TITLE: Does the integral $\int_{1}^{\infty} \sin(x\log x) \,\mathrm{d}x$ converge?
QUESTION [7 upvotes]: I tried a couple of substitutions but have so far gotten nowhere. Can anyone guide me in the right direction?
Thanks for your time.

REPLY [3 votes]: A trick that often works on such oscillatory conditionally convergent integrals.
Change variables $y=x\log x$
$$
\int_1^\infty \sin(x\log x)\,dx = \int_0^\infty \frac{\sin y}{W(y)+1}\,dy
$$
Here $W$ is the Lambert W function.  
Integrate by parts
$$
\int_0^\infty \frac{\sin y}{W(y)+1}\,dy=
1-\int_0^\infty \frac{W(y)\cos y}{(W(y)+1)^3y}\,dy
$$
But this integral is absolutely convergent.  Indeed,
$$
\varphi(y) := \frac{W(y)}{(W(y)+1)^3y} > 0\qquad\text{and}\qquad
\int_0^\infty \frac{W(y)}{(W(y)+1)^3y}\,dy = 1.
$$
This is still just barely convergent, though:  $\varphi(y) \sim 1/(y\log(y)^2)$ as $y \to \infty$.
added Do the substitution.  
$y = x \log x$, solve for $x$:
$$
y = x\log x
\\
y = (\log x) e^{(\log x)}
\\
W(y) = \log x
\\
e^{W(y)} = x
$$
For the derivative
$$
y = x\log x
\\
dy = \big(\log x + 1\big)\,dx
\\
\frac{dy}{W(y)+1} = dx
$$
For the limits of integration:
$$
x=1 \quad\Longrightarrow\quad y = x\log x = 0
\\
x \to \infty \quad \Longrightarrow \quad y = x \log x \to \infty
$$
Also: $x\log x$ is increasing in $[1,\infty)$ since its derivative is $\log x + 1$.
Therefore
$$
\int_1^\infty \sin(x\log x)\,dx = \int_0^\infty \frac{\sin y}{W(y)+1}\,dy
$$