$$I=\int_{0}^{\infty} \frac{\sin \left(x+\frac{1}{x}\right)}{x}dx=\pi J_0(2)$$
I'v found:$$I=2\int_{0}^{\infty}\frac{\sin \left(x^2+\frac{1}{x^2}\right)}{x}dx=3\int_{0}^{\infty}\frac{\sin \left(x^3+\frac{1}{x^3}\right)}{x}dx=...=n\int_{0}^{\infty}\frac{\sin \left(x^n+\frac{1}{x^n}\right)}{x}dx$$ But I still wouldn't prove the integral，can someone help me?
$$ \int_{0}^{+\infty}\sin\left(x^n+\frac{1}{x^n}\right)\frac{dx}{x} = 2\int_{1}^{+\infty}\sin\left(x^n+\frac{1}{x^n}\right)\frac{dx}{x}=2\int_{0}^{+\infty}\sin(2\cosh(nz))\,dz $$ equals, in the improper-Riemann sense, $$ \frac{2}{n}\int_{0}^{+\infty}\sin(2\cosh u)\,du = \frac{2}{n}\underbrace{\int_{1}^{+\infty}\frac{\sin(2v)}{\sqrt{v^2-1}}\,dv}_{\text{constant}}. $$ In order to derive an explicit expression for the involved constant it is practical to study the function $$ f(a) = \int_{1}^{+\infty}\frac{\sin(a v)}{\sqrt{v^2-1}}\,dv$$ by integration by parts and differentiation under the integral sign, or just through the Laplace transform. We have $$ (\mathcal{L}f)(s) = \int_{1}^{+\infty}\frac{v}{(s^2+v^2)\sqrt{v^2-1}}\,dv=\frac{\pi}{2\sqrt{1+s^2}}=g(s) $$ and by the extended binomial theorem $$ g(s) = \frac{\pi}{2}\sum_{n\geq 0}\frac{(-1)^n \binom{2n}{n}}{4^n s^{2n+1}} $$ such that $$ f(a) = (\mathcal{L}^{-1}g)(a) = \frac{\pi}{2}\sum_{n\geq 0}\frac{(-1)^n \binom{2n}{n}}{4^n (2n)!}a^{2n}=\frac{\pi}{2}\sum_{n\geq 0}\frac{(-1)^n a^{2n}}{4^n n!^2} = \frac{\pi}{2}\,J_0(a)$$ by the series definition of $J_0$. By putting everything together we get:
$$ \int_{0}^{+\infty}\sin\left(x^n+\frac{1}{x^n}\right)\frac{dx}{x} = \color{red}{\frac{\pi\,J_0(2)}{n}}\approx \frac{1105}{1571 n}. $$
Probably the simplest way to see this is to sub $x=e^t$. The result is that the integral is seen to be equal to
$$\int_{-\infty}^{\infty} dt \, \sin{(2 \cosh{t})} = 2 \operatorname{Im} \int_0^{\infty} dt \, e^{i 2 \cosh{t}} = 2 \operatorname{Im} K_0(-i 2)$$
and use the fact that
$$K_0(-i 2) = \frac{\pi}{2} \left (Y_0(2) + i J_0(2) \right ) $$
and the result follows.
