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

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.

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

$$\int_1^\infty\cos x^{2n} \,dx=\int_1^\infty\frac{(\sin x^{2n})'}{2nx^{2n-1}}\,dx=-\frac{\sin 1}{2n}+\frac{2n-1}{2n}\int_1^\infty\frac{\sin x^{2n}}{x^{2n}}\,dx\underset{n\to\infty}{\longrightarrow}0$$ (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 $$\int_{-\infty}^\infty\left[\begin{array}{c}\cos \\ \sin\end{array}\right]x^{2n}\,dx=2\Gamma\left(1+\frac{1}{2n}\right)\left[\begin{array}{c}\cos \\ \sin\end{array}\right]\frac{\pi}{4n}.$$
$$\ds{\LARGE\left. a\right)}$$ \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}