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

Question

I am rather stuck in studying the convergence of this integral:

$$ \int_{0}^{\infty} {\mathrm{d}x \over 1 + x^{2}\left\vert\,\sin\left(\,x\,\right)\,\right\vert} $$

I can't really find an equivalent, and I don't really see what I can compare it too. Any help would be appreciated.

Answer 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.

Answer 2

Note that \begin{eqnarray} \int_{0}^{\infty}\frac{dx}{1+x^{2}|\sin(x)|}&\ge&\sum_{k=1}^\infty\int_{[\sqrt k]\pi+\frac{1}{k}}^{[\sqrt k]\pi+\frac{2}{k}}\frac{dx}{1+x^{2}|\sin(x)|}\\ &\ge&\sum_{k=1}^\infty\int_{[\sqrt k]\pi+\frac{1}{k}}^{[\sqrt k]\pi+\frac{2}{k}}\frac{dx}{1+([\sqrt k]+\frac2k)^2\sin(\frac2k)}\\ &=&\sum_{k=1}^\infty\frac{1}{k}\frac1{1+([\sqrt k]\pi+\frac2k)^2\sin(\frac2k)}\\ &=&\infty \end{eqnarray} and hence the integral diverges.

Answer 3

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

Answer 4

Before stating the answer, allow me to explain the method using the function $f(x)=1+\cos(x)$ as shown in the diagram below. To study the divergent of $\int_{0}^{\infty}f(x)\,dx$ where $f(x)\ge0$, we have:


One way to do so is to compare it with a series consists of $\small\color{red}{\text{STATIC}}$ intervals. By choosing an interval width $w$ that guarantee the resulting rectangular $\left[w\cdot f(2\pi\,n+w)\right]$ to stay under the function curve, we can write (as illustrated in red): $$ \int_{0}^{\infty}f(x)\,dx \gt \sum_{n=1}^{\infty} \frac{\pi}{2}\cdot f(2\pi\,n+\frac{\pi}{2}) \\ \small \Rightarrow \int_{0}^{\infty}\left(1+\cos(x)\right)\,dx \gt \sum_{n=1}^{\infty} \frac{\pi}{2}\left(1+\cos(2\pi\,n+\frac{\pi}{2})\right) = \frac{\pi}{2}\sum_{n=1}^{\infty} 1 \rightarrow \infty $$ Another way to do it -which what we are going to use in our question- is to choose $\small\color{blue}{\text{DYNAMIC}}$ intervals $w(n)$ which should be chosen under the same condition of guarantee the rectangular $\left[w(n)\cdot f(2\pi\,n+w(n))\right]$ to stay under the function curve. In our example, we can choose $w(n)=\pi/n\colon n\ge2$, and write (as illustrated in blue): $$ \int_{0}^{\infty}f(x)\,dx \gt \sum_{n=2}^{\infty} \frac{\pi}{n}\cdot f(2\pi\,n+\frac{\pi}{n}) \\ \small \Rightarrow \int_{0}^{\infty}\left(1+\cos(x)\right)\,dx \gt \sum_{n=2}^{\infty} \frac{\pi}{n}\left(1+\cos(2\pi\,n+\frac{\pi}{n})\right) = \frac{\pi}{2}\sum_{n=2}^{\infty} \frac{1+\cos(\pi/n)}{n} \gt \frac{\pi}{2}\sum_{n=2}^{\infty} \frac{1}{n} \rightarrow \infty $$

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In the question: $$ f(x)=\frac{1}{1+x^2\,|\sin(x)|} \gt 0 \quad\colon x\ge0 $$ $f(x)$ reaches the peaks infinitely often whenever $\sin(x)=0\Rightarrow f(x)_{peaks}=1 \space\colon x=\pi k$.


Let the dynamic interval equals $\pi/k$ over each $2\pi[\sqrt{k}]$ space $\left\{2\pi[\sqrt{k}] \space\rightarrow\space 2\pi[\sqrt{k}]+\pi/k\right\}$. Thus: $$ \begin{eqnarray} \int_{0}^{\infty} \frac{dx}{1+x^{2}\,|\sin(x)|} &\ge& \sum_{k=2}^\infty \,\int_{2\pi[\sqrt{k}]}^{ 2\pi[\sqrt{k}]+\color{red}{\pi/k}} \frac{dx}{1+x^{2}\sin(x)} \\ &\ge& \sum_{k=2}^{\infty} \frac{\pi}{k}\cdot f(2\pi[\sqrt{k}]+\frac{\pi}{k}) \\ &=&\sum_{k=2}^{\infty} \frac{\pi}{k}\, \frac{1}{1+\left(2\pi[\sqrt{k}]+\pi/k\right)^{2} \sin(2\pi[\sqrt{k}]+\pi/k)} \\ &\ge&\sum_{k=2}^{\infty} \frac{\pi}{k}\, \frac{1}{1+\left(2\pi\sqrt{k}+\pi/k\right)^{2} \sin(\pi/k)} \\ &\rightarrow&\infty \end{eqnarray} $$ Hence, the integral diverges.


NB: Upon the comment regarding $\lfloor x\rfloor$ overlap intervals infinitely often, the result of this answer is WRONG. (Thanks robjohn).