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?

 A: 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.
A: Yet another approach: applying Ramanujan's master theorem which states that the Mellin transform of a function that admits the expansion of the form 
$$f(x)=\sum_{n=0}^\infty \frac{(-1)^n x^n}{n!}\phi(n) $$
is given by$$\int_0^\infty x^{s-1}f(x)dx = \Gamma(s)\phi(-s) .$$
Since $$\sin(x)= \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$
we have $$\frac{\sin(\sqrt{x})}{\sqrt{x}}=\sum_{n=0}^\infty \frac{(-1)^nx^n}{n!}\frac{\Gamma(n+1)}{\Gamma(2n+2)}$$
and therefore, for $0<s<1$: $$\int_0^\infty x^{s-1}\frac{\sin(\sqrt{x})}{\sqrt{x}}dx=\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(2-2s)}=\frac{\pi}{\sin(\pi s)\Gamma(2-2s)}$$
Substituting $t=\sqrt{x}$, $2s=\frac{1}{q}$ and multiplying by $\frac{1}{q}$ we get
$$I=\frac{1}{q}\int_0^\infty x^{\frac{1}{q}-2}\sin(x)dx=\frac{\pi}{2q\sin(\frac{\pi}{2q})\Gamma(2-\frac{1}{q})}$$
Which you can match to your proposed form by multiplying by $$\frac{\Gamma(\frac{1}{q}-1)}{\Gamma(\frac{1}{q}-1)}\frac{(\frac{1}{q}-1)}{(\frac{1}{q}-1)}$$
A: First, set $y=x^q$, to clear out the nasty argument of the sine. Then the integral is
$$ \frac{1}{q}\int_0^{\infty} y^{1/q-2}\sin{y} \, dy $$
Write the sine in terms of complex exponentials as $\frac{e^{iy}-e^{-iy}}{2i}$ in the usual way. Now, I shall run through the $e^{iy}$ one: the other can be done in the same way in the lower half-plane. Set $p=1/q-1$ to save writing.
We consider the integral
$$ \int_{\gamma} z^{p-1} e^{iz} \, dz, $$
where $\gamma$ is the contour made out of the real axis between $\varepsilon $ and $R$, the quarter-circle $z=Re^{i\theta}$, $0<\theta<\pi/2$, the imaginary axis from $iR$ to $i\varepsilon$, and finally the small quarter-circle $\varepsilon e^{i\pi/2-i\theta}$, $0<\theta<\pi/2$. It is easy to show that for $0<p<1$, the small semicircle integral tends to zero as $\varepsilon \to 0$. The large semicircle integral can be shown to converge to zero as $R \to \infty$ in the same way as Jordan's lemma is proven. Since the integrand has no singularities inside the contour, it follows that
$$ \int_0^{\infty} y^{p-1} e^{iy} \, dy = - \left(- \int_{\infty}^{0} (e^{i\pi/2}y)^{p-1} e^{-y} \, (-i) dy \right) = i e^{i (p-1) \pi/2} \Gamma(p) \\
= i e^{i \pi / (2q)} \Gamma(1/q-1) = i e^{i \pi / (2q)} \frac{q}{q-1}\Gamma(1/q). $$
Nearly there: the sensible thing to do at this point is to notice that you can take the imaginary part of the left-hand side to get the original integral, and hence
$$ \frac{1}{q} \int_0^{\infty} y^{1/q-2} \sin{y} \, dy = \frac{1}{q-1}\cos{\left(\frac{\pi}{2q}\right)} \Gamma(1/q) $$
