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One question in a complex analysis test from the past asks to find the real integral $$\int_{0}^{\infty}\frac{1-\cos x}{x^2}dx$$ Of course this should be done by contour integration, by I can't figure out how.

What I have so far in my hand:

  • Trying to integrate $\frac{1-\cos z}{z^2}$ on an arc (part of a circle) doesn't seem to yield a nice expression.
  • Note that this function has a removable singularity at $0$ so it's analytic in $\mathbb{C}$. I can also find the Taylor series but I don't think that's the way.
  • This integral equals (by trig identities) $\int_{0}^{\infty}\frac{sin^2u}{u^2}du$ but I don't know if it contirbutes in any way.

Any ideas?

35T41
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2 Answers2

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Hint

Using an integration by part, one can remark that $$\int_0^\infty \frac{1-\cos(x)}{x^2}dx=\int_0^\infty \frac{\sin(x)}{x}dx.$$

Consider $$f(z)=\frac{e^{iz}}{z}$$ on $$D=\{z\in\mathbb C\mid \varepsilon<|z|<R, \Im(z)\geq 0\}.$$

Surb
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$\int_0^{+\infty} \frac{1-cosx}{x^2} = \int_0^{+\infty}\int_0^1 \frac{sin(\alpha x)}{x}d\alpha dx = \int_0^1 \int_0^{+\infty} \frac{sin(\alpha x)}{x}dxd\alpha = \int_0^1 \frac{\pi}{2}d \alpha = \pi /2$

D F
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