#2
Post
by Grigorios Kostakos »
\begin{align*}
\displaystyle \int_{0}^{\infty}{\frac{x^{\frac{1}{n}}}{1+x^2}\,dx}&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,x^2}\\
{\frac{1}{2}t^{-\frac{1}{2}}dt\,=\,dx}\\
\end{subarray}}\,\frac{1}{2}\int_{0}^{\infty}{\frac{t^{\frac{1}{2n}-\frac{1}{2}}}{1+t}\,dt}\\
&=\frac{1}{2}\int_{0}^{\infty}{\frac{t^{\frac{1}{2n}+\frac{1}{2}-1}}{(1+t)^{\frac{1}{2n}+\frac{1}{2}+1-\frac{1}{2n}-\frac{1}{2}}}\,dt}\\
&=\frac{1}{2}{\rm{B}}\bigl({\tfrac{1}{2n}+\tfrac{1}{2},1-\bigl({\tfrac{1}{2n}+\tfrac{1}{2}}\bigr)}\bigr)\\
&=\frac{1}{2}\frac{\Gamma\bigl({\tfrac{1}{2n}+\tfrac{1}{2}}\bigr)\,\Gamma\bigl({1-\bigl({\tfrac{1}{2n}+\tfrac{1}{2}}\bigr)}\bigr)}{\Gamma(1)}\\
&=\frac{1}{2}\frac{\pi}{\sin\bigl({\pi\bigl({\tfrac{1}{2n}+\tfrac{1}{2}}\bigr)}\bigr)} \\
&=\frac{\pi}{2\sin\bigl({\tfrac{\pi}{2n}+\tfrac{\pi}{2}}\bigr)} \\
&=\frac{\pi}{2}\sec\bigl({\tfrac{\pi}{2n}}\bigr)\,, \quad n\in \mathbb{N}\,.
\end{align*}
Grigorios Kostakos