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

Solution 1:

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.

Solution 2:

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

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