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?


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 $$\frac{\Gamma(\frac{1}{q}-1)}{\Gamma(\frac{1}{q}-1)}\frac{(\frac{1}{q}-1)}{(\frac{1}{q}-1)}$$
