Integral of $\int_0^{\infty} \frac{\sin(x^n)}{x^n}dx$ where $\frac{1}{2}\lt n\lt 1$

Question

What will be the value of $$\int_0^{\infty} \frac{\sin(x^n)}{x^n}dx$$ where $$\frac{1}{2}\lt n\lt 1$$ I found the solution for $n>1$.

But how to solve this, when $$\frac1 2<n<1$$ Please help.

Thanks for all.

Answer 1

After the substitution $x^n=y$, the given integral is $f(2-1/n)/n$, where $$f(\alpha):=\int_0^\infty\frac{\sin y}{y^\alpha}\,dy=\frac\pi{2\Gamma(\alpha)\sin(\alpha\pi/2)}.\qquad(0<\alpha<2)$$ I'm sure a proof can be found here on MSE, but this formula actually holds for $\alpha\in\mathbb{C}$ with $0<\Re\alpha<2$, which can be shown first for $0<\Re\alpha<1$ (consider $\int z^{-\alpha}e^{iz}\,dz$ along the boundary of $\{z\in\mathbb{C}:r<|z|<R,0<\arg z<\pi/2\}$, take $r\to 0$ and $R\to\infty$ using Jordan's lemma, and finally apply the reflection formula for $\Gamma$), and by analytic continuation elsewhere: $$f(\alpha)=\alpha\int_0^\infty\frac{1-\cos y}{y^{\alpha+1}}\,dy$$ (after integration by parts) is clearly analytic on $0<\Re\alpha<2$.