Disproving convergence of an integeal I remember that there is a calculus statement that says the following:
Assume you have an integral of a function (analytical, 1D) with infinite upper limit - if the limit of the function is not zero - the integral diverge.

More formally:
Assume f(x) is analytical function over the interval $[1,\infty]$
If 
$\lim_{x\to \infty} f(x)$ $\neq$ 0
Then
$\int_{1}^\infty f(x)dx$
Diverge

Does this kind of statement exists? If so what is it called?

Would I be able to use the above to prove that 
$\int_{1}^\infty x^4 \cos(x^3)dx$
diverge?

Thanks in advance
The way you can proceed is by evaluating this integral for a finite limit i.e.

$$I(l) = \int_{1}^{l}x^{4}cos(x^3)dx$$

Then you can analyse 
$$L = \lim_{l\rightarrow \infty} I(l)$$
You need to distinguish between the limit existing and being nonzero, and the function not having a limit. The former is what you are talking about in the theorem, the latter is what occurs in the example.


The statement you give is true. It is akin to the result about convergent series' terms converging to zero: suppose that $ f(x) \to a \neq 0 $. Then given $\varepsilon&gt;0$, there is an $N$ so that $ \lvert f(y) - a \rvert &lt; \varepsilon $ for $y&gt;N$. In particular this is true with $\varepsilon=a/2$ But then if $y,z&gt;N$, $ \int_y^z f $ lies between $ (z-y)a/2 $ and $ 3(z-y)a/2 $. But this is not bounded as $z \to \infty$, so the improper integral does not satisfy the Cauchy criterion and so does not converge.
For the example, the function does not converge to a limit. This does not tell you anything in itself: for example, $ \int_0^{\infty} \sin(x^2) \, dx $ converges, but the integrand does not tend to a limit. You have to compute manually what is going on. What you can do is change variables to $ t = x^{1/3} $, turning the finite integral into a multiple of $ \int_1^{b^{3}} t^{2/3} \cos t \, dt $, which can be integrated by parts to produce an integral that we can recognise tends to a limit, plus a term that diverges.