Q: 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,

A: So far everything you did was correct.
I think you should consider a different approach to this question. As far as I can tell there is no closed form to this integral, but you said you just need to show that the integral is convergent.
Don't evaluate it. Do you know other ways of showing that an integral converges?

A: 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
$$ a_{k} = \int_{0}^{\pi} \frac{\cos y}{f'(g(y + k\pi))} \, dy. $$
