Investigating convergence of $\int_{0}^{\infty} \cos(x^r) dx $ and $\int_{\pi}^{\infty} \frac{\cos x}{\log x}\arctan\lfloor x\rfloor dx$ I want to investigate the convergence behavior of $$\int_{0}^{\infty} \cos(x^r)\, dx  \hspace{5mm} \textrm{and} \hspace{5mm}  \int_{\pi}^{\infty} \left(\frac{\cos x}{\log x}\right)\arctan\lfloor x\rfloor dx.$$ My theoretical tools are Abel's test and Dirichlet's test:
Say I have an integral of the form $$\int_{a}^{b}f\cdot g \hspace{1.5mm}  dx$$
with improperness (vertical or horizontal asymptote) at $b$.
Abel's test guarantees convergence if $g$ is monotone and bounded on $(a,b)$, and $\int_{a}^{b}f $ converges. Dirichlet's test guarantees convergence if $g$ is monotone on $(a,b)$ and $\displaystyle\lim_{x\to b} g(x) = 0$, and $\displaystyle\lim_{\beta \to b}$ $\int_{a}^{\beta}f $ is bounded.
For the first integral $\displaystyle\int_{0}^{\infty} \cos(x^r)\, dx $ I'm guessing a substitution $t = x^r $ will give me an expression of the form $f\cdot g$ with $\cos(t)$ as my $f$. For the second integral $\displaystyle\int_{\pi}^{\infty} \dfrac{\cos x}{\log x}\arctan\lfloor x\rfloor\, dx$, I'm (even more) clueless. Help please?
 A: Hint for the second problem: A variant of Abel/Dirichlet for sums is the following: If $a_n$ is a monotonic convergent sequence and $\sum b_n$ converges, then $\sum a_nb_n$ converges. Proof sketch: Use the Cauchy criterion for the partial sums, sum by parts,  ...
In your problem, think of $a_n = \arctan n,$ $b_n = \int_n^{n+1}\cos x/(\ln x)\, dx,$ $n=2,3,\dots $
A: For the first integral your substitution idea works. letting $t=x^r$ gives
$$ \int_0^{\infty} \cos(x^r) \, dx = \frac{1}{r} \int_0^{\infty} \frac{\cos t}{t^{1-1/r}} \, dt $$
Now as long as long as $ 1-1/r <1 $, which happens when $r>0$, we converge at zero. To converge as $x\to \infty$ we need $1-1/r>0 $ and this happens when $r>1$. To see why, write the integral as
$$ \int_0^{\frac{\pi}{2}} \frac{\cos(t)}{t^{1-1/r}} \, dt - \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \frac{\cos(t)}{(t+\pi)^{1-1/r}} \, dt + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \frac{\cos(t)}{(t+2\pi)^{1-1/r}} \, dt - \cdots \\ = \int_0^{\frac{\pi}{2}} \frac{\cos(t)}{t^{1-1/r}} \, dt - \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cos(t) \left[\frac{1}{(t+\pi)^{1-1/r}} - \frac{1}{(t+2\pi)^{1-1/r}}+\frac{1}{(t+3\pi)^{1-1/r}}-\cdots \right] dt $$
So as long as $ \frac{1}{t^{1-1/r}} $ is decreasing the part in the brackets is an alternating series and thus converges, and this happens when the exponent is positive.
