1
$\begingroup$

Like the title says, I'm curious if anyone has any insight on trying to compute these limits. Numerical investigations seem to indicate that

$$\lim_{n\to\infty} \int_{-\infty}^\infty \cos(x^{2n}) \:dx = 2$$

and

$$\lim_{n\to\infty} 2n \int_{-\infty}^\infty \sin(x^{2n}) \:dx = \pi$$

For the first one, it seems to almost follow from dominated convergence since

$$\lim_{n\to\infty} \int_{-1}^1 \cos(x^{2n}) \:dx = \int_{-1}^1 \cos(0) \:dx = 2$$

All there is left to prove is that

$$\lim_{n\to\infty} \int_1^\infty \cos(x^{2n})\:dx = 0$$

I've tried integrating by parts and a Fourier transform argument, but nothing seems to definitively pin this limit as being zero in a rigorous way.

For the other one I am completely at a loss as to where the $\pi$ would come from in a dominated convergence style argument since the usual trick would give some multiple of $\sin(1)$. Granted, the limit may not be $\pi$, but I am having even less luck with this limit than the other. Any tips are appreciated.

$\endgroup$
2
  • $\begingroup$ How the Hell can you numerically evaluate these improper integrals with an hypersocillating integrand ? $\endgroup$ – Yves Daoust Apr 5 '20 at 10:09
  • $\begingroup$ @YvesDaoust Wolfram :) $\endgroup$ – Ninad Munshi Apr 5 '20 at 10:09
1
$\begingroup$

$$\int_1^\infty\cos x^{2n} \,dx=\int_1^\infty\frac{(\sin x^{2n})'}{2nx^{2n-1}}\,dx=-\frac{\sin 1}{2n}+\frac{2n-1}{2n}\int_1^\infty\frac{\sin x^{2n}}{x^{2n}}\,dx\underset{n\to\infty}{\longrightarrow}0$$ (yes, we integrate by parts). For the sine integral, we have similarly $$2n\int_{-\infty}^\infty\sin x^{2n}\,dx=(2n-1)\int_{-\infty}^\infty\frac{1-\cos x^{2n}}{x^{2n}}\,dx=\frac{2n-1}{n}\int_0^\infty\frac{1-\cos y}{y^{2-1/(2n)}}\,dy,$$ again with the limit allowed to be taken under the integral sign (if we substitute $x^{2n}=y$ in the original integrals, it's not that easy to justify), resulting in a known integral. In fact, it's known that $$\int_{-\infty}^\infty\left[\begin{array}{c}\cos \\ \sin\end{array}\right]x^{2n}\,dx=2\Gamma\left(1+\frac{1}{2n}\right)\left[\begin{array}{c}\cos \\ \sin\end{array}\right]\frac{\pi}{4n}.$$

$\endgroup$
2
  • $\begingroup$ What's that notation in the concluding sentence? $\endgroup$ – Allawonder Apr 5 '20 at 10:48
  • 1
    $\begingroup$ @Allawonder: "this or that", just not to write out two almost identical formulas... $\endgroup$ – metamorphy Apr 5 '20 at 10:50
1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Integrals are evaluated with Ramanujan's Master Theorem. $$ \mbox{Note that}\quad \left\{\begin{array}{rcl} \ds{\sin\pars{\root{x}} \over \root{x}} & \ds{=} & \ds{\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k} \over \Gamma\pars{2 + 2k}}{\pars{-x}^{k} \over k!}} \\[1mm] \ds{\cos\pars{\root{x}}} & \ds{=} & \ds{\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k} \over \Gamma\pars{1 + 2k}}{\pars{-x}^{k} \over k!}} \end{array}\right. $$


$\ds{\LARGE\left. a\right)}$ \begin{align} &\bbox[5px,#ffd]{\lim_{n\to\infty} \int_{-\infty}^{\infty}\cos\pars{x^{2n}} \,\dd x} \\[5mm] \stackrel{x\ \mapsto\ x^{1/\pars{4n}}}{=}\,\,\,\,\,\,\, & 2\lim_{n\to\infty}\bracks{{1 \over 4n} \int_{0}^{\infty}x^{\color{red}{1/\pars{4n}} - 1}\,\,\cos\pars{\root{x}} \,\dd x} \\[5mm] = &\ 2\lim_{n\to\infty}\bracks{{1 \over 4n}\Gamma\pars{1 \over 4n}\, {\Gamma\pars{1 - 1/\bracks{4n}} \over \Gamma\pars{1 -1/\bracks{2n}}}} \\[5mm] = &\ 2\lim_{n\to\infty}\bracks{{1 \over 4n}{\pi \over \sin\pars{\pi/\bracks{4n}}}} = \bbx{2} \\ & \end{align}
$\ds{\LARGE\left. b\right)}$ \begin{align} &\bbox[5px,#ffd]{\lim_{n\to\infty} \bracks{2n\int_{-\infty}^{\infty}\sin\pars{x^{2n}} \,\dd x}} \\[5mm] \stackrel{x\ \mapsto\ x^{1/\pars{4n}}}{=}\,\,\,\,\,\,\,\,\,& \lim_{n\to\infty} \int_{0}^{\infty} x^{\color{red}{1/\pars{4n} + 1/2} - 1}\,\,\,\,\,\,{\sin\pars{\root{x}} \over \root{x}} \,\dd x \\[5mm] = &\ \lim_{n\to\infty}\braces{% \Gamma\pars{{1 \over 4n} + {1 \over 2}}\, {\Gamma\pars{1/2 - 1/\bracks{4n}} \over \Gamma\pars{1 - 1/\bracks{2n}}}} \\[5mm] = &\ \lim_{n\to\infty}\,\,\, {\pi \over \sin\pars{\pi\braces{1/2 + 1/\bracks{4n}}}} = \bbx{\pi} \\ & \end{align}
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.