Consider the function $f:\mathbb{R}\to\mathbb{R}$ given by
$$f(x)=x\sin x.$$
If $x$ is sufficiently large, I mean $R>0$ sufficiently large exists such that $x$ is in the complementar of the ball $B(0, R)$, it is possible to have a lower bound for the function $f$?
Actually I think to the limit for $x\to +\infty$ fo that function which is indeterminate, so I think that it is not possible to find a lower bound, but I am not sure about that.
${\bf EDIT:}$ So, in articular, it is possible or not to have a lower bound for the integral
$$\int_0^T f(x) dx,$$
with $T>0$, isn't it?
Could someone please help me with that?
Thank you in advance!
Best Answer
Consider the following sequences, where $k$ goes to infinity.
$$f(\pi/2 + k2\pi) = \pi/2 + k2\pi \rightarrow +\infty$$
$$f(-\pi/2 + k2\pi) = \pi/2 - k2\pi \rightarrow -\infty$$
You can compute the integral explicitely with an integration by part.
$$ \int_0^T x \sin(x) dx = [-x \cos(x)]_0^T - \int_0^T \cos(x) dx = -T \cos(T) +\sin(T).$$
Here again you can find sequences for which this last quantity goes to $-\infty$ or $+\infty$.