I was looking for a proof of the following identity
$$\int_{0}^{\infty} \sin(x^{a})dx = \Gamma\left(1 + \frac{1}{a}\right) \cdot \sin\left(\frac {\pi}{2a}\right)$$
I have tried using the Legendre duplication formula and Euler's reflection formula, but both of them have been ineffectual. Would appreciate some help.
Motivation : Fresnel integrals and integrals of the form $\int_{0}^{\infty} \frac{\sin(x)}{\sqrt [a] {x}}dx.$
With use of the Mellin transform this integral becomes straightforward. Substitute $x^a=t$:
$$ \int_{0}^{\infty} \sin(x^{a}) \; \mathrm{d}x = \frac{1}{a} \int_0^{\infty} t^{\frac{1}{a}-1} \sin(t) \; \mathrm{d}t $$ Which is the Mellin transform of $\sin(t)$ where $z=\frac{1}{a}$. Therefore, the original integral equals $$\frac{1}{a} \Gamma \left(\frac{1}{a}\right) \sin{\left(\frac{\pi}{2a}\right)}=\boxed{\Gamma \left(1+\frac{1}{a}\right) \sin{\left(\frac{\pi}{2a}\right)}}$$ Note that this result holds for $-1 < \Re \big[ \frac{1}{a} \big] <1$.