Evaluate $I=\int_0^\infty\frac{\sin x^p}{x^p}dx\,\,(p>\frac12)$.
I was able to solve it by converting the integral to a gamma function integral. We have $$I=\Im\left(\int_0^\infty\frac{e^{i x^p}}{x^p}dx\right).$$ Let $ix^p=-u$. Then, $$I=\Im\left(\frac{i^{\frac1p-1}}{p}\int_0^\infty u^{\frac1p-2}e^{-u}du\right)=\Im\left(\frac{e^{\frac\pi 2(\frac1p-1)i}}{p}\Gamma(\frac1p-1)\right)=\Im\left(-i\frac{e^{\frac\pi{2p}i}}{1-p}\Gamma(\frac1p)\right)=\frac{\Gamma(\frac1p)}{p-1}\cos(\frac\pi{2p}).$$
How can I evaluate this integral using residue theorem? And why is $p>\frac12$? (Book puts this condition). Thanks in advance.