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Prove :

$$\lim_{t\to +\infty}\int_{0}^{\infty}\frac{t}{t^2 +x} \sin\frac{1}{x} dx=0$$

Maybe dominated convergence theorem? Who can give a proof?

Thanks!

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  • $\begingroup$ did you forget an integral sign? $\endgroup$ – SalmonKiller Dec 6 '14 at 1:14
  • $\begingroup$ Sorry, I have corrected it. $\endgroup$ – Kira Yamato Dec 6 '14 at 1:24
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We can assume $t>1$ - we just throw out the beginning... Since for $y<1$, $\sin y<y$, thus for $x>1$ $$\sin\frac{1}{x}<\frac{1}{x}$$ By AM-GM: $$|\frac{t}{t^2+x}|<\frac{1}{t+\frac{x}{t}}<\frac{1}{2\sqrt{x}}$$ Also note that for $x\leq1$: $$\frac{1}{1+x}-\frac{t}{t^2+x}=\frac{t^2-t+t(1-x)}{(1+x)(t^2+x)}>0$$ So we have a dominating function for $t>1$, as we used $1\geq \sin x$: $$g(x)=\cases{ \frac{1}{1+x}& $0<x<1$\\ \frac{1}{2x\sqrt{x}}& $x>1$}$$ But $$\int_0^\infty g(x)\text{d}x=\int_0^1 \frac{1}{1+x}\text{d}x+\int_1^\infty \frac{1}{2x\sqrt{x}}\text{d}x=1+\log 2$$ And apply dominated convergence.

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Suppose $g(x)=\sup\left\{ \left|\dfrac t{t^2+x}\sin\dfrac 1 x\right| : 0\le t\right\}$. The supremum as a function of $t$ is attained when $t=\sqrt x$ and we find $g(x) = \dfrac1{2\sqrt x}\left|\sin\dfrac1 x\right|$. If one can show that $\displaystyle\int_0^\infty g(x)\,dx<\infty$ then the dominated convergence theorem will do it.

Notice that for $x>\text{some positive number}$ we have $0<\sin\dfrac1x<\dfrac1x$. So what can be said about $$ \int_{\text{some positive number}}^\infty \dfrac{1}{2x\sqrt x}\,dx\text{ ?} $$

As $x\to0$, the function being integrated remains bounded as a function of $x$, so it is integrable over any bounded interval.

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