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+\def\Re{\mathop{Re}\nolimits}
+\def\Res{\mathop{Res}}
+\let\rule\hrule
+\def\fr#1#2{{#1\over#2}}
+\def\hrule{\goodbreak\medskip\rule\medskip\goodbreak}
+
+% page 282
+\noindent{\bf 1.}
+
+With
+$$f(z) = {e^{iaz} - e^{ibz}\over z^2},$$
+there is one singularity at 0, so by Cauchy-Goursat,
+$$\int_{C_\rho} f(z)dz + \int_{C_R} f(z)dz + \int_{-\infty}^0 f(z)dz +
+\int_0^\infty f(z)dz = 0,$$
+where $C_\rho$ is the upper semicircle about $z=0,$ of radius
+$\rho\to0,$ and similar for $C_R$ where $R\to\infty.$
+
+$$\int_{C_R} f(z)dz = \int_{C_R} {e^{iaz} - e^{ibz}\over z^2}dz
+= 0,$$
+by Jordan's lemma.
+$$\int_{C_\rho} {e^{iaz} - e^{ibz} \over z^2} dz = -B_0\pi i = \pi(a-b),$$
+by the theorem in this section.
+$$\int_{-\infty}^0 f(x)dx + \int_0^\infty f(x)dx =
+-\int_0^\infty f(-x)dx + \int_0^\infty f(x)dx =$$$$
+\int_0^\infty {e^{iax} - e^{ibx} + e^{-iax} - e^{-ibx}\over x^2}dx
+% logical discontinuity! It should maybe be - e^{-iax} + e^{-ibx}.
+= 2\int{\cos(ax)-\cos(bx)\over x^2}dx,$$
+which, by transformation of the original equality, gives
+$$2\int{\cos(ax)-\cos(bx)\over x^2}dx = -\pi(a-b) \to \int f(x)dx =
+{\pi\over2}(b-a).$$
+
+\noindent{\bf 4.}
+
+$$f(z) = {z^{1/3}\over(z+a)(z+b)} = {e^{(1/3)\log z}\over(z+a)(z+b)}.$$
+$$\int_\rho^R {r^{1/3}\over (r+a)(r+b)}dr +
+\int_{C_R} f(z)dz - \int_\rho^R {r^{1/3}e^{2i\pi/3}\over (r+a)(r+b)}dr +
+\int_{C_\rho} f(z)dz = 2\pi i(\Res_{z=-a}f(z) + \Res_{z=-b}f(z)).$$
+
+These residues are at simple poles and are therefore
+$${a^{1/3}e^{\pi i/3}\over b-a}\qquad{\rm and}\qquad
+{b^{1/3}e^{\pi i/3}\over a-b},$$
+respectively.
+
+$$\left|\int_{C_\rho} f(z)dz\right| \leq
+{\rho^{1/3}\over(a+\rho)(b+\rho)}2\pi\rho \to 0$$
+$$\left|\int_{C_R} f(z)dz\right| \leq {R^{1/3}\over (R+a)(R+b)}2\pi R\to
+0.$$
+
+Therefore, rearranging the original equality,
+$$(1-e^{2i\pi/3})\int_0^\infty {r^{1/3}\over (r+a)(r+b)}dr
+= 2\pi ie^{\pi i/3}{a^{1/3}-b^{1/3}\over b-a}$$
+$$\longrightarrow \int_0^\infty {x^{1/3}\over (x+a)(x+b)}dx =
+{2\pi\over\sqrt 3}\cdot{\root 3 \of a - \root 3 \of b\over b - a}$$
+
+\hrule
+% page 287
+
+\noindent{\bf 5.}
+
+$$\int_0^\pi {d\theta\over (a+\cos\theta)^2} = {1\over 2}\int_0^{2\pi}
+{d\theta\over (a+\cos\theta)^2},$$
+by symmetry of the integrand,
+$$=\fr12\int_C {1\over (a+{z+z^{-1}\over 2})^2}{dz\over iz} =
+\fr2i\int_C {z\over (z^2+2az+1)^2}.$$
+which has two singularities, those being
+$$z^2 + 2az + 1 = 0 \to z = \pm\sqrt{a^2 - 1} - a.$$
+Only the more positive singularity $z_0$ is within $|z|<1,$ where $a >
+1.$ The integrand can be rewritten as
+$${\phi(z)\over (z-z_0)^2}$$
+where $\phi(z) = {z\over (z-z_1)^2},$ $z_1$ being the more negative
+singularity.
+The singularity has residue
+$$B_0 = \phi'(z_0) = {-z_1-z_0\over (z_0-z_1)^3}
+= {2a\over 8\sqrt{a^2-1}^3},$$
+giving the integral value
+$$\int_0^\pi {d\theta\over (a+\cos\theta)^2} = 2\pi i\fr2i\cdot{2a\over
+8\sqrt{a^2-1}^3} = {a\pi\over\sqrt{a^2-1}^3}.$$
+
+\bye