How to study the convergence of $ \int_{0}^{\infty}\frac{dx}{1+x^{2}|\sin(x)|} $?

Solution 1:

$$ \begin{align} \int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+x^2|\sin(x)|} &\le\int_0^\pi\frac{\mathrm{d}x}{1+k^2\pi^2\sin(x)}\tag{1}\\ &=2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+k^2\pi^2\sin(x)}\tag{2}\\ &\le2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+2k^2\pi x}\tag{3}\\ &=\frac1{k^2\pi}\int_0^{k^2\pi^2}\frac{\mathrm{d}x}{1+x}\tag{4}\\ &=\frac{\log\left(1+k^2\pi^2\right)}{k^2\pi}\tag{5}\\ &\le\frac{\log\left(k^2\pi^2\right)+\frac1{k^2\pi^2}}{k^2\pi}\tag{6}\\ &=\frac{2\log(\pi)+2\log(k)+\frac1{k^2\pi^2}}{k^2\pi}\tag{7} \end{align} $$ Explanation:
$(1)$: on $[k\pi,(k+1)\pi]$, $1+x^2|\sin(x)|\ge1+k^2\pi^2\sin(x-k\pi)$
$\phantom{\text{(1): }}$then subsitute $x\mapsto x+k\pi$
$(2)$: $\sin(x)=\sin(\pi-x)$
$(3)$: on $[0,\pi/2]$, $\sin(x)\ge\frac2\pi x$
$(4)$: substitute $x\mapsto\frac x{2k^2\pi}$
$(5)$: integrate
$(6)$: $\log(1+x)\le\log(x)+\frac1x$
$(7)$: expand the $\log$

Therefore, $$ \begin{align} \int_0^\infty\frac{\mathrm{d}x}{1+x^2|\sin(x)|} &\le\int_0^\pi1\,\mathrm{d}x+\sum_{k=1}^\infty\frac{2\log(\pi)+2\log(k)+\frac1{k^2\pi^2}}{k^2\pi}\\[4pt] &=\pi+\frac{2\log(\pi)}\pi\zeta(2)-\frac2\pi\zeta'(2)+\frac1{\pi^3}\zeta(4) \end{align} $$ and the integral converges.

Solution 2:

$(1)\hspace{5mm} \displaystyle 1.5<\int\limits_0^\infty\frac{dx}{1+x^2 |\sin x|} =\sum\limits_{k=0}^\infty \int\limits_{k\pi}^{(k+1)\pi}\frac{dx}{1+(-1)^k x^2 \sin x} $$\displaystyle < 1.52 + \sum\limits_{k=1}^\infty \int\limits_{k\pi}^{(k+1)\pi}\frac{dx}{1+(-1)^k x^2 \sin x}$

because of $\enspace \displaystyle 1.5<\int\limits_0^\pi\frac{dx}{1+(-1)^k x^2 \sin x}<1.52 \enspace$ . $\enspace$ Be $\enspace k\in\mathbb{N}$ .

$(2)\hspace{5mm} \displaystyle\int\limits_{k\pi}^{(k+1)\pi}\frac{dx}{1+(-1)^k x^2 \sin x} = \frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2}x+k\pi)^2 \sin(\frac{\pi}{2}x)}+ $$\displaystyle\frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2}x+(k+\frac{1}{2})\pi)^2\cos(\frac{\pi}{2}x)}$

Using $\enspace \cos(\frac{\pi}{2}x)\geq 1-x\enspace $ and $\enspace \sin(\frac{\pi}{2}x)\geq x\enspace $ for $\enspace 0\leq x\leq 1\enspace$

and because of $\enspace\displaystyle 1+(\frac{\pi}{2})^2(x+2k)^2 x \leq 1+(\frac{\pi}{2})^2(2-x+2k)^2 x \enspace$ we get

$(3)\hspace{5mm} \displaystyle\int\limits_{k\pi}^{(k+1)\pi}\frac{dx}{1+(-1)^k x^2 \sin x} \leq $$\displaystyle\frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2})^2(x+2k)^2 x}+\frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2})^2(x+2k+1)^2 (1-x)} $

$\hspace{2cm}\displaystyle =\frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2})^2(x+2k)^2 x} +\frac{\pi}{2}\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2})^2(2-x+2k)^2 x} $

$\hspace{2cm}\displaystyle\leq \pi\int\limits_0^1\frac{dx}{1+(\frac{\pi}{2})^2(x+2k)^2 x} \leq \pi\int\limits_0^1\frac{dx}{1+\pi^2 kx(k+x)} $

$(4)\hspace{5mm} \displaystyle\int\limits_0^1 \frac{dx}{1+\pi^2 kx(k+x)} = \frac{1}{ \sqrt{\pi^2 k(\pi^2 k^3-4)} } \ln\frac{2\pi^2 kx+\pi^2 k^2-\sqrt{\pi^2 k(\pi^2 k^3-4)} }{2\pi^2 kx+\pi^2 k^2+\sqrt{\pi^2 k(\pi^2 k^3-4)} } |_0^1 $

$\hspace{2cm}\displaystyle = \frac{1}{ \pi \sqrt{k(\pi^2 k^3-4)} } \ln\frac{ 2\pi^2 k^3+4k+2\pi k\sqrt{k(\pi^2 k^3-4)} }{2\pi^2 k^3+4k-2\pi k\sqrt{k(\pi^2 k^3-4)}} \leq \frac{1}{\pi\sqrt{\pi^2-4}}\frac{\ln(1+\pi^2 k^2)}{k^2}$

$\hspace{2cm}\displaystyle\leq \frac{1}{\pi\sqrt{\pi^2-4}}\frac{\ln((1+\pi^2) k^2)}{k^2}= \frac{\ln(1+\pi^2)}{\pi\sqrt{\pi^2-4}}\frac{1}{k^2}+\frac{2}{\pi\sqrt{\pi^2-4}}\frac{\ln k}{k^2}$

Therefore we get an upper bound for the integral, an estimation is:

$$1.5<\int\limits_0^\infty\frac{dx}{1+x^2 |\sin x|} < 1.52+\frac{\ln(1+\pi^2)}{\sqrt{\pi^2-4}} \zeta(2) - \frac{2}{\sqrt{\pi^2-4}}\zeta’(2)$$

Note:

It's $\enspace\displaystyle\zeta(2)=\frac{\pi^2}{6}\enspace$ and $\enspace\displaystyle -\zeta'(2)=\sum\limits_{k=1}^\infty\frac{\ln k}{k^2}\approx 0.93755 \enspace$ .