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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
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