Evaluate the integral $\int_0^{\infty} \frac{dx}{x^{\frac{\alpha}{2}}+1}$ for the general $\alpha$

Question

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.

Answer 1

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 $$t=x^{\alpha/2}$$

$$dt=\frac{\alpha}{2} x^{\alpha/2-1}dx$$

$$\mathrm {B} (a,b)=\frac{\alpha}{2}\int _{0}^{\infty }{\dfrac {x^{\alpha a/2-1}}{(1+x^{\alpha/2})^{a+b}}}\,\mathrm {d} x$$

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$$\alpha a/2-1=0$$

$$a+b=1$$

$$a=\frac{2}{\alpha}$$

$$b=1-\frac{2}{\alpha}$$

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

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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

Answer 2

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 $$ \int_0^\infty\frac1{1+x^{a/2}}\,\mathrm{d}x=\frac{2\pi}a\csc\left(\frac{2\pi}a\right) $$