For what $\alpha$ does the integral absolutely and for what conditionally converge? For what $\alpha$ does the integral absolutely and for what conditionally converge ?
$$\int_{0}^{1}\frac{\ln^{\alpha} (1+x^4)}{x^4}\cos{1 \over x}dx$$
Not sure which criteria to use to prove the convergence with a specific $\alpha$, maybe Abel's?
 A: Hint. We have, as $x \to 0$,
$$
\ln(1+x^4)=x^4+\mathcal{O}(x^5)
$$ giving
$$
\frac{\ln^{\alpha}(1+x^4)}{x^4}=\frac1{x^{4-4\alpha}}+\mathcal{O}\left(\frac1{x^{3-4\alpha}}\right)
$$ and 
$$
\frac{\ln^{\alpha}(1+x^4)}{x^4}\cos \frac1x=\frac1{x^{4-4\alpha}}\cos \frac1x+\mathcal{O}\left(\frac1{x^{3-4\alpha}}\right). \tag1
$$
Thus
$$
\int_{0}^1\frac{\ln^{\alpha}(1+x^4)}{x^4}\cos \frac1xdx \quad \text{is absolutely convergent if} \quad 4-4\alpha<1 \quad  \text{that is} \quad \alpha > \frac34.
$$ The preceding condition is necessary. 
For the conditional convergence, one may use $(1)$ and observe that, by a change of variable and by an integration by parts, we have for $b>0$,
$$
\begin{align}
\int_{0}^b\frac1{x^{4-4\alpha}}\cos \frac1xdx&=\int_{1/b}^{\infty}\frac1{u^{4\alpha-2}}\cos u \:du\\\\
&=\left. \frac1{u^{4\alpha-2}}\sin u \right|_{1/b}^{\infty}+2(2\alpha-1)\int_{1/b}^{\infty}\frac1{u^{4\alpha-1}}\sin u \: du.
\end{align}
$$ and the first bit converges for $ 4\alpha-2 >0$ and the integral converges in this case by the Dirichlet test. Then your initial initial integral is conditionally convergent for  $\dfrac12< \alpha \leq \dfrac34 $. The preceding condition is necessary.
