Convergence of $ \int_{-\infty}^\infty \cos(x\log(\lvert x\lvert ))\,dx $ Show that the improper integral
$$ \int_{-\infty}^\infty \cos(x\log(\lvert x\lvert ))\,dx $$
is convergent.
I rewrote it, using even function symmetry of cosine, as twice the integral from zero to +infinity.  Now the argument of log is simply $x$, not $|x|$.
Now, to deal with $+infty$ I replace my upper limit with $R$ and will evaluate the limit as R goes to $\infty$.
There is no issue at the origin for $x\log x$, since I think by l'hopital's rule, this is viewed as a convergent sequence - converging to $0$, as $x \to 0$.  So $\cos (x\log x)$ is well-defined for all $x$.
Does anyone have any clever ideas with how to proceed?  I saw an integration by parts method on this site to show convergence of $\cos^3(x)$ (on the positive real line), so I will try this method for $\cos(x\log x)$.  If anyone has any cool tips to offer, please feel free to share :).  
Thanks,

Notice that it suffices to consider the convergence of
$$ \int_{1}^{\infty} \cos (x \log x) \, dx. $$
Now let $f : [1, \infty) \to [0, \infty)$ by $f(x) = x \log x$. This function has an inverse $g = f^{-1}$ which is differentiable. So we have
$$ \int_{1}^{R} \cos (x \log x) \, dx
= \int_{0}^{f(R)} g'(y)\cos y \, dy
= \int_{0}^{f(R)} \frac{\cos y}{f'(g(y))} \, dy. \tag{1} $$
Notice that $f'(g(y)) = \log g(y) + 1$ is increasing and diverges to $+\infty$ as $y \to \infty$. Thus, as $R \to \infty$ it follows that (1) converges from the alternating series test.
Indeed, for large $R$ and $N = N(R) = \lfloor f(R) / \pi \rfloor$ it follows that
\begin{align*}
\int_{1}^{R} \cos (x \log x) \, dx
&= \sum_{k=0}^{N-1} \int_{k\pi}^{(k+1)\pi} \frac{\cos y}{f'(g(y))} \, dy + \int_{\pi N}^{f(R)} \frac{\cos y}{f'(g(y))} \, dy \\
&= \sum_{k=0}^{N-1} (-1)^{k} a_{k} + \mathcal{O}(a_{N}),
\end{align*}
where $a_{k}$ is defined by
$$ a_{k} = \int_{0}^{\pi} \frac{\cos y}{f'(g(y + k\pi))} \, dy. $$
