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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; $$I_{1,\beta} = \int_1^{\infty} \frac{1}{x + x^{\beta}}dx $$ $$= \int_1^{\infty} \frac{1}{x(1+x^{\beta-1})}$$ $$ = \int_1^{\infty} \frac{1}{x} - \frac{x^{\beta - 2}}{1+x^{\beta-1}}$$ $$=\ln\lvert x\rvert-\frac{1}{\beta-1}\cdot\ln\lvert 1+x^{\beta-1}\rvert\biggr\rvert_1^{\infty}$$ $$=\frac{1}{\beta - 1}\cdot\ln(\frac{x^{\beta-1}}{1+x^{\beta-1}})\biggr\rvert_1^{\infty}$$ $$=\frac{\ln(1)-\ln(1/2)}{\beta-1}$$ $$=\frac{\ln2}{\beta-1}$$

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  • $\begingroup$ Are we assuming that $\max(\alpha,\beta)-\min(\alpha,\beta)$ is an integer? $\endgroup$ – Jack D'Aurizio Jan 4 '19 at 20:49
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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.

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  • $\begingroup$ How do you transform the sum into the digamma representation? Is there an identity which relates the digamma function to an infinite summation? $\endgroup$ – Peter Foreman Jan 4 '19 at 21:19
  • $\begingroup$ +1. How about for all $\alpha$ in $\mathbb{R}$? $\endgroup$ – Gustavo Louis G. Montańo Jan 4 '19 at 21:36
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    $\begingroup$ @GustavoLouisG.Montańo The integral diverges when $\alpha$ < max(1,$\beta$) and is trivial when $\alpha=\beta$. $\endgroup$ – Peter Foreman Jan 4 '19 at 21:42
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    $\begingroup$ This answer is actually valid for $\alpha,\beta\in\mathbb{C}$ as well. As long as $\mathfrak{R}(\alpha) > $max(1,$\mathfrak{R}(\beta)$). $\endgroup$ – Peter Foreman Jan 4 '19 at 22:04
  • $\begingroup$ Robert - To use the geometric series you need to be inside its interval of convergence of $|x| < 1$. The limits of the original integral are from 1 to $\infty$, hence the reason why I made the change of $x \mapsto 1/x$ to begin with in my answer. $\endgroup$ – omegadot Jan 5 '19 at 2:28
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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.

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

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