Definite integral - closed form: $\int_{0}^{\infty}\cos\left(x^{4} + 1 \over x^{2}\right)\,{\rm d}x$

Question

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.

Answer 1

$$ I=\int_{0}^{\infty}\cos\left(x^{2} + \frac{1}{x^{2}}\right)\,{\rm d}x =\int_{0}^{1}\cos\left(x^{2} + \frac{1}{x^{2}}\right)\,{\rm d}x +\int_{1}^{\infty}\cos\left(x^{2}+\frac{1}{x^{2}}\right)\,{\rm d}x $$

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 $$\int_{0}^{\infty}\cos\left(x^{2} + \frac{1}{x^{2}}\right)\,{\rm d}x =\left[\cos\left(2\right) - \sin\left(2\right)\right]\, \frac{\sqrt{\,\pi\,}}{2\,\sqrt{\,2\,}\,}$$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}\cos\pars{x^{4} + 1 \over x^{2}}\,\dd x:\ {\large ?}}$

\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}

Answer 3

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 $$ \int_{0}^{\infty}\cos\left(x^{4} + 1 \over x^{2}\right)\,dx=\color{blue}{\frac{1}{2}\sqrt{\frac{\pi}{2}}(\cos2-\sin2)}. $$

Answer 4

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.