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Can someone please help me to solve this integral using hypergeometric, beta, or other well known function ?

$$ I= \int_{R}^\infty \frac{1}{u+x^{-a/2}} \phantom. dx $$

with: u is positive real number. and $ 2<a<6 $.

We can work with $ R=0 $, if it's hard to integrate it with every real R.

Many thanks in advance.

Jean Marie
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adil
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  • We have already discussed this issue in question (http://math.stackexchange.com/q/2067630): you have a divergent integral as @Olivier Oloa has rightly pointed it. – Jean Marie Dec 27 '16 at 14:18

1 Answers1

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We assume $u>0,\,R\ge0$. One has, as $x \to \infty$, $$ \frac{1}{u+x^{-a/2}} \sim \frac{1}{u} $$ then by comparison $\displaystyle \int_{R}^M \frac{dx}{u+x^{-a/2}} $ and $\displaystyle \int_{R}^M \frac{dx}{u}$ are of the same nature, that is divergent as $M \to \infty$.

Olivier Oloa
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  • Thank you very much for your response. Alternatively, if I fix the M in a big value (e.g: 1000 km, as x is a distance), so the integral is convergent, and we can calculate the exacte value of the area ? please what's its Primitive ? – adil Dec 28 '16 at 12:22
  • By setting $I(a):=\int_{R}^M \frac{dx}{u+x^{-a/2}}$ we have a closed form in terms of the hypergeometric function, $I(a)=F(M)-F(R)$ where $F(x)=\frac{2x^{1+\frac{a}{2}}}{a+2}:_2F_1\Big(1,\frac{2+a}{a},2+\frac{2}{a},-x^{a/2} u\Big)$. – Olivier Oloa Dec 28 '16 at 12:38
  • Thanks a lot for your useful response. Just on small thing, I got in an online integral calculator as a premitive of this integral the following: $; \frac{x._2F_1 (1,-2/a,\frac{a-2}{a}; \frac{-x^{-a/2}}{u})} {u}$ . I think, it's the same, but there is some kind of simplification. Can you please, help me to find the relationship between the both solutions, or advice me if you know some online procedures or calculators. Because, I need this kind of hypergeometric transformations rules in other contexts. Many thanks in advance? – adil Jan 03 '17 at 12:40