Evaluating $\int\limits_0^\infty{\frac{1}{1+x^2+x^\alpha}dx}$

I'm trying to evaluate$$f(\alpha)=\int\limits_0^\infty{\frac{1}{1+x^2+x^\alpha}dx}$$ I proved:
$f(\alpha)$ converges when $\alpha\in\mathbb{R}$
$f(2-\alpha)=f(\alpha)$
$f(0)=f(2)=\frac{\pi}{2\sqrt{2}}$
$f(1)=\frac{2\pi}{3\sqrt{3}}$
$f(-\infty)=f(\infty)=\frac{\pi}{4}$
Similar question:$$\int\limits_0^\infty{\frac{1}{1+x^\alpha}dx}=\frac{\pi}{\alpha}\csc\frac{\pi}{\alpha}$$ I tried all of the techniques can be used in evaluating this integral, but I still cannot get the answer. When I was using complex analysis, I found that the poles of $\frac{1}{1+x^2+x^\alpha}$ is hard to be found.

Solution 1:

Not a full answer, some considerations.

First, we make a substitution:

$$x=\tan t$$

The integral becomes (I use $a$ instead of $\alpha$ for convenience):

$$f(a)=\int_0^{\pi/2} \frac{dt}{1+\cos^{2-a} t~ \sin^a t}$$

Which by the way, makes the functional equation from the OP very clear.

For the values of $a$ satisfying $|\cos^{2-a} t~ \sin^a t|\leq 1$, we can expand the integrated function as a geometric series:

$$f(a)=\sum_{n=0}^\infty (-1)^n \int_0^{\pi/2} \cos^{2n-an} t~ \sin^{an} t ~dt$$

However, the range of $a$ allowing this representation is quite restrictive: $0 \leq a \leq 2$.

The integral in the general terms is easy to recognize as Beta function:

$$\int_0^{\pi/2} \cos^{2n-an} t~ \sin^{an} t ~dt=\frac{1}{2} B \left(n-\frac{an-1}{2},\frac{an+1}{2} \right)$$

Now the original integral becomes:

$$f(a)=\frac{1}{2} \sum_{n=0}^\infty (-1)^n B \left(n-\frac{an-1}{2},\frac{an+1}{2} \right)$$

Let's use the Gamma function representation:

$$f(a)=\frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{\Gamma \left(n-\frac{an-1}{2} \right) \Gamma \left(\frac{an+1}{2} \right)}{\Gamma (n+1)} $$

For some $a$ this might be a Hypergeometric function. To find its form, let us consider the ratio of the terms:

$$\frac{c_{n+1}}{c_n}=\frac{\Gamma \left(n+1-\frac{an}{2}-\frac{a}{2}+\frac{1}{2} \right) \Gamma \left(\frac{an}{2}+\frac{a}{2}+\frac{1}{2} \right)}{\Gamma \left(n-\frac{an}{2}+\frac{1}{2} \right) \Gamma \left(\frac{an}{2}+\frac{1}{2} \right)} \frac{-1}{n+1}$$

If $a$ is even, the ratio will be a polynomial, and we will have a generalized Hypergeometric function.

But because of the condition on $a$ for which we derived the series, the only allowed even values are $a=0$ and $a=2$, for which cases the OP already provides a closed form.