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\begin{document}

{\bf Question}

Show that, for $z$ in the upper half plane, $|\exp iz|\leq1$.

Show that the function

$$f(z)=\frac{1+iz-\exp iz}{z^2}$$

has a removable singularity at $z=0$.

Apply Cauchy's theorem to $f(z)$ using the contour formed by the
real axis from $-R$ to $R$ and the upper half of the circle
$|z|=R$ and, by letting $R\to\infty$, prove that

$$\int_0^\infty\frac{1-\cos x}{x^2}dx=\frac{\pi}{2}.$$


\vspace{0.25in}

{\bf Answer}

$\exp(iz)=\exp ix\exp-y$

So $|\exp(iz)|=\exp-y\leq1$ for $y\geq0$

$\ds f(z)=\frac{1+iz-\exp iz}{z^2}=
\frac{-z^2-iz^3+z^4\cdots}{z^2}$

$\hspace{0.5in}=-1-iz+z^2\cdots\to-1$ as $z\to0$

So $f(z)$ when defined at $z=0$, by $f(0)=-1$ is analytic.

Thus $\ds\int_Cf(z)dz=0$

$\ds\int_{-R}^Rf(x)dx=-\int_{{\rm
semicircle}}f(z)dz=-\int\frac{1}{z^2}-\int\frac{iz}{z^2}-\int\frac{\exp
iz}{z^2}=I_1+I_2+I_3$

$\ds|I_1|\leq\frac{1}{R^2}\pi R\to0$ as $R\to\infty$

$\ds|I_3|\leq\frac{1}{R^2}\pi R\to0$ as $R\to\infty$ as $|\exp
iz|\leq1$ on $C$.

$\ds I_2=-i\int\frac{1}{z}dz=
-i\int_0^\pi\frac{iRe^{i\theta}}{Re^{i\theta}}=\pi$

$\ds\int_{-R}^Rf(x)dx=\int_{-R}^R\frac{1+ix-\cos x-i\sin
x}{x^2}dx=2\int_0^R\frac{1-\cos x}{x^2}dx$

So $\ds\int_0^\infty\frac{1-\cos x}{x^2}dx=\frac{\pi}{2}$




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