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Given that $alpha,betainmathbb{R}$ 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;
$$I_{1,beta} = int_1^{infty} frac{1}{x + x^{beta}}dx $$
$$= int_1^{infty} frac{1}{x(1+x^{beta-1})}$$
$$ = int_1^{infty} frac{1}{x} - frac{x^{beta - 2}}{1+x^{beta-1}}$$
$$=lnlvert xrvert-frac{1}{beta-1}cdotlnlvert 1+x^{beta-1}rvertbiggrrvert_1^{infty}$$
$$=frac{1}{beta - 1}cdotln(frac{x^{beta-1}}{1+x^{beta-1}})biggrrvert_1^{infty}$$
$$=frac{ln(1)-ln(1/2)}{beta-1}$$
$$=frac{ln2}{beta-1}$$

Let's say $alpha > max(1,beta)$.
$$ frac{1}{x^alpha + x^beta} = frac{1}{x^alpha (1 + x^{beta-alpha})} =
sum_{k=0}^infty frac{(-1)^k x^{kbeta}}{x^{(1+k)alpha}}$$
and integrating term-by-term
$$I_{alpha,beta} = sum_{k=0}^infty frac{(-1)^{k}}{(1+k)alpha - k beta -1}
= frac{1}{beta-1}+frac{Psileft(frac{beta - 1}{2alpha-2beta}right) - Psileft(frac{alpha-1}{2alpha-2beta}right)}{2alpha-2beta}$$
where $Psi$ is the digamma function.

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.

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
$$I_{alpha, beta} = frac{1}{2beta - 2 alpha} left [psi left (frac{2 beta - alpha - 1}{2 beta - 2alpha} right ) - psi left (frac{beta - 1}{2 beta - 2alpha} right ) right ], qquad beta > max (1,alpha).$$