# Is there an elementary proof that $\int_0^{\infty}|\sin(x)|^{x}\ dx$ converges or diverges?

Is there an elementary proof that $$\displaystyle\int_0^{\infty}|\sin(x)|^{x}\ dx$$ converges or diverges?

I tried the following: $$|\sin(x)|\leq 1-\frac{1}{3}\left( x-\frac{\pi}{2} \right)^2$$on $$[0,\pi]$$.

Therefore, $$\int_0^{\infty}|\sin(x)|\ dx\leq\sum_{n=0}^{\infty}\left(\int_0^{\pi} \left(1-\frac{1}{3}\left( x-\frac{\pi}{2} \right)^2\right)^{\pi n}\ dx \right).$$

But now WolframAlpha says that those integrals require the hypergeometric function which I know nothing about.

As usual, I've got a feeling that I'm missing an easier method...

• Why do you have an improper integral after the sum? Commented Jun 24, 2021 at 14:55
• I also suggest to simplify the integral in the sum as far as possible. It can help you get the answer Commented Jun 24, 2021 at 15:04
• It should be noted that you actually want $\int_0^{\infty}|\sin(x)|\ dx\geq\cdots$ as $\int_0^{\infty}|\sin(x)|\ dx\leq\infty$ is inconclusive. Commented Jun 25, 2021 at 6:47
• I thought it was convergent going into the question. Obviously now I can see it is divergent… Commented Jun 25, 2021 at 10:03

Note that \begin{align}\int_0^\infty|\sin x|^x\,dx&=\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}|\sin x|^x\,dx\\&\ge\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}|\sin x|^{(k+1)\pi}\,dx=\sum_{k=0}^\infty\int_0^\pi|\sin(x+k\pi)|^{(k+1)\pi}\,dx\\&=\sum_{k=0}^\infty\int_0^\pi\sin^{(k+1)\pi}x\,dx\\&\ge\sum_{k=0}^\infty\int_0^{\pi/2}\sin^{(k+1)\pi}x\,dx=\sum_{k=0}^\infty\frac{\sqrt\pi\Gamma((k+1)\pi/2+1/2)}{2\Gamma((k+1)\pi/2+1)}\end{align} where the integral can be evaluated using contour integration. Now Stirling gives \begin{align}\frac{\Gamma(z)}{\Gamma(z+1/2)}&\sim\sqrt e\exp\left(\left(z-\frac12\right)\log z-z\log\left(z+\frac12\right)\right)\\&=\sqrt e\exp\left(-\frac12-\frac12\log z+{\cal O}\left(\frac1z\right)\right)\sim\frac1{\sqrt z}\end{align} so the integral must be divergent.

• It is better to change the power $\pi(k+1)$ on $2(k+1)$ getting a bigger integral or on $4(k+1)$ getting a smaller integral with the same properties Commented Jun 24, 2021 at 15:15
• I think the formula you use is not true for irrational powers Commented Jun 24, 2021 at 15:24
• @TimurBakiev Aha, we can get round that by writing $\lceil(k+1)\pi\rceil$. Commented Jun 24, 2021 at 15:26
• WolframAlpha says $t!!^2/(t+1)!$ actually goes as $1/\sqrt{t}$. This is sufficient to show divergence, but it means $t!!^2$ grows slower than $(t+1)!$. Commented Jun 24, 2021 at 15:38
• @Gary I've removed all the extra jazz about the double factorial and have gone straight to gamma asymptotics for convenience. Commented Jun 24, 2021 at 16:13

\begin{align}\text{Integral} &=\sum_{n=0}^{\infty} \int_{0}^{\pi} |\sin(x)|^{x +n\pi}dx \ge \sum_{n=0}^{\infty} \int_{0}^{\pi} |\sin(x)|^{\pi +n\pi}dx \\ & \stackrel{\text{Fubini}}{=} \int_0^{\pi} \frac{\sin(x)^{\pi}}{1-\sin(x)^{\pi}}dx= -\pi+2\int_0^{\pi/2}\frac{1}{1-\sin(x)^{\pi}}dx\\ \\ &\ge -\pi+2\int_0^{\pi/2}\frac{1}{1-\sin(x)^{4}}dx =-\pi+\int_0^{\pi/2}\frac{2}{(1-\sin(x)^{2})(1+\sin(x)^2)}dx \\ &\ge -\pi+\int_0^{\pi/2}\frac{1}{1-\sin(x)^{2}}dx \\ &=-\pi+ \left( \tan(x)\right)\bigg\vert_0^{\pi/2} =\infty \end{align}

https://en.wikipedia.org/wiki/Fubini%27s_theorem

• It seems that $C=\frac \pi 2$. Commented Jun 24, 2021 at 15:45
• @ClaudeLeibovici: I think you are right about that. Commented Jun 24, 2021 at 15:50
• The switch to the approximation is a little neater if you write the integral as $\int_0^{\pi/2}dx/[1-\cos(x)^\pi]$, so you can use $1 - \cos(x)^\pi \le \pi/(2x^2)$. Commented Jun 24, 2021 at 15:50
• @eyeballfrog You're right. Commented Jun 24, 2021 at 15:51
• I think I follow this method, and I came up with something similar. One nitpick though. In the second line, instead of $\ = -1+2\int_0^{\pi/2}\frac{1}{1-\sin(x)^{\pi}}dx\$ shouldn't it be $\ = -\pi+2\int_0^{\pi/2}\frac{1}{1-\sin(x)^{\pi}}dx\$ ? Also, there is a more direct/alternative way of doing the final step (although I'll admit that I used WA to find this integral): $$\large{\ \int_0^{\pi/2}\frac{1}{1-\sin(x)^{\pi}}dx \geq\ \int_0^{\pi/2}\frac{1}{1-\sin(x)^{4}}dx = \lim_{a\to\frac{\pi}{2}^-}\frac14 \left[ \sqrt{2}\arctan\left(\sqrt{2}\tan(x)\right) + 2\tan(x) \right]_0^{a} = \infty}$$ Commented Jun 24, 2021 at 18:23

Informally, if we take $$\log f(x)$$ for $$x$$ near $$\pi(n+1/2)$$, then \begin{align}\log f(x) &= x \log|\sin x| \\&\sim \pi(n+1/2) \log|1 - (x - \pi(n+1/2))^2 + O((x-\pi(n+1/2))^4)| \\&\sim -\pi(n+1/2) (x - \pi(n+1/2))^2\end{align} as $$x \to \pi(n+1/2)$$. Therefore, by an argument similar to the one in Laplace's method, we expect that \begin{align}\int_{n\pi}^{(n+1)\pi} f(x)\,dx &\sim \int_{-\infty}^\infty \exp(-\pi (n+1/2) (x-\pi(n+1/2))^2 )\,dx \\&= \sqrt{\frac{\pi}{\pi(n+1/2)}} = (n+1/2)^{-1/2}\end{align} as $$n \to \infty$$. Since $$\sum_{n=1}^\infty (n+1/2)^{-1/2}$$ diverges, by the limit comparison test, it will follow that $$\sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi} f(x)\,dx$$ also diverges. Therefore, the improper integral $$\int_0^\infty f(x)\,dx$$ diverges as well.

It is left as a (nontrivial, admittedly) exercise for the reader to formalize this argument.

Trying to make things as elementary as possible, note that the intervals $$[\pi/2+k\pi,1+\pi/2+k\pi]$$ for $$k=0,1,2,\ldots$$ are disjoint, and thus

\begin{align} \int_0^\infty|\sin x|^x\,dx &\ge\sum_{k=0}^\infty\int_0^1|\sin(x+\pi/2+k\pi)|^{x+\pi/2+k\pi}\,dx\\ &=\sum_{k=0}^\infty\int_0^1|\cos x|^{x+\pi/2+k\pi}\,dx\\ &\ge|\cos1|^{1+\pi/2}\sum_{k=0}^\infty\int_0^1|\cos x|^{k\pi}\,dx\\ &\ge|\cos1|^{1+\pi/2}\sum_{k=0}^\infty\int_0^1(1-x)^{k\pi}\,dx\\ &=|\cos1|^{1+\pi/2}\sum_{k=0}^\infty{1\over k\pi+1}\\ &=\infty \end{align}

We were able to replace $$|\cos x|^{x+\pi/2}$$ with the constant $$|\cos1|^{1+\pi/2}\approx0.20543$$ in the second inequality step because the cosine function is decreasing (and less than or equal to $$1$$) on the interval $$[0,1]$$. The inequality $$\cos x\ge1-x$$ for $$x\in[0,1]$$ is easy to check; the final infinite series diverges by comparison to the harmonic series -- if you like, we've shown that

$$\int_0^{(N+1)\pi}|\sin x|^x\,dx\gt{1\over5\pi}\sum_{k=1}^N{1\over k}$$

Integral Recursion \begin{align} \int_0^\pi\sin^{2k}(x)\,\mathrm{d}x &=-\int_0^\pi\sin^{2k-1}(x)\,\mathrm{d}\cos(x)\tag{1a}\\ &=(2k-1)\int_0^\pi\cos^2(x)\sin^{2n-2}(x)\,\mathrm{d}x\tag{1b}\\ &=(2k-1)\int_0^\pi\left(\sin^{2k-2}(x)-\sin^{2k}(x)\right)\mathrm{d}x\tag{1c}\\ &=\frac{2k-1}{2k}\int_0^\pi\sin^{2k-2}(x)\,\mathrm{d}x\tag{1d} \end{align} Explanation:
$$\text{(1a)}$$: prepare to integrate by parts
$$\text{(1b)}$$: integrate by parts
$$\text{(1c)}$$: $$\cos^2(x)=1-\sin^2(x)$$
$$\text{(1d)}$$: add $$\frac{2k-1}{2k}$$ times $$\text{(1a)}$$ to $$\frac1{2k}$$ times $$\text{(1c)}$$

Product of the Factors \begin{align} \prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2 &\ge\frac14\prod_{k=2}^n\frac{2k-1}{2k}\frac{2k-2}{2k-1}\tag{2a}\\ &=\frac1{4n}\tag{2b} \end{align} Explanation:
$$\text{(2a)}$$: cross multiplication shows that $$\frac{2k-1}{2k}\ge\frac{2k-2}{2k-1}$$
$$\text{(2b)}$$: telescoping product

Integral Estimate

Applying the square root of $$(2)$$ to the recursion in $$(1)$$ yields, by induction, $$\int_0^\pi\sin^{2n}(x)\,\mathrm{d}x\ge\frac\pi{\sqrt{4n}}\tag3$$ Therefore, \begin{align} \int_{(n-1)\pi}^{n\pi}|\sin(x)|^x\,\mathrm{d}x &\ge\int_{(n-1)\pi}^{n\pi}|\sin(x)|^{n\pi}\,\mathrm{d}x\tag{4a}\\ &\ge\int_{(n-1)\pi}^{n\pi}\sin^{2\left\lceil\frac{n\pi}2\right\rceil}(x)\,\mathrm{d}x\tag{4b}\\ &=\int_0^\pi\sin^{2\left\lceil\frac{n\pi}2\right\rceil}(x)\,\mathrm{d}x\tag{4c}\\[3pt] &\ge\frac\pi{\sqrt{2\pi(n+1)}}\tag{4d} \end{align} Explanation:
$$\text{(4a)}$$: $$|\sin(x)|\le1$$ and $$x\le n\pi$$
$$\text{(4b)}$$: $$|\sin(x)|\le1$$ and $$n\pi\le 2\left\lceil\frac{n\pi}2\right\rceil$$
$$\text{(4c)}$$: periodicity of $$|\sin(x)|$$
$$\text{(4d)}$$: $$\left\lceil\frac{n\pi}2\right\rceil\le\frac{n\pi}2+1\le\frac{\pi(n+1)}2$$; apply $$(3)$$

Conclusion

Breaking up the integral and applying $$(4)$$ gives \begin{align} \int_0^\infty|\sin(x)|^x\,\mathrm{d}x &=\sum_{n=1}^\infty\int_{(n-1)\pi}^{n\pi}|\sin(x)|^x\,\mathrm{d}x\tag{5a}\\ &\ge\sqrt{\frac\pi2}\,\sum_{n=1}^\infty\frac1{\sqrt{n+1}}\tag{5b} \end{align} which diverges.