# Integral of bounded function with limit zero at $\pm \infty$

Lets say we have a function bounded between $0$ and $1$. This function is high dimensional:
$0<f(X) \le1, ~~~ X \in \mathbb{R}^D$

Now, we calculate the limit for all elements of $X$ going to plus and minus infinity. We find out that they are zero.

Can we say that the integral of the function over the entire domain of $X$ is finite?

Can we say that if we get even non-zero limit?

Finally, if the zero limit is insufficient, is there some other condition that suffices?

Imagine a simple example in 1 dimension $$f(x)=\left\{\begin{array}{c}1/|x|,\quad x>1\\ 0,\quad \mbox{otherwise}\end{array}\right.$$

This function is bounded, and its limit at infinity is zero. However, $\int_{-\infty}^{\infty}f(x)dx=\infty$. the condition $\lim_{x\to\pm\infty}=0$ is a necessary condition for the converging of the integral, but it is not sufficient.

In order to get a convergence, you need an extra assymptotic behavior of $f(x)$: it has to go to zero faster than $1/x$. For example, the function

$$f(x)=\left\{\begin{array}{c}1/|x|^p,\quad x>1\\ 0,\quad \mbox{otherwise}\end{array}\right.$$ with converges $\forall p\in(1,\infty)$

• That messes my brain up a lot. I changed the question slightly, is that ok? Apr 29, 2016 at 7:25
• Well, I would assume (with great caution) that this is not very important. I mean this function does have limit zero and is defined over $R^D$, in a bounded way. Apr 29, 2016 at 7:28
• @gebruiker Try $f(x)=\min\left(1,\frac{1}{||X||}\right)$ to extend to the multidimensions of the original question Apr 29, 2016 at 7:28
• Great example, I really like how you gave the marginal case as a counter example. Apr 29, 2016 at 7:35

No, you cannot conclude that the integral is finite.

As an example take a function like $$f(x,y)=\min\{1, 1/\sqrt{xy}\}$$ This will be bounded between $0$ and $1$, but for large $x$ and $y$ the function will be $1/\sqrt {xy}$ and its integral behaves like $\sqrt{xy}$ and thus is not finite.

• Is there an additional condition that will suffice? Apr 29, 2016 at 7:33
• See the edit in seoanes answer. Apr 29, 2016 at 7:35