<s>[INST] 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!
 [/INST] 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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