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Prove that:
$$\int_{0}^{\infty}\cos(x^a)\,dx=\frac{\pi\csc\left ( \frac{\pi}{2a} \right )}{2a\Gamma (1-a)},\,\,\,\,\,a\in \mathbb{N}$$
without contours or complex analysis.

Tolaso J Kos wrote:Prove that:
$$\int_{0}^{\infty}\cos(x^a)\,dx=\frac{\pi\csc\left ( \frac{\pi}{2a} \right )}{2a\Gamma (1-a)},\,\,\,\,\,a\in \mathbb{N}$$
without contours or complex analysis.

$\displaystyle \int\limits_{0}^{+\infty }{\cos \left( x^{a} \right)dx}\underbrace{=}_{\sqrt[a]{y}=x}\frac{1}{a}\int\limits_{0}^{+\infty }{y^{\frac{1}{a}-1}\cos ydy}=\frac{1}{a}\int\limits_{0}^{+\infty }{y^{\frac{1}{a}-1}\cdot \frac{e^{iy}+e^{-iy}}{2}dy}$

$\displaystyle =\frac{\Gamma \left( \frac{1}{a} \right)}{2a}\left( i^{\frac{1}{a}}+i^{-\frac{1}{a}} \right)=\frac{\Gamma \left( \frac{1}{a} \right)}{2a}\left( \left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)^{\frac{1}{a}}+\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)^{-\frac{1}{a}} \right)$

$\displaystyle =\frac{\Gamma \left( \frac{1}{a} \right)}{2a}\left( \cos \frac{\pi }{2a}+i\sin \frac{\pi }{2a}+\cos -\frac{\pi }{2a}+i\sin -\frac{\pi }{2a} \right)=\frac{\cos \frac{\pi }{2a}\Gamma \left( \frac{1}{a} \right)}{a}$

$\displaystyle =\frac{\cos \frac{\pi }{2a}}{a}\frac{\pi }{\Gamma \left( 1-\frac{1}{a} \right)\sin \frac{\pi }{a}}=\frac{\pi }{2a}\frac{1}{\Gamma \left( 1-\frac{1}{a} \right)\sin \frac{\pi }{2a}}=\frac{\pi }{2a}\frac{\csc \frac{\pi }{2a}}{\Gamma \left( 1-\frac{1}{a} \right)}$

Una forma es con transformada de fourier para $f\left( t \right)=\left| t \right|^{a}$

whitexlotus wrote:
Una forma es con transformada de fourier para $f\left( t \right)=\left| t \right|^{a}$

I would love to see the Fourier approach.