Compute $\int_1^{\infty} \frac{1}{x^\alpha + x^\beta}dx$ Given that $\alpha,\beta\in\mathbb{R}$ such that the following integral converges I would like to find a closed form for: $$I_{\alpha,\beta} = \int_1^{\infty} \frac{1}{x^\alpha + x^\beta}dx$$
I have found ways to represent the integral for cetain fixed cases, but is there a general representation of the integral for all $\alpha$ and $\beta$ ?
For example; 
$$I_{1,\beta} = \int_1^{\infty} \frac{1}{x + x^{\beta}}dx $$
$$= \int_1^{\infty} \frac{1}{x(1+x^{\beta-1})}$$
$$ = \int_1^{\infty} \frac{1}{x} - \frac{x^{\beta - 2}}{1+x^{\beta-1}}$$
$$=\ln\lvert x\rvert-\frac{1}{\beta-1}\cdot\ln\lvert 1+x^{\beta-1}\rvert\biggr\rvert_1^{\infty}$$
$$=\frac{1}{\beta - 1}\cdot\ln(\frac{x^{\beta-1}}{1+x^{\beta-1}})\biggr\rvert_1^{\infty}$$
$$=\frac{\ln(1)-\ln(1/2)}{\beta-1}$$
$$=\frac{\ln2}{\beta-1}$$
 A: Let's say $\alpha > \max(1,\beta)$.
$$ \frac{1}{x^\alpha + x^\beta} = \frac{1}{x^\alpha (1 + x^{\beta-\alpha})} = 
\sum_{k=0}^\infty \frac{(-1)^k x^{k\beta}}{x^{(1+k)\alpha}}$$
and integrating term-by-term 
$$I_{\alpha,\beta} = \sum_{k=0}^\infty \frac{(-1)^{k}}{(1+k)\alpha - k \beta -1}
= \frac{1}{\beta-1}+\frac{\Psi\left(\frac{\beta - 1}{2\alpha-2\beta}\right) - \Psi\left(\frac{\alpha-1}{2\alpha-2\beta}\right)}{2\alpha-2\beta}$$
where $\Psi$ is the digamma function.
A: For $\alpha \neq \beta $. Put $$t={1 \over 1+ x^{\beta - \alpha}} .$$ Then you integral becomes 
\begin{align} {1 \over \beta -\alpha } \int_0^{1 \over 2 } t^{\alpha -1 \over \beta - \alpha }(1-t)^{1-\beta \over \beta -\alpha} \, dt,\end{align}
which is ${1 \over \beta -\alpha} B(1/2; {\beta -1 \over \beta -\alpha},   {1-\alpha \over \beta -\alpha}) $,
where $B(x;\mu,\nu)$ is the incomplete Beta function.
A: Assuming $\beta > \max (1,\alpha)$. Enforcing a substitution of $x \mapsto 1/x$ in the integral to begin with gives
$$I_{\alpha, \beta} = \int_0^1 \frac{x^{\beta - 2}}{1 + x^{\beta -\alpha}} \, dx.$$
Inside the interval of convergence, by exploiting the geometric series, namely
$$\frac{1}{1 + x^{\beta - \alpha}} = \sum_{n = 0}^\infty (-1)^n x^{n \beta - n \alpha}, \qquad |x| < 1$$
the integral can be rewritten as
\begin{align}
I_{\alpha, \beta} &= \sum_{n = 0}^\infty (-1)^n \int_0^1 x^{n \beta - n \alpha + \beta - 2} \, dx\\
&= \sum_{n = 0}^\infty (-1)^n \frac{1}{n \beta - n \alpha + \beta - 1}. \qquad (*)
\end{align}
To handle the infinity sum that arises we will make use of the following result (see Eq. (6) in the link)
$$\sum_{n = 0}^\infty \frac{(-1)^n}{z n + 1} = \frac{1}{2z} \left [\psi \left (\frac{z + 1}{2z} \right ) - \psi \left (\frac{1}{2z} \right ) \right ]. \qquad (**)$$
Here $\psi (x)$ is the digamma function.
Rewriting the sum in ($*$) in the form of ($**$), namely
$$I_{\alpha, \beta} = \frac{1}{\beta - 1} \sum_{n = 0}^\infty \frac{(-1)^n}{\left (\dfrac{\beta - \alpha}{\beta - 1} \right ) n + 1},$$
as we have $z = (\beta - \alpha)/(\beta - 1)$, we finally arrive at
$$I_{\alpha, \beta} = \frac{1}{2\beta - 2 \alpha} \left [\psi \left (\frac{2 \beta - \alpha - 1}{2 \beta - 2\alpha} \right ) - \psi \left (\frac{\beta - 1}{2 \beta - 2\alpha} \right ) \right ], \qquad \beta > \max (1,\alpha).$$
