Evaluating $ \int^{\infty}_{-\infty}\sin\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right) dx$ $$\int^{\infty}_{-\infty}\sin\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right) dx$$
This is a problem from the Pi Mu Epsilon Journal, and I'm having great trouble answering it. I've tried some substitutions and any trick I could think of to find some multiple of the integral, but everything has led to a dead end. Perhaps it's not even solvable with methods of real analysis?
Hints and other comments will be much appreciated!
 A: Not a different answer but rather generalizing the technique Peter Tamaroff used in his answer. Notice the fact that $(u - u^{-1})^2 + 2 = u^2 + u^{-2}$, we can generalize this for any convergent definite integral. 
Say if we want to compute:
$$
\int^{\infty}_{-\infty} f(x)\,dx, 
$$
knowing this is convergent. Using the trick we can see the integral is:
$$
\int^{\infty}_{-\infty} f(x)\,dx =  \int^{\infty}_{-\infty} f\left(x - \frac{a}{x}\right)\,dx \tag{1}
$$
Proof: We can make the substitution: 
$$ u = x - \frac{a}{x}, \quad \text{for some } a>0.$$
Notice $u$ is again from $-\infty$ to $\infty$ for both $x>0$ and $x<0$. Then we have:
$$
x = \frac{u + \sqrt{u^2 + 4a}}{2} > 0  \;\text{ or }\; \frac{u - \sqrt{u^2 + 4a}}{2} < 0
$$
hence
$$
\frac{dx}{du} = \left\{\begin{aligned}
\frac{1}{2}+ \frac{u}{2\sqrt{u^2 + 4a}} \quad \text{when }x>0
\\
\frac{1}{2} - \frac{u}{2\sqrt{u^2 + 4a}} \quad \text{when }x<0
\end{aligned}\right.
$$
Hence: 
$$
\int^{\infty}_{-\infty} f(u)\left(\frac{1}{2}+ \frac{u}{2\sqrt{u^2 + 4a}}\right)  du = \int^{\infty}_{0} f\left(x - \frac{a}{x}\right) \,dx \tag{2}
$$
and
$$
\int^{\infty}_{-\infty} f(u)\left(\frac{1}{2} -  \frac{u}{2\sqrt{u^2 + 4a}}\right)  du = \int^{0}_{-\infty} f\left(x - \frac{a}{x}\right)\,dx \tag{3}
$$
(2)+(3) yields (1). 

The integral in OP is 
$$
\int^{\infty}_{-\infty}\sin\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right) dx
= \int^{\infty}_{-\infty}\sin\left({\pi}^4\Big(x - \frac{1}{\pi^2 x}\Big)^2 + 2\pi^2  \right) dx.
$$
Now using (1):
$$
\int^{\infty}_{-\infty}\sin\left({\pi}^4\Big(x - \frac{1}{\pi^2 x}\Big)^2 + 2\pi^2  \right) dx = \int^{\infty}_{-\infty}\sin ({\pi}^4 x^2 + 2\pi^2) \,dx
$$ 
We can easily see this is the same integral as Peter Tamaroff got in his answer:
$$
\frac{2}{\pi }\int_0^\infty  {\sin } \left( {{{\left( {\pi x} \right)}^2} + 2{\pi ^2}} \right)dx,
$$
and then Fresnel integral kicks in.
A: By evenness and $\pi x\mapsto u$ we get it is $$\frac{2}{\pi }\int_0^\infty  {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)du$$
Now, split at $x=1$, $$\frac{2}{\pi }\int_0^1 {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)du + \frac{2}{\pi }\int_1^\infty  {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)du$$ By $u\mapsto u^{-1}$, we get $$\frac{2}{\pi }\int_1^\infty  {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)\frac{{du}}{{{u^2}}} + \frac{2}{\pi }\int_1^\infty  {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)du$$ or $$\frac{2}{\pi }\int_1^\infty  {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)\left( {1 + \frac{1}{{{u^2}}}} \right)du$$
Since $u^2+u^{-2}=(u-u^{-1})^2+2$ and $(u-u^{-1})'=1+u^{-2}$ we get $$\frac{2}{\pi }\int_0^\infty  {\sin } \left( {{{\left( {\pi x} \right)}^2} + 2{\pi ^2}} \right)dx$$
Now use the sine sum formula and the values of the Fresnel integrals to conclude, the value is: $$\frac{{1 }}{{\sqrt 2{\pi ^{3/2}}}}\left( {\cos 2{\pi ^2} + \sin 2{\pi ^2}} \right)$$
See this for some generalized formulas
