Find a general solution for $\int_{0}^{\infty} \sin\left(x^n\right)\:dx$

Some Hints:

$$I=\int_0^{\infty} \sin (x^n)dx $$ On substitution $x^n=t$ and using the series of $\sin$ we get $$I=\frac 1n \int_0^{\infty} t^{\frac 1n} \left(\sum_{k=0}^{\infty} (-1)^k \frac {t^{2k}k!}{(2k+1)!k!} \right) dt$$

On substituting $t^2=u$ we get $$ I= \frac {1}{2n} \int_0^{\infty} u^{\frac {1-n}{2n}}\left(\sum_{k=0}^{\infty} \frac {\frac {k!}{(2k+1)!}}{k!} (-u)^k \right) du$$

Now by Ramanujan's Master Theorem

$$I=\frac {1}{2n} \Gamma(s)\phi(-s)$$ where $\phi(k)=\frac {k!}{(2k+1)!}$ and $s=\frac {n+1}{2n}$

Hence along with properties of Gamma function, Mellin Transform and the Euler's reflection formula we get $$I=\frac {\pi}{2n\cos \left(\frac {\pi}{2n}\right)\Gamma \left(1-\frac 1n\right)}=\sin \left(\frac {\pi}{2n}\right)\frac {\Gamma\left(\frac 1n\right)}{n}$$

With a special case of $n=2$ we get the value of special integral popularly known as Fresnel integral with limit as $x$ tends to infinity

Start out with a couple of integration by parts: $$ \begin{align} \int_0^\infty\sin(x)\,e^{-xy}\,\mathrm{d}x &=-\frac1y\int_0^\infty\sin(x)\,\mathrm{d}e^{-xy}\tag1\\ &=\frac1y\int_0^\infty\cos(x)\,e^{-xy}\,\mathrm{d}x\tag2\\ &=-\frac1{y^2}\int_0^\infty\cos(x)\,\mathrm{d}e^{-xy}\tag3\\ &=\frac1{y^2}-\frac1{y^2}\int_0^\infty\sin(x)\,e^{-xy}\,\mathrm{d}x\tag4\\ &=\frac1{y^2+1}\tag5 \end{align} $$ Explanation:
$(1)$: prepare to integrate by parts
$(2)$: integrate by parts
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: add $\frac{y^2}{y^2+1}$ times $(4)$ to $\frac1{y^2+1}$ times the LHS of $(1)$

Now write $$ \begin{align} \int_0^\infty\sin\left(x^n\right)\,\mathrm{d}x &=\frac1n\int_0^\infty\sin(x)\,x^{\frac1n-1}\,\mathrm{d}x\tag6\\[3pt] &=\frac1{n\,\Gamma\!\left(1-\frac1n\right)}\int_0^\infty\sin(x)\int_0^\infty y^{-\frac1n}e^{-xy}\,\mathrm{d}y\,\mathrm{d}x\tag7\\ &=\frac1{n\,\Gamma\!\left(1-\frac1n\right)}\int_0^\infty y^{-\frac1n}\int_0^\infty\sin(x)\,e^{-xy}\,\mathrm{d}x\,\mathrm{d}y\tag8\\ &=\frac1{n\,\color{#C00}{\Gamma\!\left(1-\frac1n\right)}}\color{#090}{\int_0^\infty\frac{y^{-\frac1n}}{y^2+1}\,\mathrm{d}y}\tag9\\ &=\color{#C00}{\frac{\Gamma\!\left(\frac1n\right)\sin(\frac\pi{n})}{\color{#000}{n}\pi}}\color{#090}{\frac\pi2\sec\left(\frac\pi{2n}\right)}\tag{10}\\[9pt] &=\Gamma\!\left(1+\frac1n\right)\sin\left(\frac\pi{2n}\right)\tag{11} \end{align} $$ Explanation:
$\phantom{1}(6)$: substitute $x\mapsto x^{1/n}$
$\phantom{1}(7)$: $\int_0^\infty y^{-\frac1n}e^{-xy}\,\mathrm{d}y=x^{\frac1n-1}\Gamma\!\left(1-\frac1n\right)$
$\phantom{1}(8)$: Fubini
$\phantom{1}(9)$: apply $(5)$
$(10)$: $(4)$ from this answer for the green, and $(2)$ from the same answer for the red
$(11)$: simplify