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

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.

• 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 $\infty$ . Commented Dec 15, 2016 at 9:14
• @user90369 The integral in question isn't equal to the one you wrote, so how can it help to study it? Commented Dec 15, 2016 at 19:03
• 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 $$\sum_{k=0}^\infty \int_{k\pi}^{(k+1)\pi} \frac{dx}{1 + (-1)^k x^2 \sin x} = \int_0^\infty \frac{dx}{1+x^2| \sin x|}$$ Commented Dec 15, 2016 at 21:35
• $$\frac{1}{1+x^{2}|\sin(x)|}\gt0 \Rightarrow \\ \int_{0}^{\infty}\frac{dx}{1+x^{2}|\sin(x)|} \gt \lim_{N\rightarrow\infty}\,\sum_{n=0}^{N}\frac{1}{1+(\pi\,n)^{2}|\sin(\pi\,n)|} = \lim_{N\rightarrow\infty}\,\sum_{n=0}^{N}\frac{1}{1} = \lim_{N\rightarrow\infty}N= \infty$$ Commented Dec 16, 2016 at 11:44
• @Hazem Orabi : It would be better if you explain your steps detailed as an answer. :-) Commented Dec 16, 2016 at 11:52

\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.

• 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. Commented Dec 17, 2016 at 13:46

$(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 . • 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. Commented Dec 17, 2016 at 0:13 • @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. Commented Dec 17, 2016 at 0:15 • 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? Commented Dec 17, 2016 at 0:22 • @Hazem Orabi : Yes, of course, but I am sorry that I cannot answer you before monday. Till then ... :-) Commented Dec 17, 2016 at 0:26 • U r correct and the 1st to notice the convergence. Bravo! (+1). Commented Dec 17, 2016 at 20:42 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. • I don't really see how your result shows that it diverges. Could you explain? Commented Dec 16, 2016 at 21:34 • 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? Commented Dec 16, 2016 at 23:14 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 $$In the question:$$ 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). • Does$[x]=\lfloor x\rfloor\$? If so, many of the intervals overlap.
– robjohn
Commented Dec 17, 2016 at 9:40
• @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. Commented Dec 17, 2016 at 13:31