0

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}

Guy Fsone
  • 23,903
  • @Masacroso I mean absolute value – Guy Fsone Aug 21 '17 at 13:26
  • 1
    I deleted my answer because the infinite integral converges, and i could have a proof for Riemman integrability, but not for Lebesgue integrability. – Brethlosze Aug 21 '17 at 13:33
  • just an idea: you can write the integral as the sum of two series using the fact that the cosine is periodic. Or maybe exists some change of variable that make the limits of integration finite. – Masacroso Aug 21 '17 at 13:36
  • What did you try? – Did Aug 21 '17 at 17:01
  • 1
    Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Aug 22 '17 at 05:34

3 Answers3

5

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*}

Robert Z
  • 145,942
5

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 $\int_{-\infty}^{+\infty}\left(\frac{2+\cos x}{3}\right)^{x^4}\,dx\leq \color{blue}{7}.$

Jack D'Aurizio
  • 353,855
  • Hello, @Jack please how do you prove your Lemma? – Guy Fsone Aug 23 '17 at 13:58
  • @GuyFsone: it is enough to consider a Padé approximant of $\log\frac{2+\cos\sqrt{x}}{3}$ on the interval $[0,\pi^2]$. But you do not really need such sharp bound, $\frac{2+\cos x}{3}\leq e^{-x^2/18}$ is simpler to prove and works just as well for proving the convergence of the given integral; even $\frac{2+\cos x}{3}\leq e^{-x^2/1000}$ does the job. – Jack D'Aurizio Aug 23 '17 at 16:00
0

\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}

Guy Fsone
  • 23,903