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I need to prove that:

$$\int_0^\infty \frac{\cos x}{1+x} \ dx = \int_0^\infty \frac{\sin x}{(1+x)^2}\ dx$$

but, one converges absolutely whereas the other is not.

I've tried few things like subtitution, integration by parts. I've also tried to use linearty and substract one from the other, hoping the result would be zero. Yet, none of those worked for me. I guess it's probably demanding some creative subtitution I can't see but I'm not sure.

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    $\begingroup$ Integration by parts is good. $\endgroup$ – André Nicolas Nov 18 '15 at 22:00
  • $\begingroup$ Right! Thank you @AndréNicolas! $\endgroup$ – hibye Nov 18 '15 at 22:03
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You may integrate by parts, obtaining $$ \int_0^M \frac{\cos x}{1+x} \ dx =\frac{\sin M}{1+M} + \int_0^M \frac{\sin x}{(1+x)^2}\ dx, $$ then let $M \to \infty$ to get the announced result.

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  • $\begingroup$ Oh right. I just figured this too! Thanks! $\endgroup$ – hibye Nov 18 '15 at 22:02
  • $\begingroup$ You are welcome! $\endgroup$ – Olivier Oloa Nov 18 '15 at 22:03
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$$\int_{0}^{N}\frac{\cos x}{1+x}\,dx = \left.\frac{\sin x}{1+x}\right|_{0}^{N}+\int_{0}^{N}\frac{\sin x}{(1+x)^2}\,dx $$ and the first term of the RHS is $O\left(\frac{1}{N}\right)$.

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  • $\begingroup$ Missing a $dx$ on the LHS integral. $\endgroup$ – Brian Tung Nov 18 '15 at 22:04

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