Show that: $\int_{0}^{\infty} \frac{\sin{x^{q}}}{x^{q}} dx = \frac{\Gamma{\frac{1}{q}}}{q-1}\cos{\frac{\pi}{2q}} \mbox{, q > 1}$ 
How do you show that:
  $$
\int_{0}^{\infty} \frac{\sin{x^{q}}}{x^{q}} dx = \frac{\Gamma{\frac{1}{q}}}{q-1}\cos{\frac{\pi}{2q}} \mbox{, q > 1}
$$
  Without using Gamma function?


As said by Chappers in the comments, to prove a result involving a $\Gamma$ function without even mentioning it is quite pointless. Anyway assuming $q>1$,
$$I=\int_{0}^{+\infty}\frac{\sin(x^q)}{x^q}\,dx = \frac{1}{q}\int_{0}^{+\infty}\frac{\sin z}{z^{2-\frac{1}{q}}}\,dz \tag{1}$$
and by exploiting the following Laplace (direct and inverse) transforms:
$$\mathcal{L}^{-1}\left(\frac{1}{z^{2-\frac{1}{q}}}\right)=\Gamma\left(2-\frac{1}{q}\right)^{-1} s^{\frac{q-1}{q}},\qquad\mathcal{L}(\sin z)=\frac{1}{1+s^2}\tag{2}$$
we get:
$$ I = \frac{1}{q\cdot\Gamma\left(2-\frac{1}{q}\right)}\int_{0}^{+\infty}\frac{s^{1-\frac{1}{q}}}{s^2+1}\,ds = \frac{1}{\Gamma\left(2-\frac{1}{q}\right)}\int_{0}^{+\infty}\frac{t^{2q-2}}{t^{2q}+1}\,dt\tag{3} $$
and the last integral can be easily computed through the residue theorem or the substitution $\frac{1}{t^{2q}+1}=u$, leading to a value of the Beta function that can be written as a cosine in virtue of the $\Gamma$ duplication formula.
