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
 A: 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 
$$L = \lim_{l\rightarrow \infty} I(l)$$
A: 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 $\varepsilon>0$, there is an $N$ so that $ \lvert f(y) - a \rvert < \varepsilon $ for $y>N$. In particular this is true with $\varepsilon=a/2$ But then if $y,z>N$, $ \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.
