Prove the fact that two integrals are equal 
How can I prove that the following two integrals are equal (when they converge of course):
$$\int\limits_0^\infty\frac{x^\frac{1}{2n}}{1+x^2}\space\text{d}x=\Gamma\left(\frac{1}{2n}\right)\cdot\int\limits_0^\infty\frac{\cos\left(x\right)}{x^\frac{1}{2n}}\space\text{d}x\tag1$$


My first option was to use Laplace transform, but I do not exactly see how I can use that, so any approach is welcome.
 A: $$
\begin{align}
\int_0^\infty\frac{x^\frac1{2n}}{1+x^2}\,\mathrm{d}x
&=\frac12\int_0^\infty\frac{x^{\frac{1-2n}{4n}}}{1+x}\,\mathrm{d}x\tag1\\
&=\frac12\Gamma\!\left(\frac{2n+1}{4n}\right)\Gamma\!\left(\frac{2n-1}{4n}\right)\tag2\\[3pt]
&=\frac\pi2\csc\left(\frac\pi2-\frac\pi{4n}\right)\tag3\\[6pt]
&=\frac\pi2\sec\left(\frac\pi{4n}\right)\tag4
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x^{1/2}$
$(2)$: Beta Function integral
$(3)$: Euler's Reflection Formula
$(4)$: trig identity

$$\newcommand{\Re}{\operatorname{Re}}
\begin{align}
\Gamma\!\left(\frac1{2n}\right)\int_0^\infty\frac{\cos(x)}{x^\frac1{2n}}\,\mathrm{d}x
&=\Gamma\!\left(\frac1{2n}\right)\Re\left(\int_0^\infty x^{-\frac1{2n}}e^{ix}\,\mathrm{d}x\right)\tag5\\
&=\Gamma\!\left(\frac1{2n}\right)\Re\left(e^{i\pi\frac{2n-1}{4n}}\int_0^\infty x^{-\frac1{2n}}e^{-x}\,\mathrm{d}x\right)\tag6\\
&=\cos\left(\frac\pi2-\frac\pi{4n}\right)\,\Gamma\!\left(\frac{2n-1}{2n}\right)\Gamma\!\left(\frac1{2n}\right)\tag7\\[4pt]
&=\pi\sin\left(\frac\pi{4n}\right)\csc\left(\frac\pi{2n}\right)\tag8\\[4pt]
&=\frac\pi2\sec\left(\frac\pi{4n}\right)\tag9
\end{align}
$$
Explanation:
$(5)$: Euler's Formula
$(6)$: Cauchy's Integral Theorem using $[0,R]\cup Re^{i[0,\pi/2]}\cup[iR,0]$ as $R\to\infty$
$(7)$: Gamma Function integral and Euler's Formula
$(8)$: trig identity and Euler's Reflection Formula
$(9)$: trig identity
A: Laplace transform works here:
$$\color{red}{\Gamma \left(\frac{1}{2 n}\right) \int_0^{\infty } \frac{\cos (x)}{x^{\frac{1}{2 n}}} \, dx}=\\\Gamma \left(\frac{1}{2 n}\right) \int_0^{\infty }
   \left(\mathcal{L}_x[\cos (x)](s)\right) \left(\mathcal{L}_x^{-1}\left[\frac{1}{x^{\frac{1}{2 n}}}\right](s)\right) \, ds=\\\Gamma
   \left(\frac{1}{2 n}\right) \int_0^{\infty } \frac{s s^{-1+\frac{1}{2 n}}}{\left(1+s^2\right) \Gamma \left(\frac{1}{2 n}\right)} \,
   ds=$$
Change s to x
$$\\\int_0^{\infty } \frac{\Gamma \left(\frac{1}{2 n}\right) x x^{-1+\frac{1}{2 n}}}{\left(1+x^2\right) \Gamma \left(\frac{1}{2 n}\right)} \,
   dx=\\\color{red}{\int_0^{\infty } \frac{x^{\frac{1}{2 n}}}{1+x^2} \, dx}$$
