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I'm reading through Stein and Shakarchi's Complex Analysis textbook, but I'm a bit confused by their proof that $$ \int_{0}^{\infty} \frac{1-\cos(x)}{x^2} dx = \frac{\pi}{2}$$

They consider the function $f(z) = \frac{1-e^{iz}}{z^2}$ and an indented semicircle as their contourenter image description here

The part where I'm confused is the integral of $f(z)$ over $\gamma_{\epsilon}^+$. The way they evaluate this integral is by first noting that $$ f(z) = \frac{-iz}{z^2} + E(z) $$ where $E(z)$ is bounded as $z\rightarrow 0$. I'm fine with the rest of the proof but I'm puzzled by $E(z)$.

My question: How is this function bounded as $z \rightarrow 0$?

It seems like $E(z)$ would just be $$ E(z) = \frac{1+iz+e^{iz}}{z^2}$$

Is it because we're integrating over $\gamma_{\epsilon}^+$ and so $|z| = \epsilon$ and so

$$ \left| E(z) \right| = \left| \frac{1}{z} + \frac{i}{z} + \frac{e^{iz}}{z^2} \right| \leq \left|\frac{1}{z} \right| + \left| \frac{i}{z} \right| + \left| \frac{e^{iz}}{z^2} \right| = \frac{1}{\epsilon} + \frac{1}{\epsilon^2} + \frac{1}{\epsilon^2} $$?

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    $\begingroup$ For $z \neq 0$, $f(z)+{iz \over z^2} = {1+iz - e^{iz} \over z^2} = \sum_{k=2}^\infty { i^k \over k!}z^{k-2}$, the latter is analytic everywhere. $\endgroup$
    – copper.hat
    Mar 2 at 19:45
  • $\begingroup$ @copper Sorry your comment beat my answer. It took me 5 minutes to find a } that should have been a ). $\endgroup$ Mar 2 at 20:01
  • $\begingroup$ @TedShifrin :-) np at all. $\endgroup$
    – copper.hat
    Mar 2 at 20:13

1 Answer 1

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No, \begin{align*} E(z) &= \frac{1-e^{iz}}{z^2} - \frac{-iz}{z^2} = \frac{1-\left(1+iz+\frac{(iz)^2}2 + \frac{(iz)^3}6 +\dots\right)}{z^2} - \frac{-iz}{z^2} \\ &= \frac{-\left(\frac{(iz)^2}2 + \frac{(iz)^3}6 +\dots\right)}{z^2} = \frac12 + O(z). \end{align*} It appears you lost a negative sign in writing down your formula for $E(z)$.

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