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How to prove that \begin{equation}\nonumber \int_0^\infty \sin^2\left[\pi\left(x + \frac{1}{x}\right)\right]dx \end{equation} does not exist?

Amey Joshi
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    For large $x$ we have $x+1/x\approx x$. Thus we have something like $\int_M^\infty \sin^2 u,du$ at the upper end. This contributes $\pi$ for each period of $u$ of length $2\pi$, and those contributions add up. I would try $u=x+1/x$, when you can bound $du/dx$ away from zero for large enough $x$. – Jyrki Lahtonen Jun 11 '14 at 09:32
  • Do you mean "not exist" or "not converge"? – draks ... Jun 11 '14 at 10:00
  • @Draks: Yes, I meant "does not converge". – Amey Joshi Jun 11 '14 at 10:05

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Write it as $\int_0^\infty \frac12 (1- \cos(\pi(2x+2/x)))dx$. Even the constant factor $\frac12$ results in an infinite integral...

draks ...
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  • Thanks. However, do we need to show that $\smallint_0^\infty \cos(\pi(2x + 2/x))dx$ is convergent? It should not happen that we write the integral as a sum of two terms each of which is separately divergent but the sum ends up being convergent. – Amey Joshi Jun 11 '14 at 10:10
  • To my knowlegde even $\int_0^\infty \cos(x)$ doesn't have an explicite value, so I don't expect one for your integral. Jyrki's approach seems to be a good one... – draks ... Jun 11 '14 at 10:14