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.
On
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.
On
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 $$\, _2F_3\left(-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2},\frac{3}{2};-\frac{1} {4}\right)+\frac{1}{8} (2 \gamma -3) \pi \simeq0.1094982566845155420432374$$
On
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
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 $(x_0,+\infty)$.