This is a question from an old exam qualifier:
Show that the improper integral $\int_{-\infty}^{\infty}\cos(x\log\left|x\right|)dx$ is convergent.
I first notice that \begin{equation*} \int_{-\infty}^{\infty}\cos(x^{2}) \le \int_{-\infty}^{\infty}\cos(x\log\left|x\right|) \le \int_{-\infty}^{\infty}\cos(x) \end{equation*} where $\int_{-\infty}^{\infty}\cos(x)$ diverges and $\int_{-\infty}^{\infty}\cos(x^{2}) < \infty$.
Next, a quick check on Wolfram alpha suggests this integral will diverge.
Regardless, I attempt to explore the integral further to see if I can determine for myself whether or not the integral converges.
My attempt is to use integration by parts as done in this post: Using, \begin{equation*} \int_{-\infty}^{\infty}\cos(x\log\left|x\right|)dx = \int_{-\infty}^{-1}\cos(x\log\left|x\right|)dx + \int_{-1}^{1}\cos(x\log\left|x\right|)dx + \int_{1}^{\infty}\cos(x\log\left|x\right|)dx \end{equation*}
I find that \begin{alignat*}{2} \int_{1}^{\infty}\cos(x\log\left|x\right|)dx &= \int_{1}^{\infty}\frac{\log\left|x\right|+1}{\log\left|x\right|+1}\cos(x\log\left|x\right|)dx \\ &= \frac{\sin(x\log\left|x\right|)}{\log\left|x\right|+1}\Big{\vert}_{1}^{\infty} + \int_{1}^{\infty}\frac{\sin(x\log\left|x\right|)}{x(\log\left|x\right|+1)^{2}}dx \\ &= 0 + \int_{1}^{\infty}\frac{\sin(x\log\left|x\right|)}{x(\log\left|x\right|+1)^{2}}dx \end{alignat*} From here, I am not sure how to determine whether this integral converges or diverges. Thanks in advance.
A direct comparison shows that the final integral is convergent, using the fact that $|\sin(x \log |x|) \le 1$; for we have
$$\int_1^{\infty} \left|\frac{\sin{x \log |x|}}{x (\log |x| + 1)^2}\right| dx \le \int_1^{\infty} \frac{1}{x (\log x + 1)^2} dx$$
Setting $u = \log x + 1$ and integrating, the right-hand integral is equal to $1$.