Divergence of Improper integral with trigonometric functions

Question

I need to test the convergence of the following improper integral:

$$ \int_1^{+\infty} |\sin x|^x dx $$

I tried to find a minorant, but I did not find it immediately. Furthermore, I tried to split the integral into a series, by dividing the interval x>1 in intervals of the form $[1,\pi]$, $[k\pi,(k+1)\pi]$, with $k \ge 1$. Can you help me, please?

Answer 1

We begin by noting that, for $n \geq 0$,

$$ \int_{0}^{\pi} \sin^n x \, \mathrm{d}x = 2 \int_{0}^{\frac{\pi}{2}} \sin^n x \, \mathrm{d}x \geq 2 \int_{0}^{\frac{\pi}{2}} \left(\frac{2x}{\pi} \right)^n \, \mathrm{d}x = \frac{\pi}{n+1}. \tag{1} $$

Here, the inequality $\sin x \geq 2x/\pi$, valid for $0 \leq x \leq \frac{\pi}{2}$, is utilized. Next, we bound the integral from below by

\begin{align*} \int_{1}^{\infty} \left| \sin x \right|^x \, \mathrm{d}x &\geq \int_{\pi}^{\infty} \left| \sin x \right|^x \, \mathrm{d}x \\ &= \sum_{n=1}^{\infty} \int_{n\pi}^{(n+1)\pi} \left| \sin x \right|^{x} \, \mathrm{d}x \\ &\geq \sum_{n=1}^{\infty} \int_{n\pi}^{(n+1)\pi} \left| \sin x \right|^{(n+1)\pi} \, \mathrm{d}x \\ &= \sum_{n=1}^{\infty} \int_{0}^{\pi} (\sin x)^{(n+1)\pi} \, \mathrm{d}x. \end{align*}

Now this lower bound can be further bounded from below by using $\text{(1)}$ to prove the desired divergence.

Answer 2

If $x=(k-1/2)\pi+t$ with $k>2$ and $|t|\leqslant t_k:=\sqrt{2(1-2^{-1/(k\pi)})}$, then $$|\sin x|^x=(\cos t)^x\geqslant(1-t_k^2/2)^{k\pi}=1/2,$$ hence a lower bound of $(1/2)\sum_{k>2}(2t_k)=\sum_{k>2}t_k$ for the integral.

But $t_k\sqrt{k}\to\sqrt{(2\log 2)/\pi}\neq 0$ as $k\to\infty$, and $\sum_{k>2}1/\sqrt{k}$ diverges.

Answer 3

We have that $$ \int_{k\pi}^{(k+1)\pi} |\sin x|^x \, dx \geq \int_0^\pi (\sin x)^{4(k+1)}\, dx =: J_{2(k+1)}. $$ As for the values of $J_n$, we have that $J_0 = \pi$ and the recurrence relation $$ J_{n} = \frac{2n-1}{2n}\, J_{n-1}\,, $$ so that $$ J_n = \pi \, \frac{(2n-1)!!}{(2n)!!} = \sqrt{\pi}\, \frac{\Gamma(n+1/2)}{\Gamma(n+1)}. $$ Since $$ \lim_{n\to +\infty} \frac{\Gamma(n-\alpha) n^\alpha}{\Gamma(n)} = 1, $$ we finally get $$ J_n \sim \sqrt{\frac{\pi}{n}} $$ so that $\sum J_{2n}$ diverges and, by comparison, also the integral diverges.