Does the integral $\int_{1}^{\infty} \sin(x\log x) \,\mathrm{d}x$ converge? I tried a couple of substitutions but have so far gotten nowhere. Can anyone guide me in the right direction?
Thanks for your time.
 A: It depends on what we mean by saying that the integral converges. In some contexts (typically basic analysis courses), we might mean that
$$\exists\lim_{R\to\infty}\int_0^R \sin(x\log x)dx.$$
Under that interpretation, the answer is yes. Denote $x\mapsto x\log x$ as $\phi$. Then $\phi$ is an increasing diffeomeomorphism $[1,\infty]\to[0,\infty]$. If we let $x_n=\phi^{-1}(\pi n)$, then $I_n = \int_{x_n}^{x_{n+1}}\sin\phi(x)dx$ are sign-alternating and decreasing to zero in their absolute value, implying that $\sum_{n\in\Bbb N}I_n$ converges (hence, the above limit exists and euqals the value of the series; why?)
However, $x\mapsto\sin(x\log x)$ isn't absolutely integrable (compare the above construction to a similar one for $x\mapsto\sin(x^2)$), implying that $\int_1^\infty\sin(x\log x)dx$ also doesn't exist in the sense of partitions of unity.
A: 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
$$
\int_1^\infty \sin(x\log x)\,dx = \int_0^\infty \frac{\sin y}{W(y)+1}\,dy
$$
