Does the integral $\int ^{+\infty}_{0}\frac{x}{1+x^6\sin^2(x)}dx$ + converge? Does the improper integral $$\int ^{+\infty}_{0}\frac{x}{1+x^6\sin^2(x)}dx$$ converge?
Most of the basic methods (substituiton, integration by parts...) probably don't work
What I tried :
Since the function $f(x)= \frac{x}{1+x^6\sin^2(x)}$ is positive the Function $F(t)=\int ^{t}_{0}\frac{x}{1+x^6\sin^2(x)}\,dx$ has a limit when $t$ tends to infinity.
Then \begin{align}
\int ^{n\pi}_{0}\frac{x}{1+x^6\sin^2(x)}dx & = \sum ^{n-1}_{k=0}\int ^{(k+1)\pi}_{k\pi}\frac{x}{1+x^6\sin^2(x)}dx
 \\
 & = \sum ^{n-1}_{k=0}\int ^{\pi}_{0}\frac{u+k\pi}{1+(u+k\pi)^6\sin^2(u)}dx 
 \end{align}
This is a trick we usually have improper integrals with $\sin$ or $\cos$.
However now I don't know how to proceed. I tried looking for upper-bounds (to prove that it converges) or lower bounds(to prove that it diverges) but I couldn't.
Thanks
 A: First, we need a lemma (proof at end):

Let $f : [0,\infty) \to [0,\infty)$ be any non-negative intergrable function such that for some $c > 0$, $f(x)$ is increasing on $[0,c]$ and decreasing on $[c,\infty)$.  We have
$$\sum_{k=0}^\infty f(k) \le \int_0^\infty f(x) dx + f(c)$$

As a corollary, for such a $f$ and any $a,h > 0$, we have
$$\sum_{k=0}^\infty f(a+kh) \le \frac1h \int_0^\infty f(x) dx + \sup_{x\ge 0} f(x)$$
For any $u \in (0,\pi)$, let $\rho = \sin^2 u$. In above lemma, substitute $a$ by $u$, $h$ by $\pi$ and $f(x)$ by $\frac{x}{1 + x^6\rho}$, we obtain:
$$\begin{align}
\sum_{k=0}^\infty \frac{u+k\pi}{1 + (u+k\pi)^6\sin^2 u}
&\le \frac1\pi \int_0^\infty \frac{x}{1+x^6\rho}dx + \sup_{x\ge 0} \frac{x}{1+x^6\rho}\\
&= \frac{1}{3\sqrt{3}}\rho^{-1/3} + \frac{5^{5/6}}{6} \rho^{-1/6}
\end{align}$$
As a function of $u$, RHS is singular at $u \sim 0$ and $u \sim \pi$.
The leading divergences are

*

*$\frac{1}{3\sqrt{3}} u^{-2/3}$ for $u \sim 0$.

*$\frac{1}{3\sqrt{3}} (\pi - u)^{-2/3}$ for $u \sim \pi$.

This sort of singularities are tame enough and RHS is Lebesgue intergrable over $[0,\pi]$. Let
$$\mathcal{I}(t) = \int_0^t \frac{xdx}{1+x^6\sin^2 x}
\quad\text{ and }\quad
\varphi_n(u) = \sum_{k=0}^{n-1} \frac{u+k\pi}{1+(u+k\pi)^6\sin^2u}$$
Since $\varphi_n(u)$ is a sum of non-negative functions
whose pointwise limit $\varphi_{\infty}(u) \stackrel{def}{=} \lim\limits_{n\to\infty}\varphi_n(u)$ is bounded from above by a Lebesgue  intergrable function. By DCT, $\varphi_\infty(u)$ is also Lebesgue intergrable over $[0,\pi]$.
Translate this to the Riemann definite integral $\mathcal{I}(t)$, we have
$$\lim_{n\to\infty}\mathcal{I}(n\pi) = \lim_{n\to\infty}\int_0^\pi \varphi_n(u) du$$
exist and equal to the Lebesgue integral $\int\limits_0^\pi \varphi_\infty(u) du$.
Since $\mathcal{I}(t)$ is an increasing function in $t$. The existence of limit for $t \in \{ \pi, 2\pi, \ldots \}$ implies $\lim\limits_{t\to \infty}\mathcal{I}(t)$ also exists and converges to same limit. In other words, the Riemann improper integral
$$\int_0^\infty \frac{xdx}{1 + x^6\sin^2 x}$$
converges.

Proof of Lemma
Let $N = \lfloor c \rfloor$.

*

*$f(x)$ is increasing on $[0,N]\implies \sum\limits_{k=0}^{N-1} f(k) \le \int_0^N f(x)dx$

*$f(x)$ is decreasing on $[N+1,\infty) \implies
\sum\limits_{k=N+2}^\infty f(k) \le \int_{N+1}^\infty f(x)dx$

*$f(x) \ge \begin{cases}f(N), & x \in [N,c]\\ f(N+1), & x \in [c,N+1]\end{cases}$ $\implies \min(f(N),f(N+1)) \le \int\limits_N^{N+1} f(x)dx$

*By definition of $c$, $\max(f(N),f(N+1)) \le f(c)$.

Combine all these, our lemma follows.
