Definite integral - closed form: $\int_{0}^{\infty}\cos\left(x^{4} + 1 \over x^{2}\right)\,{\rm d}x$ I'm struggling with this definite integral:
$$ 
\int_{0}^{\infty}\cos\left(x^{4} + 1 \over x^{2}\right)\,{\rm d}x.
$$
Any help will be greatly appreciated.
 A: Rewrite:
$$
\int_{0}^{\infty}\cos\left(x^{4} + 1 \over x^{2}\right)\,dx=\int_{0}^{\infty}\cos\left(x^{2} + \frac1{x^2}\right)\,dx=\Re\left[\int_{0}^{\infty}e^{\Large-i\left(x^{2} + \frac1{x^2}\right)}\,dx\right].\tag1
$$
Consider my answer that I posted on Math SE
$$
\begin{align}
\int_{0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx
&=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{\large-2\sqrt{ab}}.
\end{align}
$$
Taking $a=i$ and $b=i$, where $i=\sqrt{-1}$, then $(1)$ turns out to be
$$
\begin{align}
\Re\left[\int_{0}^{\infty}e^{\Large-i\left(x^{2} + \frac1{x^2}\right)}\,dx\right]
&=\frac{1}{2}\Re\left[\sqrt{\frac{\pi}{i}}e^{\large-2\sqrt{i\cdot i}}\right]\\
&=\frac{1}{2}\Re\left[\sqrt{\pi}\cdot i^{-\large\frac12} \cdot\ e^{\large-2i}\right],\tag2
\end{align}
$$
where
$$
i^{-\large\frac12}=\left(\cos\left(\frac\pi2\right)+i\sin\left(\frac\pi2\right)\right)^{-\large\frac12}=e^{\Large-\frac\pi4i}=\cos\left(\frac\pi4\right)-i\sin\left(\frac\pi4\right)=\frac{1}{\sqrt2}-\frac{i}{\sqrt2}
$$
and
$$
e^{\large-2i}=\cos2-i\sin2.
$$
Taking the real part of $(2)$, we obtain
$$
\int_{0}^{\infty}\cos\left(x^{4} + 1 \over x^{2}\right)\,dx=\color{blue}{\frac{1}{2}\sqrt{\frac{\pi}{2}}(\cos2-\sin2)}.
$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}\cos\pars{x^{4} + 1 \over x^{2}}\,\dd x:\ {\large ?}}$

\begin{align}
&\color{#00f}{\large\int_{0}^{\infty}\cos\pars{x^{4} + 1 \over x^{2}}\,\dd x}=\
\overbrace{\int_{0}^{\infty}\cos\pars{x^{2} + {1 \over x^{2}}}\,\dd x}
^{\ds{\mbox{Set}\ x \equiv \expo{\theta}}}\
\\[3mm]&=\ \int_{-\infty}^{\infty}\cos\pars{2\cosh\pars{2\theta}}\,\expo{\theta}\,\dd\theta
=\int_{-\infty}^{\infty}\cos\pars{2\cosh\pars{2\theta}}\,
\bracks{\cosh\pars{\theta} + \sinh\pars{\theta}}\,\dd\theta
\\[3mm]&=2\int_{0}^{\infty}\cos\pars{2\cosh\pars{2\theta}}\,\cosh\pars{\theta}
\,\dd\theta
\\[3mm]&=2\ \overbrace{%
\int_{0}^{\infty}\cos\pars{2\bracks{2\sinh^{2}\pars{\theta} + 1}}\,
\cosh\pars{\theta}\,\dd\theta}^{\ds{\mbox{Set}\ t \equiv \sinh\pars{\theta}}}
=2\int_{0}^{\infty}\cos\pars{4t^{2} + 2}\,\dd t
\\[3mm]&=\int_{0}^{\infty}\cos\pars{t^{2} + 2}\,\dd t
=\cos\pars{2}\int_{0}^{\infty}\cos\pars{t^{2}}\,\dd t
-\sin\pars{2}\int_{0}^{\infty}\sin\pars{t^{2}}\,\dd t
\\[3mm]&=\cos\pars{2}\lim_{\xi \to \infty}{\rm C}\pars{\xi}
-\sin\pars{2}\lim_{\xi \to \infty}{\rm S}\pars{\xi}
\end{align}
  where $\ds{{\rm C}\pars{\xi}}$ and $\ds{{\rm S}\pars{\xi}}$ are the
  Fresnel Integrals and the above limits are equal to $\ds{\root{\pi \over 8}}$.

\begin{align}
&\color{#00f}{\large\int_{0}^{\infty}\cos\pars{x^{4} + 1 \over x^{2}}\,\dd x}=
\color{#00f}{\large\bracks{\cos\pars{2} - \sin\pars{2}}\root{\pi \over 8}}
\approx -0.8306
\end{align}
A: $$
I=\int_{0}^{\infty}\cos\left(x^{2} + \frac{1}{x^{2}}\right)\,{\rm d}x
=\int_{0}^{1}\cos\left(x^{2} + \frac{1}{x^{2}}\right)\,{\rm d}x
+\int_{1}^{\infty}\cos\left(x^{2}+\frac{1}{x^{2}}\right)\,{\rm d}x
$$
Substituting $x=1/t$ on the second integral and adding up yields
\begin{align}
&\int_{0}^{1}
\left(1 + \frac{1}{t^{2}}\right)\cos\left(t^{2} + \frac{1}{t^{2}}\right)\,{\rm d}t
=\int_{0}^{1}\cos\left(\left[t-\frac{1}{t}\right]^{2}+2\right)
\,{\rm d}\left(t-\frac{1}{t}\right)
\\[3mm]&=\int_{-\infty}^{0}\cos\left(u^{2}+2\right)\,{\rm d}u
=\int_{0}^{\infty}\cos\left(u^{2} + 2\right)\,{\rm d}u
\\[3mm]&=\cos\left(2\right)
\int_{0}^{\infty}\cos\left(u^{2}\right)\,{\rm d}u
-\sin\left(2\right)\int_{0}^{\infty}\sin\left(u^{2}\right)\,{\rm d}u
\end{align}
Feel free to look up the Fresnel integrals, i.e
$$\int_{0}^{\infty}\cos\left(u^{2}\right)\,{\rm d}u
=\int_{0}^{\infty}\sin\left(u^{2}\right)\,{\rm d}u
=\frac{\sqrt{\,\pi\,}\,}{2\,\sqrt{\,2\,}\,}
$$
Adding up we finally arrive at
$$\int_{0}^{\infty}\cos\left(x^{2} + \frac{1}{x^{2}}\right)\,{\rm d}x
=\left[\cos\left(2\right) - \sin\left(2\right)\right]\,
\frac{\sqrt{\,\pi\,}}{2\,\sqrt{\,2\,}\,}$$
A: Note
\begin{align} 
\int_{0}^{\infty}\cos\left(x^{4} + 1 \over x^{2}\right)\,{d}x
&= \frac12 \int_{-\infty}^{\infty}\cos\left((x-\frac1x)^2+2\right)\>dx= \frac12 \int_{-\infty}^{\infty}\cos\left(x^2+2\right)\>dx\\
&= \cos2\int_{0}^{\infty}\cos x^2\>dx - \sin2\int_{0}^{\infty}\sin x^2\>dx\\
&
= \frac{\sqrt\pi}2\cos(2+\frac\pi4)
\end{align}
where
$$\int_{-\infty}^{\infty}f\left(x-\frac1x\right)dx= \int_{-\infty}^{\infty}f(x)dx, \>\>\>
 \int_{0}^{\infty}\cos x^2dx = \int_{0}^{\infty}\sin x^2dx = \frac{\sqrt\pi}{2\sqrt2} $$
are used.
