Does this integral converge or diverge ? $\int_\Bbb R \left (\frac{2+\cos x}{3}\right )^{x^4}dx?$

Question

I would like to prove that the function

$$f(x) = \left (\frac{2+\cos x}{3}\right )^{x^4}$$ is Lebesgue-integrable on $\mathbb R. $ Namely I would like to show that,

$$\int_{\Bbb R} \left|f(x)\right|dx<\infty$$

My answer: \begin{split}\int_{\Bbb R}\left|f(x)\right|dx&=&2\sum_{n=0}^{\infty}\int_{2n\pi}^{(2n+2)\pi}\left (\frac{2+\cos (x)}{3}\right )^{x^4}\,dx\\ &=&2\sum_{n=0}^{\infty}\int_0^{2\pi}\left (\frac{2+\cos (x+2n\pi)}{3}\right )^{(x+2n\pi)^4}\,dx\\ &\leq& 2\sum_{n=0}^{\infty}\int_0^{2\pi}\left (\frac{2+\cos (x)}{3}\right )^{(2n\pi)^4}\,dx<\infty?. \end{split}

Answer 1

Hint. Note that $0<\frac{1}{3}\leq \frac{2+\cos x}{3}\leq 1$ and \begin{align*}\int_{-\infty}^{\infty}\left|f(x)\right|dx&= 2\sum_{k=0}^{\infty}\int_0^{2\pi}\left (\frac{2+\cos (t+2k\pi)}{3}\right )^{(t+2k\pi)^4}\,dt\\ &\leq 4\sum_{k=0}^{\infty}\int_0^{\pi}\left (\frac{2+\cos (t)}{3}\right )^{(2k\pi)^4}\,dt. \end{align*}

Answer 2

Lemma. For any $x\in[-\pi,\pi]$ we have $\frac{2+\cos x}{3}\leq e^{-x^2/9}$.

It follows that: $$\begin{eqnarray*} \int_{0}^{+\infty}\left(\frac{2+\cos x}{3}\right)^{x^4}\,dx &\leq&\int_{0}^{+\infty}e^{-x^6/9}\,dx+2\sum_{n\geq 1}\int_{0}^{+\infty}\exp\left[-\frac{x^2}{9}(x+2n\pi)^4\right]\,dx\\&\leq& \frac{\Gamma\left(\frac{1}{6}\right)}{3^{2/3}}+2\sum_{n\geq 1}\int_{0}^{+\infty}\exp\left[-\frac{16n^4 \pi^4}{9}x^2\right]\,dx\\&=&\frac{\Gamma\left(\frac{1}{6}\right)}{3^{2/3}}+\frac{3}{4\pi^{3/2}}\sum_{n\geq 1}\frac{1}{n^2}\\&=&\frac{\Gamma\left(\frac{1}{6}\right)}{3^{2/3}}+\frac{3\sqrt{\pi}}{8}\end{eqnarray*}$$ so $\int_{-\infty}^{+\infty}\left(\frac{2+\cos x}{3}\right)^{x^4}\,dx\leq \color{blue}{7}.$

Answer 3

\begin{split}\int_{\Bbb R}\left|f(x)\right|dx&=&2\sum_{n=0}^{\infty}\int_{2n\pi}^{(2n+2)\pi}\left (\frac{2+\cos (x)}{3}\right )^{x^4}\,dx\\ &=&2\sum_{n=0}^{\infty}\int_0^{2\pi}\left (\frac{2+\cos (x+2n\pi)}{3}\right )^{(x+2n\pi)^4}\,dx\\ &\leq& 2\sum_{n=0}^{\infty}\int_0^{2\pi}\left (\frac{2+\cos (x)}{3}\right )^{(2n\pi)^4}\,dx<\infty. \end{split}

Let us justify this: $$ \frac13 \le \left (\frac{2+\cos x}{3}\right )^{x^4}\le 1 $$ implies $$\left (\frac{2+\cos (x)}{3}\right )^{(x+2n\pi)^4} \le\left (\frac{2+\cos (x)}{3}\right )^{(2n\pi)^4}~~x\in(0,2\pi) $$ on the other hand, $a(x) = \ln\left(\frac{2+\cos (x)}{3}\right )< 0 ~~a.e$ then, $$ \lim_{n\to \infty }n^4 \left(\frac{2+\cos (x)}{3}\right )^{(2n\pi)^4}= \lim_{n\to \infty }n^4\exp\left(a(x)(2n\pi)^4 \right) =0$$ hence, for $n$ large enough we have $$\int_0^{2\pi}\left (\frac{2+\cos x}{3}\right )^{(2n\pi)^4}\,dx \le \frac{C}{n^{4}}$$ this entails that \begin{split}\int_{\Bbb R}\left|f(x)\right|dx&\le & 2\sum_{n=0}^{\infty}\int_0^{2\pi}\left (\frac{2+\cos (x)}{3}\right )^{(2n\pi)^4}\,dx\\ &\le& C\sum_{n =0}^{\infty}\frac{1}{n^4} \end{split}