Find a general solution for $\int_{0}^{\infty} \sin\left(x^n\right)\:dx$ So, I was recently working on the Sine Fresnal integral and was curious whether we could generalise for any Real Number, i.e. 
$$I = \int_{0}^{\infty} \sin\left(x^n\right)\:dx$$
I have formed a solution that I'm uncomfortable with and was hoping for qualified eyes to have a look over. 
So, the approach I took was to employ Complex Numbers (I forget the name(s) of the theorem that allows this). 
But 
$$\sin\left(x^n\right) = \Im\left[-e^{-ix^n}\right]$$
And so, n
$$ I = \int_{0}^{\infty} \sin\left(x^n\right)\:dx = \Im\left[\int_{0}^{\infty} -e^{-ix^n}\:dx \right]= -\Im\left[\int_{0}^{\infty} e^{-\left(i^{\frac{1}{n}}x\right)^{n}}\:dx \right]$$
Applying a change of variable $u = i^{\frac{1}{n}}x$ we arrive at:
\begin{align}
I &= -\Im\left[i^{-\frac{1}{n}}\int_{0}^{\infty} e^{-u^{n}}\:du \right] \\
  &= -\Im\left[i^{-\frac{1}{n}}\frac{\Gamma\left(\frac{1}{n}\right)}{n} \right]\\
  &= \sin\left(\frac{\pi}{2n}\right)\frac{\Gamma\left(\frac{1}{n}\right)}{n}
\end{align}
My area of concern is in the substitution. As $i^{-\frac{1}{n}} \in \mathbb{C}$, I believe the limits of the integral should have been from $0$ to $i^{-\frac{1}{n}}\infty$. Is that correct or not?
I'm also struggling with bounds on $n$ for convergence. Is this expression valid for all $n\in\mathbb{R}$
Any guidance would be greatly appreciated
 A: 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
A: 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
A: Another approach substitutes $y=x^n$ and writes $y^{1/n-1}$ in terms of a Gamma integral, viz. $$I=\Im\int_0^\infty\frac{1}{n}y^{1/n-1}\exp(iy) dy=\Im\int_0^\infty\int_0^\infty\frac{1}{n\Gamma(1/n)}z^{-1/n}\exp(-y(z-i))dydz.$$By Fubini's theorem, and using $\Im\frac{1}{z-i}=\frac{1}{1+z^2}$,$$I=\int_0^\infty\frac{1}{n\Gamma(1/n)}\frac{z^{-1/n}}{1+z^2}dz.$$Then the substitution $z=\tan u$ obtains a Beta integral, which can be rewritten in terms of Gamma functions, and the result you've claimed is proven true, by the reflection formula of the Gamma function. Alternatively, we can a keyhole contour, as when proving the reflection formula; this doesn't use the above substitution, but benefits from another: $v=z^2$.
A: Here is an alternative approach that avoids complex numbers and series altogether. To get round these two obstacles I will use a property of the Laplace transform.
Let
$$I = \int_0^\infty \sin (x^n) \, dx, \qquad n > 1.$$
We begin by enforcing a substitution of $x \mapsto x^{1/n}$. This gives
$$I = \frac{1}{n} \int_0^\infty \frac{\sin x}{x^{1 - 1/n}} \, dx.$$
The following useful property (does this result have a name? It would be so much nicer if it did!) for the Laplace transform will be used:
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$
Noting that
$$\mathcal{L} \{\sin x\}(t) = \frac{1}{1 + t^2},$$
and
$$\mathcal{L}^{-1} \left \{\frac{1}{x^{1-1/n}} \right \} (t)= \frac{1}{\Gamma (1 - \frac{1}{n})} \mathcal{L}^{-1} \left \{\frac{\Gamma (1 - \frac{1}{n})}{x^{1-1/n}} \right \} (t) = \frac{t^{-1/n}}{\Gamma (1 - \frac{1}{n})},$$
then
\begin{align}
I &= \frac{1}{n} \int_0^\infty \sin x \cdot \frac{1}{x^{1 - \frac{1}{n}}} \, dx\\
&= \frac{1}{n} \int_0^\infty \mathcal{L} \{\sin x\} (t) \cdot \mathcal{L}^{-1} \left \{\frac{1}{x^{1 - \frac{1}{n}}} \right \} (t) \, dt\\
&= \frac{1}{n\Gamma (1 - \frac{1}{n})} \int_0^\infty \frac{t^{-1/n}}{1 + t^2} \, dt.
\end{align}
Enforcing a substitution of $t \mapsto \sqrt{t}$ yields
\begin{align}
I &= \frac{1}{2 n \Gamma \left (1 - \frac{1}{n} \right )} \int_0^\infty \frac{t^{-\frac{1}{2} - \frac{1}{2n}}}{t + 1} \, dt\\
&= \frac{1}{2 n \Gamma \left (1 - \frac{1}{n} \right )} \operatorname{B} \left (\frac{1}{2} - \frac{1}{2n}, \frac{1}{2} + \frac{1}{2n} \right )\\
&= \frac{1}{2 n \Gamma \left (1 - \frac{1}{n} \right )} \Gamma \left (\frac{1}{2} - \frac{1}{2n} \right ) \Gamma \left (\frac{1}{2} + \frac{1}{2n} \right ). \tag1
\end{align}
Applying Euler's reflexion formula we have
$$\Gamma \left (\frac{1}{2} - \frac{1}{2n} \right ) \Gamma \left (\frac{1}{2} + \frac{1}{2n} \right ) = \frac{\pi}{\sin \left (\frac{\pi}{2n} + \frac{\pi}{2} \right )} = \frac{\pi}{\cos \left (\frac{\pi}{2n} \right )},$$
and
$$\Gamma \left (1 - \frac{1}{n} \right ) = \frac{\pi}{\sin \left (\frac{\pi}{n} \right ) \Gamma \left (\frac{1}{n} \right )}.$$
So (1) becomes
$$I = \frac{\sin (\frac{\pi}{n} ) \Gamma (\frac{1}{n})}{2n \cos (\frac{\pi}{2n} )},$$
or
$$I = \sin \left (\frac{\pi}{2n} \right ) \frac{\Gamma \left (\frac{1}{n} \right )}{n}, \qquad n > 1$$
where in the last line the double angle formula for sine has been used. 
