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

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. 
