i have made some experiments in maple evaluating the integral $$\int_0^\infty\sin(x^{1+a})dx$$ and the computer give me the following result
$$\int_0^\infty\sin(x^{1+a})dx=\frac{\sqrt{\pi}2^{\frac{2}{2+2a}}\Gamma(\frac{1}{2}+\frac{1}{2+2a})}{(2+2a)\Gamma(1-\frac{1}{2+2a})}$$
and i realy want to know how can i prove that. i hope some of you can help me. An interesting thing is that de right side converges to $1$ when $a$ goes to $0$.
Presumably $a > 0$. The change of variable $t = x^{1+a}$ leads to $$ \dfrac{1}{1+a} \int_0^\infty \sin(t)\; t^{-a/(1+a)}\ dt$$
Now let $a/(1+a) = b$, and note that $0 < b < 1$. Now for $\rm{Re}(r) > 0$
$$ \int_0^\infty e^{-r t}\ t^{-b}\ dt = r^{b-1} \Gamma(1-b)$$ Taking the limit as $r \to i$ (and not worrying about convergence) we get $$ \int_0^\infty e^{ -i t} t^{-b}\ dt = (i)^{b-1} \Gamma(1-b) = e^{i\pi (b-1)/2} \Gamma(1-b)$$ and taking $-$ the imaginary part of this $$ \int_{0}^\infty \sin(t) t^{-b}\ dt = - \sin(\pi (b-1)/2) \Gamma(1-b) $$ Now you want to apply some Gamma function identities...