Does $\int_1^\infty \frac{\log x}{x^{3}} \sin x \,dx $ exist? I want to determine if the following indefinite integral exists:
$$\int_{1}^{\infty} \frac{\log x}{x^{3}} \sin x dx.$$
I tried to solve the integral then calculate the limit
$$ \lim_{\lambda \to \infty} ( \int_{1}^{\lambda} \frac{\log x}{x^{3}} \sin x dx ) $$
but I couldn't come to any easy way to solve the integral $\int \frac{\log x}{x^{3}} \sin x dx$ in order to calculate its limit.
 A: Since
$$|\sin x|\le1$$
and
$$\lim_{x\to\infty}\frac{\log x}{x}=0$$
then for $x$ large enough 
$$\left|\frac{\log x}{x^3}\sin x\right|\le \frac{\log x}{x^3}=_\infty o\left(\frac1{x^2}\right)$$
hence the given integral is convergent by comparison.
A: Pick $x_0 \gg 1$ such that $|\log x| \leq x$ for $x > x_0$, and split $\int_1^\infty = \int_1^{x_0}+\int_{x_0}^\infty$. The first is an ordinary Riemann integral; for the second, remark that
$$
\left| \frac{\log x}{x^3} \sin x \right| \leq \frac{|\sin x|}{x^2} \leq \frac{1}{x^2},
$$
which is integrable on $(x_0,+\infty)$.
A: This is just for your curiosity since you already received good answers from other participants.  
Surprizingly (at least to me) or not, the antiderivative has a closed form (it is not a nightmare but close to; so I shall not give it here). 
Concerning the integral from $1$ to $\infty$, its value is given by 
$$\,
   _2F_3\left(-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2},\frac{3}{2};-\frac{1}
   {4}\right)+\frac{1}{8} (2 \gamma -3) \pi \simeq0.1094982566845155420432374$$
A: You can use the direct comparison test which states that given 2 functions $f(x)$ and $g(x)$  that are both continuous on $[a, \infty)$, if:
$$ 0 < f(x) < g(x), \; \forall x \in [a, \infty) $$
Then
$$ \int_{a}^{\infty} f(x)\,dx \quad \text{converges if} \quad \int_{a}^{\infty} g(x)\,dx \quad \text{converges} $$
In this case, notice that on $[1, \infty)$, $\sin x \leq 1$ and $\ln x < x$, so:
$$ \frac{\ln x \sin x}{x^3} < \frac{1}{x^2}, \; \forall x \in [1, \infty) $$
And:
$$ \int_{1}^{\infty} \frac{1}{x^2}dx \quad \text{converges}$$
Therefore, the original integral converges
