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I think this integral does not converge. I want to estimate downward the integral, but don't know how to.

Daniel Fischer
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Scheider
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    Substitute $u = \frac{1}{x}$. – Daniel Fischer Oct 29 '14 at 13:15
  • $-\int_{0}^{\infty} \frac{\sin^2 (u)}{u^2}\mathrm{d}u$ and then? – Scheider Oct 29 '14 at 13:21
  • And then you see that the integral is finite. (And you have a sign error.) – Daniel Fischer Oct 29 '14 at 13:24
  • http://latex.codecogs.com/gif.latex?%5Cleft%20%7C%5Cint_%7B0%7D%5E%7B1%7D%5Cfrac%7B%5Csin%20%5E2%28u%29%7D%7Bu%5E2%7D%20%5Cright%20%7C%3C%5Cleft%20%7C%5Cint_%7B0%7D%5E%7B1%7D%5Cfrac%7B%5Csin%20%28u%29%7D%7Bu%7D%20%5Cright%20%7C%3C%5Cleft%20%7C%20%5Cint_%7B0%7D%5E%7B1%7D%20%5Cfrac%7B1%7D%5Csqrt%7B%7Bu%7D%7D%20%5Cright%20%7C is this correct? – Scheider Oct 29 '14 at 13:36

2 Answers2

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Set $u=1/x$, then $du=-dx/x^2$, and $$ \int_0^\infty \sin^2\left(\frac{1}{x}\right)\,dx=\int_0^\infty\frac{\sin^2u\,du}{u^2}, $$ which converges, as $$ \frac{\sin^2u}{u^2}\le \max\left\{\frac{1}{u^2},1\right\}. $$

Note. Using residue calculus one could obtain that $$ \int_0^\infty\frac{\sin^2u\,du}{u^2}=\frac{\pi}{2}. $$

Yiorgos S. Smyrlis
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Hint. Using Daniel's suggestion, you are led to evaluate $$ \int_0^{+\infty}\frac{\sin^2 x}{x^2}dx $$ which is equal to $$ \int_0^{+\infty}\frac{\sin x}{x}dx=\frac{\pi}{2} $$ by an integration by parts applied on $[\epsilon,M]$ (then $\epsilon \rightarrow 0$, $M \rightarrow \infty.$)

Olivier Oloa
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