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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.

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The numerical value seems to be around $-.83$. It most definitely converges since your integral is highly oscillatory. I would like to see if anyone can get a closed form expression for this though. – Cameron Williams May 14 '14 at 16:44
The closed form expression is $\int_0^{\infty } \cos \left(\frac{x^4+a^2}{x^2}\right) \, dx = \frac{1}{2} \sqrt{\frac{\pi }{2}} \left(\cos \left(2 a\right)-\sin \left(2 a \right)\right) $, for a > 0. I will try to show it. – user111187 May 14 '14 at 16:46

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

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

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Fast and very nice! – user111187 May 14 '14 at 16:48
That is not what Mathematica is getting. So maybe you should check. – bobbym May 14 '14 at 16:49
It's the same as Mathematica told me, so I think it's correct. – user111187 May 14 '14 at 16:50
Sorry, it is okay now. – bobbym May 14 '14 at 16:52
+1. It is a nice answer and it is pretty short. I hope you don't mind for my edition. – Felix Marin May 14 '14 at 18:35

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

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Awesome! +1 d(>‿◠)b – Anastasiya-Romanova 秀 May 15 '14 at 4:58
The computations in the other answer apply to $a$ and $b$ positive real numbers. Does the result apply to $a$ and $b$ complex? Certainly not without some more work, as witnessed by the awkward treatment of $\sqrt{i}$ in the answer here (one might recall at this point that there is NO square root function defined on the complex plane)... – Did May 15 '14 at 6:20
@Did I guess it's also work for complex numbers since use the similar technique to answer a few of integration questions here. I know the fact that: "there is no square root function defined on the complex plane", but I often see this treatment of imaginary number to solve many math questions (I forgot where) like I used in this answer. It's just like the math formula that is considering $0^0=1$, although it looks awkward but it works. – Tunk-Fey May 15 '14 at 8:06
If your point is that you or other users apply without precaution some identities a priori valid in the real domain to some complex values of the parameters to answer some questions on MSE (and that these answers may even garner upvotes and/or be accepted), then yes, I am aware of this unfortunate situation and I maintain that no valid proof can emerge from such moves without further justifications. In the present answer, there are none. – Did May 15 '14 at 8:21
The comment I left was incorrect. It would lead to an evaluation of $\int_{0}^{\infty} \cos \left(x^{2}- \frac{1}{x^{2}} \right) \ dx$, not $\int_{0}^{\infty} \cos \left(x^{2}+\frac{1}{x^{2}} \right) \ dx$. – Random Variable May 15 '14 at 8:48

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

\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}

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Very creative answer. +1 – Random Variable May 17 '14 at 17:50
@RandomVariable Thanks a lot. – Felix Marin May 17 '14 at 22:19

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