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Show that $\displaystyle \int\limits_{0}^{\infty} \frac{x}{1+x^6\sin^2x}\,dx$ is convergent

Since we need to check if the given improper integral is convergent, I considered Abels/Dirichlet test for convergence, but for that, although I got monotone bounded or monotone functions converging to zero, the integral of the other function was not bounded.

Certainly though we can break the domain into intervals, $[n\pi, (n+1)\pi]$, thereby bounding the integral over each interval, and the integrand being continuous over each interval is thereby integrable. Then performing summation of the integrals over each interval we check whether its bounded and if so the convergence follows.

BUT, I was in search for a function of form $\frac{1}{x^p}$ with $p>1$ such that it is always larger than the given integrand over the domain $[a,\infty]$, $a>0$, but with $\sin^2x$ in the denominator, I couldn't. I also tried to think of some function larger than the integrand over the domain $[a, \infty]$, upon which I could apply the Abels/Dirichilet tests, but that didn't worked also.

So any solutions to the given problem considering comparison tests or any other approach that would make the solution brief would be highly appreciated.

Sangchul Lee
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Subham Karmakar
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    This is just an idea (I did not fully performed the calculations): divide the intervals of length $\pi$ into two, based on where $\sin^2 x =1/x^2$. You can then get one of the integrals to be smaller than $\int_D x dx$, where $D$ is a shrinking domain, such that summing up these integrals you get a finite number. – Andrei Jun 27 '22 at 15:09

1 Answers1

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You can proceed as in this answer:

Note that $\left| \sin x \right| \geq \frac{2}{\pi}|x|$ for $|x| \leq \frac{\pi}{2} $. Using this, we get:

$$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+a \sin^2 x} \leq \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+a\bigl(\frac{2}{\pi}x\bigr)^2} \leq \int_{-\infty}^{\infty} \frac{\mathrm{d}x}{1+a\bigl(\frac{2}{\pi}x\bigr)^2} = \frac{C}{\sqrt{a}} $$

for some constant $C$. From this, it follows that

\begin{align*} &\int_{0}^{\infty} \frac{x}{1+x^6\sin^2 x} \, \mathrm{d}x \\ &= \int_{0}^{\frac{\pi}{2}} \frac{x}{1+x^6\sin^2 x} \, \mathrm{d}x + \sum_{n=1}^{\infty} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x+n\pi}{1+(x+n\pi)^6\sin^2 x} \, \mathrm{d}x \\ &\leq \int_{0}^{\frac{\pi}{2}} \frac{x}{1+x^6\sin^2 x} \, \mathrm{d}x + \sum_{n=1}^{\infty} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(n+\frac{1}{2})\pi}{1+(n-\frac{1}{2})^6\pi^6\sin^2 x} \, \mathrm{d}x \\ &\leq \int_{0}^{\frac{\pi}{2}} \frac{x}{1+x^6\sin^2 x} \, \mathrm{d}x + \sum_{n=1}^{\infty} \frac{C(n+\frac{1}{2})\pi}{(n-\frac{1}{2})^3\pi^3} \\ &< \infty. \end{align*}

Sangchul Lee
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