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Let $a>2$ be a real variable. My objective is to find an approximation of the integral defined as \begin{equation} \int_b^{ + \infty } {\frac{x}{{1 + {x^a}}}}. \end{equation} Here $b$ is a positive real number.

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  • $\begingroup$ In what regime? $b\to 0$ and $b\to\infty$ are easy, but the answers are different. The intermediate range is when life gets hard. Also, what would you consider a "good approximation"? What precision are you aiming at and what computation time can you afford? $\endgroup$
    – fedja
    Feb 24 '21 at 1:05
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    $\begingroup$ To the (potential) voters to close: I would wait until the OP has a chance to explain what he or she is really after here. $\endgroup$
    – fedja
    Feb 24 '21 at 1:10
  • $\begingroup$ Do you mean computing the integral numerically for given $a,b$? Softwares can do this, for example Pari/GP with the function intnum, by specifying the behavior $\sim x^{1-a}$ at $+\infty$. $\endgroup$ Feb 24 '21 at 7:47
  • $\begingroup$ For example in the case $a=5/2$ and $b=1$, intnum(x=1, [+oo, -3/2], x/(1+x^(5/2))) gives 1.7766271453035772760815104540407586926. $\endgroup$ Feb 24 '21 at 7:50
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    $\begingroup$ Integral does not converge for $a<2$, otherwise it is given by $\frac{b^{2-a} \, _2F_1\left(1,\frac{a-2}{a};2-\frac{2}{a};-b^{-a}\right)}{a-2}$ $\endgroup$
    – yarchik
    Feb 24 '21 at 8:10
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Write $\int_b^\infty = \int_0^\infty - \int_0^b$. The first integral gives a Beta function, which evaluated yields $\frac{\pi}{a\sin(2\pi/a)}$. If $b^a\ll 1$, you can get a good approximation to the second integral by converting the integral to a weighted geometric series and integrating term by term, obtaining $$b^2\sum_{k=0}^\infty (-1)^k \frac{b^{ak}}{ak+2}.$$ In this mentioned regime, your integral is $$\frac{\pi}{a\sin(2\pi/a)}+b^2/2-\frac{b^{a+2}}{a+2}+\mathcal{O}(b^{2a+2}).$$

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