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...
 A: $$\begin{align}\text{Integral} &=\sum_{n=0}^{\infty} \int_{0}^{\pi} |\sin(x)|^{x +n\pi}dx \ge \sum_{n=0}^{\infty} \int_{0}^{\pi} |\sin(x)|^{\pi +n\pi}dx \\
& \stackrel{\text{Fubini}}{=}  \int_0^{\pi} \frac{\sin(x)^{\pi}}{1-\sin(x)^{\pi}}dx= -\pi+2\int_0^{\pi/2}\frac{1}{1-\sin(x)^{\pi}}dx\\
\\
&\ge -\pi+2\int_0^{\pi/2}\frac{1}{1-\sin(x)^{4}}dx =-\pi+\int_0^{\pi/2}\frac{2}{(1-\sin(x)^{2})(1+\sin(x)^2)}dx 
\\
&\ge -\pi+\int_0^{\pi/2}\frac{1}{1-\sin(x)^{2}}dx 
\\
&=-\pi+ \left( \tan(x)\right)\bigg\vert_0^{\pi/2} =\infty
\end{align}$$
https://en.wikipedia.org/wiki/Fubini%27s_theorem
A: 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.
A: 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.
A: 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.
A: 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
$$\int_0^{(N+1)\pi}|\sin x|^x\,dx\gt{1\over5\pi}\sum_{k=1}^N{1\over k}$$
