From the case $\alpha = 0$ we know that $\int_1^\infty \cos(x^2) \, \mathrm{d}x$ exists. Therefore the function $(1,\infty) \to \mathbb{R}, y \mapsto \int_1^y \cos(x^2) \, \mathrm{d}x \, ,$ is bounded. The function $x \mapsto \left[\frac{\pi}{2} - \arctan(\sqrt{x})\right]^\alpha$ is decreasing on $(1,\infty)$ with limit $0$ for $\alpha >0$ . We can now apply Dirichlet's test again to conclude that the improper Riemann integral converges for every $\alpha > 0$ .
Note that we have absolute convergence if and only if $\alpha > 2$ . For $\alpha = 2$ the integrand is asymptotically equivalent to $\frac{|\cos(x^2)|}{x}$ and the integral of this function diverges.
Make a substitution $x=e^u$, the integral becomes
$$\int^\infty_{\ln 3} \exp(u-u^a+1)du$$
As an alternative form of p-test,
$$\int^\infty_1 \exp(f(x))dx$$
converges only if $\lim_{x\to\infty}\frac{f(x)}{\ln x}<-1$.
We need $a>1$ for convergence, because when $a>1$, $\lim_{u\to\infty}\frac{u-u^a+1}{\ln u}=-\infty$.
In addition, when $a=1$, $\lim_{u\to\infty}\frac{u-u^a+1}{\ln u}=0\not< -1$.
When $a<1$, $\lim_{u\to\infty}\frac{u-u^a+1}{\ln u}=+\infty\not<-1$.
Why does the alternative p-test work?
I think it is better to give an intuitive explanation. Firstly, for $f(x)=k\ln x$, p-test implies convergence only for $k<-1$.
If another $f(x)$ diverges to $-\infty$ quicker than $-\ln x$, then $e^{f(x)}$ drops to zero quicker, and by comparison test the integral will converge. In this case, As $f$ goes negative more quickly, $f(x)<-\ln x$ for large $x$, meaning that $$\lim_{x\to\infty}\frac{f(x)}{-\ln x}>1$$
Similarly, if another $f(x)$ goes negative slower than/at the same rate as $-\ln x$, by comparison test the integral diverges. In this case, $f(x)\ge-\ln x$ for large $x$, meaning that $$\lim_{x\to\infty}\frac{f(x)}{-\ln x}\le 1$$
It can be rigorously proved that $$\int^\infty_1\exp{f(x)}dx\text{ converges iff }\lim_{x\to\infty}\frac{f(x)}{-\ln x}>1$$ provided that $f$ is continuous in $[1,\infty)$.
Best Answer
$$\int_{0}^{1}\frac{dx}{\sqrt[3]{e^x-1}}\stackrel{x\mapsto \log t}{=}\int_{1}^{e}\frac{dt}{t\sqrt[3]{t-1}}\stackrel{t\mapsto s+1}{=}\int_{0}^{e-1}\frac{ds}{(s-1)\sqrt[3]{s}}\stackrel{s\mapsto u^3}{=}\int_{0}^{\sqrt[3]{e-1}}\frac{3u}{u^3-1}\,du$$ where $$ \frac{3u}{u^3-1} \stackrel{\text{Residues}}{=} \frac{1}{u-1}+\frac{1-u}{u^2+u+1} $$ and $$ \frac{1-u}{u^2+u+1} = \frac{4}{(2u+1)^2+3}-\frac{4u}{(2u+1)^2+3} $$ lead to the answer in less than a page. This is a trick you are probably going to apply many times: turn the integrand function into a rational function, then perform a partial fraction decomposition. Objects like $\frac{1}{(u+\alpha)^n}$ and $\frac{1}{(u^2+\alpha^2)^n}$ always have simple primitives, that's the core of symbolic integration algorithms.