Evaluate the integral $\int_0^{\infty} \frac{dx}{x^{\frac{\alpha}{2}}+1}$ for the general $\alpha$ I want to find the solution to the following integral $$\int_0^{\infty} \frac{dx}{x^{\frac{\alpha}{2}}+1}$$ where $\alpha$ can be any value greater than 2 such that $\alpha /2 >1$ but can be any real value above 1. I have looked at the book "Integral series and products" but the result provided depends on the particular value of $\alpha /2$ that is whether it is even or odd will dictate the result. I want to know if there is some general result for this integral. Thanks in advance.
 A: Using one of the definitions of the Beta function:
$${\displaystyle \mathrm {B} (a,b)=\int _{0}^{\infty }{\dfrac {t^{a-1}}{(1+t)^{a+b}}}\,\mathrm {d} t,\qquad \mathrm {Re} (a)>0,\ \mathrm {Re} (b)>0\!}$$
We set $$t=x^{\alpha/2}$$
$$dt=\frac{\alpha}{2} x^{\alpha/2-1}dx$$
$$\mathrm {B} (a,b)=\frac{\alpha}{2}\int _{0}^{\infty }{\dfrac {x^{\alpha a/2-1}}{(1+x^{\alpha/2})^{a+b}}}\,\mathrm {d} x$$

$$\alpha a/2-1=0$$
$$a+b=1$$
$$a=\frac{2}{\alpha}$$
$$b=1-\frac{2}{\alpha}$$


$$\int_0^{\infty} \frac{dx}{x^{\frac{\alpha}{2}}+1}=\frac{2}{\alpha} \mathrm {B} \left(\frac{2}{\alpha}, 1-\frac{2}{\alpha} \right)$$


Using the property of Beta function:
$$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
We can write:
$$\mathrm {B} \left(\frac{2}{\alpha}, 1-\frac{2}{\alpha} \right)=\Gamma \left(\frac{2}{\alpha} \right) \Gamma \left( 1-\frac{2}{\alpha} \right)$$
And the reflection formula for the Gamma function will get us the result of @robjohn
A: In this answer, it is shown that for $m>0$ and $-1<n<m-1$,
$$
\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x
$$
Let $n=0$ and $m=a/2$, to get
$$
\int_0^\infty\frac1{1+x^{a/2}}\,\mathrm{d}x=\frac{2\pi}a\csc\left(\frac{2\pi}a\right)
$$
