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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.

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    $\begingroup$ You can study $\displaystyle\int\limits_{k\pi}^{(k+1)\pi}\frac{dx}{1+(-1)^k x^2 \sin x}$ and add the estimations of the integrals from $k=0$ to $\infty$ . $\endgroup$ – user90369 Dec 15 '16 at 9:14
  • $\begingroup$ @user90369 The integral in question isn't equal to the one you wrote, so how can it help to study it? $\endgroup$ – John Mayne Dec 15 '16 at 19:03
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    $\begingroup$ When $x \in (k\pi, (k+1)\pi)$ and $k$ is odd, then $\sin x < 0$ so $(-1)^k \sin x > 0$. When $k$ is even, then $\sin x > 0$ so $(-1)^k\sin > 0$. Thus $$\sum_{k=0}^\infty \int_{k\pi}^{(k+1)\pi} \frac{dx}{1 + (-1)^k x^2 \sin x} = \int_0^\infty \frac{dx}{1+x^2| \sin x|}$$ $\endgroup$ – CodeLabMaster Dec 15 '16 at 21:35
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    $\begingroup$ $$ \frac{1}{1+x^{2}|\sin(x)|}\gt0 \Rightarrow \\ \int_{0}^{\infty}\frac{dx}{1+x^{2}|\sin(x)|} \gt \lim_{N\rightarrow\infty}\,\sum_{n=0}^{N}\frac{1}{1+(\pi\,n)^{2}|\sin(\pi\,n)|} = \lim_{N\rightarrow\infty}\,\sum_{n=0}^{N}\frac{1}{1} = \lim_{N\rightarrow\infty}N= \infty $$ $\endgroup$ – Hazem Orabi Dec 16 '16 at 11:44
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    $\begingroup$ @Hazem Orabi : It would be better if you explain your steps detailed as an answer. :-) $\endgroup$ – user90369 Dec 16 '16 at 11:52
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$$ \begin{align} \int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+x^2|\sin(x)|} &\le\int_0^\pi\frac{\mathrm{d}x}{1+k^2\pi^2\sin(x)}\tag{1}\\ &=2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+k^2\pi^2\sin(x)}\tag{2}\\ &\le2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+2k^2\pi x}\tag{3}\\ &=\frac1{k^2\pi}\int_0^{k^2\pi^2}\frac{\mathrm{d}x}{1+x}\tag{4}\\ &=\frac{\log\left(1+k^2\pi^2\right)}{k^2\pi}\tag{5}\\ &\le\frac{\log\left(k^2\pi^2\right)+\frac1{k^2\pi^2}}{k^2\pi}\tag{6}\\ &=\frac{2\log(\pi)+2\log(k)+\frac1{k^2\pi^2}}{k^2\pi}\tag{7} \end{align} $$ 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.

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  • $\begingroup$ The inequality $\left\{x^2|\sin(x)| \ge (k\pi)^2\sin(x-k\pi)\space\colon x\in[k\pi ,\, k\pi+\pi]\right\}$ is the key to solve the case, And Case Closed. (+1) as always. $\endgroup$ – Hazem Orabi Dec 17 '16 at 13:46
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$(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 $\enspace\displaystyle -\zeta'(2)=\sum\limits_{k=1}^\infty\frac{\ln k}{k^2}\approx 0.93755 \enspace$ .

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  • $\begingroup$ Allow me to Thank you for two things: (1st) your respond to my initial comment regarding the divergent of the integral, and (2nd) the effort to write this answer. Unfortunately, my previously comment was partially wrong (I mean the method)! But the result is correct (I mean the integral is divergent). I will share an answer shortly, then we will discuss. $\endgroup$ – Hazem Orabi Dec 17 '16 at 0:13
  • $\begingroup$ @Hazem Orabi : If you are right then there is a mistake with my calculations ... but where ? :-) --- Because of time I have to finish now and can continue on monday. $\endgroup$ – user90369 Dec 17 '16 at 0:15
  • $\begingroup$ Let me share my answer first, maybe I am wrong! I promise I will check your answer if I fail to convenes you about the divergent! Deal? $\endgroup$ – Hazem Orabi Dec 17 '16 at 0:22
  • $\begingroup$ @Hazem Orabi : Yes, of course, but I am sorry that I cannot answer you before monday. Till then ... :-) $\endgroup$ – user90369 Dec 17 '16 at 0:26
  • $\begingroup$ U r correct and the 1st to notice the convergence. Bravo! (+1). $\endgroup$ – Hazem Orabi Dec 17 '16 at 20:42
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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.

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  • $\begingroup$ I don't really see how your result shows that it diverges. Could you explain? $\endgroup$ – John Mayne Dec 16 '16 at 21:34
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    $\begingroup$ For most large $k$, we have $\lfloor\sqrt k\rfloor=\lfloor\sqrt{k+1}\rfloor$, so don't the intervals in your sum typically overlap? $\endgroup$ – Barry Cipra Dec 16 '16 at 23:14
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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:

enter image description here
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): $$ \int_{0}^{\infty}f(x)\,dx \gt \sum_{n=2}^{\infty} \frac{\pi}{n}\cdot f(2\pi\,n+\frac{\pi}{n}) \\ \small \Rightarrow \int_{0}^{\infty}\left(1+\cos(x)\right)\,dx \gt \sum_{n=2}^{\infty} \frac{\pi}{n}\left(1+\cos(2\pi\,n+\frac{\pi}{n})\right) = \frac{\pi}{2}\sum_{n=2}^{\infty} \frac{1+\cos(\pi/n)}{n} \gt \frac{\pi}{2}\sum_{n=2}^{\infty} \frac{1}{n} \rightarrow \infty $$


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 $\sin(x)=0\Rightarrow f(x)_{peaks}=1 \space\colon x=\pi k$.

enter image description here
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).

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  • $\begingroup$ Does $[x]=\lfloor x\rfloor$? If so, many of the intervals overlap. $\endgroup$ – robjohn Dec 17 '16 at 9:40
  • $\begingroup$ @robjohn: YES indeed, you are right. My answer is wrong. Thanks for the comment and the right answer! (the integral converges, true). I will keep the answer in for the benefit of applicable methods. Thanks again Rob. $\endgroup$ – Hazem Orabi Dec 17 '16 at 13:31

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