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+\def\Res{\mathop{\rm Res}}
+\def\Re{\mathop{\rm Re}\nolimits}
+\def\Im{\mathop{\rm Im}\nolimits}
+\let\rule\hrule
+\def\hrule{\goodbreak\medskip\rule\medskip\goodbreak}
+
+%page 253
+
+\noindent{\bf 4.}
+
+\noindent{\it (a)}
+
+$$(\sec z)^{-1} = \cos z = q(z).$$
+$$p(z) = z.$$
+$$p(\pi/2+n\pi) = \pi/2 + n\pi = z_n\neq 0, \qquad q(\pi/2+n\pi) = 0,
+\qquad q'(\pi/2+n\pi) = -\sin(\pi/2+n\pi) = (-1)^{n+1} \neq 0,$$
+so this is a simple pole with residue
+$$\Res_{z=z_n}(z\sec z) = (-1)^nz_n.$$
+
+\noindent{\it (b)}
+
+Similarly,
+$$(\tanh(z)) = {\sinh z\over\cosh z},$$
+giving $p(z) = \sinh z$ and $q(z) = \cosh z,$
+which forms a simple pole because
+$$p(z_n) = -\sin(\pi/2 + n\pi) = (-1)^n, \qquad q(z_n) = \cos(\pi/2 +
+n\pi) = 0, \qquad q'(z_n) = \sin(\pi/2 + n\pi) = (-1)^n.$$
+
+$$p(z_n)/q'(z_n) = (-1)^n/(-1)^n = 1 = \Res_{z=z_n}\tanh z$$
+
+\noindent{\bf 6.}
+
+The value of this integral is equal to $2\pi i$ times the sum of the
+residues of $$f(z) = {1\over z^2\sin z}$$ within the square.
+These are $z^2 = 0 \to z = 0$ and $\sin z = 0 \to z = z_n = \pi n,$
+where $-N\leq n\leq N.$
+This set includes $z = 0.$
+
+$$f(z) = {p(z)\over q(z)}$$
+where $p(z) = 1$ and $q(z) = z^2\sin z.$ $p(z_n)\neq 1,$ $q(z_n) = 0,$
+and $q(z_n) = z_n^2\cos z_n + 2z_n\sin z_n = z_n^2(-1)^n \neq 0$ (except
+at $z_n = 0.$)
+The sum of residues within the range of $-N\leq n\leq N$ and $n\neq 0$
+is $$\sum_{n=1}^N (-1)^nz_n^{-2}+z_{-n}^{-2} = \sum_{n=1}^N {2(-1)^n\over(\pi
+n)^2}.$$
+
+At $z_n = 0,$
+$$f(z) = {\phi(z)\over z^3},$$
+where $\phi(z) = {z\over\sin z},$
+This gives a residue at 0 of ${\phi''(0)\over 3!}$.
+$$\phi''(z) = -{\cos z\over(\sin z)^2} - {\cos z - z\sin z\over(\sin z)^2}
++ {2z(\cos z)^2\over(\sin z)^3} =
+{-2\cos z\sin z + z(\sin z)^2 + 2z(\cos z)^2 \over (\sin z)^3}.$$
+$$\phi''(0) = 1,$$
+found by noting that, near 0, $\sin z\to z$ and $\cos z\to 1.$
+This gives a 1/6 residue.
+
+Together, this makes
+$$\int_{C_N} {dz\over z^2\sin z} = 2\pi i\left[{1\over 6}+2\sum_{n=1}^N
+{(-1)^n\over\pi^2 n^2}\right].$$
+
+If the integral tends to 0 as $N\to\infty,$
+$$0 = {1\over 6}+2\sum_{n=1}^N{(-1)^n\over \pi^2 n^2} \to 
+-{\pi^2\over 12} = \sum_{n=1}^N{(-1)^n\over n^2},$$
+meaning the series tends to $\pi^2/12.$
+
+\noindent{\bf 10.}
+
+Because $p(z)/q(z)$ has a pole of order $m,$
+$${p(z)\over q(z)} = {\phi(z)\over (z-z_0)^m},$$
+where $\phi(z)$ is analytic and non-zero at $z_0,$ further giving
+$$q(z) = {p(z)\over\phi(z)}(z-z_0)^m,$$
+and because $\phi(z)$ is nonzero, $p(z)/\phi(z)$ is analytic, and this
+tells us that $q(z)$ has a zero of order $m$ because $q(z) =
+g(z)(z-z_0)^m$ where $g(z)$ is a nonzero analytic function.
+
+\hrule
+%page 264
+
+\noindent{\bf 1.}
+
+$$z^2+1 = (z-i)(z+i).$$
+The only singularity above the real axis is $z = i.$
+This point is a simple pole, so its residue is $1/(i+i) = -i/2.$
+This gives the value of the integral as $\pi i*(-i/2) = \pi/2$ (since
+the integrand is even and $\pi R/(R^2-1)$ vanishes to 0 as $R$ tends to
+infinity).
+
+\noindent{\bf 4.}
+
+The singularities of this integrand are, like in the example, the sixth
+roots of $-1,$ and they are simple poles, but the residues $B_k$
+evaluate instead to $$B_k = \Res_{z=c_k} {z^2\over z^6+1} = {c_k^2\over
+6c_k^5} = {c_k^3\over 6c_k^6} = -{c_k^3\over 6}.$$
+$$c_k = e^{i(\pi/6+k\pi/3)} \to c_k^3 = e^{i(\pi/2+k\pi)} = i(-1)^k.$$
+Since $k\in\{0,1,2\},$ the residues sum to $-i/6,$ giving the value of
+the integral to be $\pi i*(-i/6) = -\pi/6,$ (again valid because the
+integrand is even and $\pi R^3/(R^6-1)$ vanishes to 0 as $R$ tends to
+infinity).
+
+\noindent{\bf 8.}
+
+$$q(z)=(z^2+1)(z^2+2z+2) = (z+i)(z-i)(z+(1+i))(z+(1-i)).$$
+The singularities above the real axis are simple poles $z = i, -1+i.$
+At these singularities, the residues $p(z)/q'(z)$ are respectively
+$${i\over (2i)(1+2i)(1)} = {1-2i\over10}$$ and $${-1+i\over
+(-1+2i)(-1)(2i)} = {-1+3i\over 10}.$$
+
+$M_R = {R\over (R^2-1)(R^2-2)} \to R\pi M_R = {R^2\over
+(R^2-1)(R^2-2)},$
+so $\int_{C_R}f(x)dx = 0$ as $R$ tends to infinity.
+This allows us to determine the Cauchy principal value of the integral
+to be $2\pi i\cdot i/10 = -\pi/5.$
+
+\hrule
+%page 273
+
+\noindent{\bf 1.}
+
+$$\int_{-\infty}^\infty {\cos x\thinspace dx\over (x^2+a^2)(x^2+b^2)} =
+\Re \int_{-\infty}^\infty {e^{ix}dx\over (x^2+a^2)(x^2+b^2)}.$$
+The singularities in the upper half of the plane are simple poles
+$x=ai, x=bi.$
+These have residues $${e^{-a}\over (2ai)(-a^2+b^2)}$$ and $$e^{-b}\over
+(2bi)(-b^2+a^2),$$ respectively.
+$$\int_{-\infty}^\infty {e^{ix}dx\over (x^2+a^2)(x^2+b^2)}
+= 2\pi iB - \int_{C_R} f(z)e^{iz} dz.$$
+This last integral tends to $0$ by Jordan's lemma because $f(z)$ has
+maximum value ${1\over (R^2-a^2)(R^2-b^2)},$ which tends to 0 as $R$
+tends to infinity where $R>a$ and $R>b.$
+$$2\pi iB = 2\pi i\left({e^{-a}/2a-e^{-b}/2b \over i(a^2-b^2)}\right) =
+\pi({e^{-a}/a - e^{-b}/b \over (a^2-b^2)}),$$
+proving the integral formula.
+
+\noindent{\bf 5.}
+
+$$\int_{-\infty}^\infty {x^3\sin ax\over x^4+4}dx = \Im
+\int_{-\infty}^\infty {x^3e^{iax}\over x^4+4}dx.$$
+
+$$\int_{-\infty}^\infty {x^3e^{iax}\over x^4+4}dx = 2\pi iB - \int_{C_R}
+f(x)e^{iax}dx.$$
+Where $|x|>R>\sqrt 2,$ the maximum value of $f(x)$ is
+$$M_R<{R^3\over R^4-4},$$
+which tends to $0$ as $R\to\infty,$ telling us the integral evaluates to
+$0.$
+The singularities in the upper half of the plane are $c_k = \sqrt
+2e^{i(\pi/4+k\pi/2)},$ where $k\in\{0,1\}.$
+These are simple poles, so they evaluate to $$p(c_k)/q'(c_k) = {c_k^3
+e^{iac_k}\over 4c_k^3} = e^{iac_k}/4 =
+e^{ia\sqrt2 e^{i(\pi/4+k\pi/2)}}/4 =
+$$$$ e^{ia\sqrt2(\cos(\pi/4+k\pi/2)+i\sin(\pi/4+k\pi/2))}/4
+= e^{ia((-1)^k+i)}/4
+= e^{a(i(-1)^k-1)}/4
+= e^{-a}(\cos(a(-1)^k)+i\sin(a(-1)^k))/4,$$
+summing to $2e^{-a}\cos(a)/4.$
+Therefore, the integral evaluates to $2\pi ie^{-a}\cos a/2,$
+giving imaginary component (and value of the original integral)
+$\pi e^{-a}\cos a.$
+
+\noindent{\bf 8.}
+
+$$\int_{-\infty}^\infty {\sin x dx\over x^2+4x+5} =
+\Im\int_{-\infty}^\infty {e^{ix} dx\over (x+(2+i))(x+(2-i))}$$
+
+The only singularity in the upper half of the plane is $-2+i,$
+with a residue of
+$$e^{-2i-1}\over 2i,$$
+giving the integral a value
+${2\pi ie^{-2i-1}\over 2i} = \pi e^{-1-2i} = {\pi\over e}(\cos(-2)+i\sin(-2)),$
+giving an imaginary component $-\pi\sin(2)/e.$
+
+This answer is valid because $f(z)$ tends to zero on $C_R$ as
+$R\to\infty,$ where $$f(z) = {1\over z^2+4z+5}.$$
+
+\bye