Proof for divergence of $\int_1^\infty \cos(x^\frac{3}{4})$ As the title says, I am unable to find a proper proof for the divergence of
$$\int_1^\infty \cos(x^{3/4})$$
As its is not positive, I can't use any of the divergence tests, nor compare it to a sum.
Opening the integral and trying to solve it is resulting in something very ugly, and I doubt that it is the right way to do it.
Could use some help, thank you.
 A: HINT:
Enforce the substitution $x=t^{4/3}$.  Then, $dx=\frac43 t^{1/3}\,dt$. Therefore, we have
$$\int_1^\infty \cos(x^{3/4})\,dx=\frac43\int_1^\infty t^{1/3}\cos (t)\,dt$$
Can you finish from here?
A: Here's a very long approach.
$$(1) \quad e^{ix^p}=\cos(x^p)+\sin(x^p) \cdot i$$
So we can learn a lot from the integral,
$$(2) \quad \int_{0}^{\infty} e^{-x^{p}} \ dx$$
Let's set up a contour integral that goes from $[0,\infty)$; then moves over a circle arc from $R$ to some $\theta \cdot R$ then comes back to the origin. In mathematical form have,
$$(3) \quad \int_C e^{-z^{p}} \ dz=\int_{0}^{R} e^{-x^{p}} \ dx+\color{red}{\int_{C_R} e^{-x^{p}} \ dx}-\color{blue}{\int_{0}^{R \cdot \theta} e^{-x^{p}} \ dx}$$
We'd like to pick $\theta$ so that the red integral goes away and the blue one is equivalent to an integral over $(1)$. If substitute $x=u \cdot \theta$ into the blue integral, we get,
$$(4) \quad \int_C e^{-z^{p}} \ dz=\int_{0}^{R} e^{-x^{p}} \ dx+\color{red}{\int_{C_R} e^{-x^{p}} \ dx}-\theta \cdot \color{blue}{\int_{0}^{R} e^{-\theta^{p} \cdot x^{p}} \ dx}$$
So we need $-\theta^{p}=i$. One solution is $\theta=e^{-\cfrac{\pi}{2p} \cdot i}$. This makes the red integral zero as long as $p \gt 1$. If $p \lt 1$ then the red integral doesn't go to zero, and thus the contour integral doesn't converge, and as well see, the real integral is finite. This would imply that the integral in question diverges. Since there are no singularities enclosed by the contour, the integral is $0$. So we have,
$$0=\int_{0}^{R} e^{-x^{p}} \ dx-\theta \cdot \color{blue}{\int_{0}^{R} e^{i \cdot x^{p}} \ dx}$$
Which implies that,
$$\int_{0}^{R} e^{i \cdot x^{p}} \ dx=\cfrac{1}{\theta} \cdot \int_{0}^{R} e^{-x^{p}} \ dx$$
If we take the real part of this, and take the limit as $R$ approaches infinity, we have the relation,
$$\int_0^{\infty} \cos(x^p) \ dx=\operatorname{Re}\left(\cfrac{1}{\theta} \cdot \int_{0}^{\infty} e^{-x^{p}} \ dx \right)$$
Recall that, 
$$(5) \quad \int_{0}^{\infty} e^{-x^{p}} \ dx=\cfrac{\Gamma\left(\cfrac{1}{p}\right)}{p}$$
So we have,
$$(6) \quad \int_0^{\infty} \cos(x^p) \ dx=\cfrac{1}{p} \cdot \Gamma\left(\cfrac{1}{p}\right) \cdot \cos\left(\cfrac{\pi}{2p}\right) \ , \quad p \gt 1$$
Since $\cos(x^p)$ has a finite range on $[0,1]$ the results that follow hold equally for your integral.
$$\int_1^{\infty} \cos(x^{3/4}) \ dx$$
diverges because $p=3/4 \lt 1$.
A: Denote $f(x)=\cos x^{3/4}$.
Let $a_k = (2k\pi+\pi/2)^{4/3}$ and $b_k = (2k\pi+3\pi/2)^{4/3}$.
These are the zeros of the integrand, and are order in the following way: $a_0 < b_0 < a_1 < b_1 < \dotsb$. What can we say about
$$A_{k} := \int_{a_{k}}^{b_{k}} f(x)dx \text{ and } B_{k} := \int_{b_{k}}^{a_{k+1}} f(x) dx$$
Hint: $A_k>0$ is an increasing sequence and $B_k<0$ is decreasing, and
$$ \int_1^\infty f(x) dx = \int_1^{a_0} f(x) dx + A_0 + B_0 + A_1 + B_1 + \dotsm$$
(has to be taken properly, this holds if the integral was convergent; since the sum on the right is divergent, the integral is divergent as well.)
