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I am trying to show that $$\int_{0}^\infty \frac{1-\cos x}{x^2}dx=\pi/2.$$The hint is "try simple substitution", and not incidentally, the previous problem has shown that $\int_0^\infty \frac{\sin^2(xu)}{u^2}du=\frac{\pi}{2}|x|$. This looks an awful lot like we'd like to reduce it to the earlier case, for $x=1$.

What shall we try to substitute for? I think we'd have some problems subbing for cosine, since it does not approach a limit at infinity (correct me if there is a way to make this substitution). Subbing for $x^2$ hasn't gotten me anywhere.

We might want to try to split it up, and see if anything better comes out of trying to integrate $\int_0^\infty \frac{\cos x}{x^2}dx$. No luck there so far.

Any ideas?

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  • $\begingroup$ Thank you for editing my typo! $\endgroup$ – Eric Auld Jul 13 '13 at 22:42
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Hint: Use the identity

$$ \cos(x)=1-2\sin^2(x/2). $$

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    $\begingroup$ Thank you very much. That problem was really bothering me. $\endgroup$ – Eric Auld Jul 13 '13 at 22:41
  • $\begingroup$ @EricAuld: You are very welcome. $\endgroup$ – Mhenni Benghorbal Jul 14 '13 at 3:13
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    $\begingroup$ @MhenniBenghorbal: +1. straightforward hint. $\endgroup$ – Mikasa Jul 17 '13 at 12:41

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