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In this post - Integrating $\int_0^\infty \frac{1-\cos x }{x^2}dx$ via contour integral. - I understood upto the point where $|\frac{1-e^{iz}}{z^2}| \leq \frac{2}{|z^2|}$. I didn't understand the next sentence equating $\int_{|x|>\epsilon} \frac{1-e^{x^2}}{x^2} dx$.

I think it should be 1/2 the quantity on the right, since there are two values of $\int_{|x|>\epsilon} \frac{1-e^{x^2}}{x^2} dx$ in the original equation. Can someone explain this?

I am doing a self-study of complex-analysis by Shakarchi.

Bernard
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u_any_45
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  • All they are doing is to let $R \to \infty$ in the previous equation. – Kavi Rama Murthy Sep 07 '20 at 23:25
  • @KaviRamaMurthy can you be more specific? There are two integrals where the integrand is $\frac{1-e^{ix}}{x^2}$. One goes from -R to $-\epsilon$ and the other goes from $\epsilon$ to R. – u_any_45 Sep 07 '20 at 23:48
  • If you add the integral from $-\infty$ to $-\epsilon$ and the integral from $\epsilon$ to $\infty$ you get the integral over $|x| \geq \epsilon$. – Kavi Rama Murthy Sep 07 '20 at 23:55
  • Thank you @KaviRamaMurthy! I have one more question in the same problem - $\int_{\gamma \epsilon +} \frac{1-e^{iz}}{z^2} $ -> $\int_{0}^{\pi} ii d\theta$ where $z=e^{i\theta}$ and $dz=i\epsilon e^{i \theta d\theta}$. How?or how does $1-e^{iz}$ = i – u_any_45 Sep 08 '20 at 00:04

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