\pm iN$ and $-N-\frac{1}{2}\pm iN$. On the vertical sides one has $\cot (\pi (x+iy))=\mp i\tanh (\pi y)$ because $x=\pm (N+\frac{1}{2})$, so $\mathrm{\mid }\cot \mathrm{\mid }$ is uniformly bounded there. On the horizontal sides $\cot (\pi (x\pm iN))$ tends uniformly to $\mp i$ as $N\to \mathrm{\infty }$. Since $f(z)=O(z^{-2})$, the integrand is $O(z^{-2})$ on every side, and the boundary integral tends to $0$.

By [L3], the sum of all residues of $F$ inside the rectangle is therefore $0$. Those residues are the integer residues from step 1.1 together with the nonintegral poles of $f$. Letting $N\to \mathrm{\infty }$ yields ${\sum }_{n\in \mathbb{Z}}f(n)+{\sum }_{a\mathrm{\notin }\mathbb{Z}}\mathrm{Res}(\pi \cot (\pi z)f(z),a)=0,$ which is the stated identity.

Let $F(z)=\pi \csc (\pi z)f(z)$. By [L1], $\sin (\pi z)$ has a simple zero at [L1, L2, algebra] each integer $n$, and gives $\mathrm{Res}(F,n)=\frac{\pi f(n)}{\pi \cos (\pi n)}=(-1{)}^{n}f(n)\mathrm{.}$

On the same rectangles used for the cotangent theorem, the factor $\csc (\pi z)$ is uniformly bounded on the vertical sides because $\mathrm{\mid }\sin (\pi (N+\frac{1}{2}+iy))\mathrm{\mid }=\mathrm{\mid }\cosh (\pi y)\mathrm{\mid }\ge 1$, and on the horizontal sides it decays exponentially like $e^{-\pi \mathrm{\mid }N\mathrm{\mid }}$. Since $f(z)=O(z^{-2})$, the boundary integral of $F$ tends to $0$.

Applying [L3] and letting the rectangle expand gives ${\sum }_{n\in \mathbb{Z}}(-1{)}^{n}f(n)+{\sum }_{a\mathrm{\notin }\mathbb{Z}}\mathrm{Res}(\pi \csc (\pi z)f(z),a)=0.$ Rearranging yields the stated alternating summation formula.

At every nonzero integer $n$, the function $\pi \cot (\pi z)$ has residue $1$, so $F$ has residue $1\mathrm{/}{n}^2$ there.

Near $0$ one has $\sin (\pi z)=\pi z-\frac{{\pi }^3{z}^3}{6}+O(z^5),\cos (\pi z)=1-\frac{{\pi }^2{z}^2}{2}+O(z^4),$ so $\pi \cot (\pi z)=\frac{1}{z}-\frac{{\pi }^2}{3}z+O(z^3)\mathrm{.}$ Therefore $F(z)=\frac{1}{z^3}-\frac{{\pi }^2}{3z}+O(z),$ and the residue of $F$ at $0$ is $-{\pi }^2\mathrm{/}3$.