$\lim_{n\to\infty} \int_{-\infty}^\infty \cos(x^{2n}) \:dx$ 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.
 A: $$\int_1^\infty\cos x^{2n} \,dx=\int_1^\infty\frac{(\sin x^{2n})'}{2nx^{2n-1}}\,dx=-\frac{\sin 1}{2n}+\frac{2n-1}{2n}\int_1^\infty\frac{\sin x^{2n}}{x^{2n}}\,dx\underset{n\to\infty}{\longrightarrow}0$$ (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 $$\int_{-\infty}^\infty\left[\begin{array}{c}\cos \\ \sin\end{array}\right]x^{2n}\,dx=2\Gamma\left(1+\frac{1}{2n}\right)\left[\begin{array}{c}\cos \\ \sin\end{array}\right]\frac{\pi}{4n}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
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}
