$X\to\infty$ and $Y\to\infty$ independently I was reading this definition
Definition Suppose $\int_{-X}^{Y} f(t) \, dt $ exists for all $Y>-X$. If $\int_{-X}^{Y} f(t) \, dt $ tends to a finite limit as $X\to\infty$ and $Y\to\infty$ independently, then that  $\int_{-\infty}^{\infty} f(t) \, dt $ exists.
My question is
what does it mean that  $X\to\infty$ and $Y\to\infty$ independently? Also $X\to\infty$ and $Y\to\infty$ dependently?
 A: I'm assumming that $X,Y>0$. Here, $X\to \infty, Y\to\infty$ 'independently' means that $X,Y$ cannot have any relationships with one another. For example, $Y=X^2$ or $Y=X$ means that they are not independent (and thus are dependent). It is important that the integral converges to a unique limit as $X,Y\to\infty$ independently because we want to rule out certain cases where the integral will converge to a finite limit when $X,Y$ are in a certain relationship. For example, if $Y=X$ then
$$\int_{-X}^Y \sin(t) \, dt =0$$
and thus, as $X,Y\to\infty$ (dependently in this case), it looks as though
$$\int_{-\infty}^\infty \sin(t) \, dt =0$$
For the rigorous definition, the integral $$\int_{-X}^Y f(t) \,dt$$ tends to a finite limit as $X\to \infty, Y\to \infty$ independently means that there is a number $L$ such that for any $\epsilon >0$, there will be a $M\in\mathbb{N}$ such that if $X,Y>M$ then $$\left| \int_{-X}^Y f(t) \, dt - L \right|<\epsilon$$
Notice that the definition only requires $X,Y$ to be larger than $M$, not by how much. Thus, $X$ can be much larger, smaller or just equal to $Y$ and so they are independent in this sense.
