I am investigating the convergence of the integral
$$\int_1^\infty \frac{1}{x^{2+\sin(x)}}\mathrm dx.$$
I know that $\int_1^\infty \frac{1}{x^a}$ converges iff $a>1$, so ultimately this comes down to the question of where $2+\sin(x)$ spends "enough time close enough to $1$" to cause the integral to diverge.
I think the integral diverges, although in some sense "just barely," and I cannot think of a simple proof. Regardless, here is my idea, and I would appreciate any feedback/suggestions.
For brevity, let $g(x)=\frac{1}{x^{2+\sin(x)}}$. Define the sequence $\\{f_n\\}$ such that
$$f_n(x)=\begin{cases}
g(x) &\text{if $2+\sin(x) < \tfrac 1n$} \\\\
0 &\text{otherwise}
\end{cases}$$
In particular, note that $E_n := \bigcup_{k=0}^\infty [(2k+\tfrac 32)\pi-\tfrac 1n,(2k+\tfrac 32)\pi-\tfrac 1n]$ is contained within the support of $f_n$ for all $n$. Clearly the sequence of $f_n(x)$ is dominated by $g(x)$ and converges a.e. to the constant $0$ function, so by the Dominated Convergence Theorem, if the original integral converges, then the limit of the integrals of the $f_n$'s should approach $0$ as $n\to\infty$.
However, I claim
$$\int_1^\infty f_n(x) \mathrm dx \geq \int_{E_n} \frac{1}{x^{1+1/n}} \mathrm dx \approx \frac Cn\int_1^\infty \frac{1}{x^{1+1/n}} \mathrm dx = C$$
for some $C>0$. I have no idea how to argue this formally, but the general idea is that $E_n$ takes up $\tfrac 1n$ of each period of $\sin$, and as such we would expect the integral over $E_n$ to be $\tfrac 1n$ of the integral over all of $(1,\infty)$, with an appropriate nonzero constant fudge factor for translation. I am well aware this is not at all rigorous, but does it seem plausible?
Assuming this is correct, this would show that the limit of the integrals of the $f_n$'s does not approach $0$, and as such the original integral does not converge.
Does this seem reasonable? I'm admittedly not the best at analysis so I may be stating one or more incorrect things, but in any case any feedback or alternate approaches would be appreciated.
Best Answer
Here's a reasonably clean (I hope) solution that certainly shares ideas with the remarks of the OP and of Edward H.
Let $x_k = (2k-\frac12)\pi$ be the $k$th positive minimum of $2+\sin x$. One can check that $2+\sin(x_k+\delta) \le 1+\frac12\delta^2$ for all $\delta\in\Bbb R$ (this is the sine function lying below its quadratic Taylor polynomials at those points). In particular, $2+\sin x \le 1+1/\log x_k$ for all $x\in[x_k-\sqrt{2/\log x_k},x_k]$. Therefore for $k\ge2$, \begin{align*} \int_{2(k-1)\pi}^{2k\pi} \frac1{x^{2+\sin x}}\,dx &\ge \int_{x_k-\sqrt{2/\log x_k}}^{x_k} \frac1{x^{2+\sin x}}\,dx \\ &\ge \int_{x_k-\sqrt{2/\log x_k}}^{x_k} \frac1{x_k^{1+1/\log x_k}}\,dx \\ &= \frac{\sqrt{2/\log x_k}}{ex_k} \ge \frac1{2x_k\sqrt{\log x_k}} = \frac1{2(2k-\frac12)\pi\sqrt{\log (2k-\frac12)\pi}}. \end{align*} Summing over $k\ge2$ yields $$ \int_1^\infty \frac1{x^{2+\sin x}}\,dx \ge \sum_{k=2}^\infty \frac1{2(2k-\frac12)\pi\sqrt{\log (2k-\frac12)\pi}}, $$ which diverges by the integral test since $\displaystyle\int\frac1{2(2u-\frac12)\pi\sqrt{\log (2u-\frac12)\pi}}\,du = \frac{\sqrt{\log (2u-\frac12)\pi}}{2\pi}+C$ which tends to infinity.