Understanding the limit $\lim_{t \to \infty} \int_0^\infty \frac{t}{t^2 + x} \sin(1/x)dx$ 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!
 A: 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.
A: 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.
