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Can you please help me to solve or write these 2 integrals with some well known functions as gamma, hypergeometric, beta functions,etc...?

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

with: n is positive integer, u is positive real number. and 2< a <6.

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

Many thanks in advance.

PS: the first integral is already solved, $ J= \int_{0}^\infty (\frac{x^{-a/2}}{1+x^{-a/2}})^m .(\frac{1}{1+x^{-a/2}})^n .dx $.

adil
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1 Answers1

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Concerning integral $J$, where we have set $b:=a/2$:

$$J= \displaystyle\int_{0}^\infty \left(\frac{x^{-b}}{1+x^{-b}}\right)^m \left(\frac{1}{1+x^{-b}}\right)^n dx$$

Multiplying numerators and denominators by $x^b$ :

$$J= \displaystyle\int_{x=0}^\infty \left(\frac{1}{1+x^{b}}\right)^m \left(\frac{x^b}{1+x^{b}}\right)^n dx$$

Now, by change of variable $x^b=y \ \iff x=y^{1/b}$:

$$\tag{1}J= \displaystyle\frac{1}{b}\int_{y=0}^\infty \frac{y^{c-1}}{(1+y)^{m+n} } dy$$

with $c$ defined by $c:=n+\frac{1}{b}$.

We recognize in $(1)$ a form of the beta integral ; see formula (10) in (http://homepage.tudelft.nl/11r49/documents/wi4006/gammabeta.pdf):

$$B(u,v)=\displaystyle\int_0^{\infty} \dfrac{s^{u-1}}{(s+1)^{u+v}}ds=\dfrac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}.$$

giving:

$$J=\frac{1}{b}\dfrac{\Gamma \left(n+\frac{1}{b}\right)\Gamma\left(m-\frac{1}{b}\right)}{\Gamma(m+n)}$$

Jean Marie
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  • @Rohan Thanks for your editing. – Jean Marie Dec 22 '16 at 10:47
  • No problem at all. –  Dec 22 '16 at 10:47
  • Thank you very much JeanMarie & Rohan, you are gentleman. I want to cry from joy :)) It's very helpful for me, I hope that other gentlemen like you support me fro the first integral, I thought taht it's easier than the one which you solved, we can work with R=0, if that's simplify the integration. many thanks again :) – adil Dec 22 '16 at 13:16
  • There is a problem with the first integral: as the fraction to be integrated is equivalent to $1$ at infinity, the integral is divergent, even if we take $R \neq 0$. – Jean Marie Dec 22 '16 at 14:06
  • I think that the fraction inside the first integral is equivalent to $\frac{-1}{u^{n -1}}$, not to 1. thanks – adil Dec 22 '16 at 15:46
  • Sorry, I know that u and n are constants, but I don't understand why the limite of the fraction is equal to 1; For me, I got: $ \lim_{x\to \infty} \frac{(u^{n -1}+1-x^{a/2})}{(1+u^{n -1}x^{a/2})} = \frac{-x^{a/2}}{u^{n -1}x^{a/2}} = \frac{-1}{u^{n -1}} $ . Sorryme if I'm wrong. – adil Dec 22 '16 at 17:15
  • I'm sorry. I'm wrong. Your limit is the right one. Nevertheless, the fact that the limit is not zero implies the divergence of the integral, whatever the value of $R$. – Jean Marie Dec 22 '16 at 18:23
  • No problem. So, there is no simplified form of this integral, even with gamma function or a well known integral as Gauss, It's not necessary that's integreble. – adil Dec 22 '16 at 18:43
  • It might be interesting to say in which context you have met these integrals. It happens that some integrals can be given a meaning in the distributional sense for example (principal value PV, or finite part FP integrals) – Jean Marie Dec 22 '16 at 18:46
  • Thank a lot again for your reactivity. I'm doing my reserch work in the field of Wireless communications, and I find this result by myself, and in the final step I need just to write these integrals in well known form in order to do my simulations. Note that all steps before reaching these 2 integrals, are 100% corrects. in the other hand, I don't understand what do you mean by the second sentence " It happens that some integrals can be given a meaning in the distributional sense for example (principal value PV, or finite part FP integrals) " . Regards :) – adil Dec 22 '16 at 22:46
  • I am puzzled by the fact that your first integral, which, surely, has a physical meaning, is divergent. 2) This is why I was wondering if there is a more general context in which it might be considered as convergent in a certain sense. You surely have met Dirac distribution $\delta$, but there is a whole bunch of distributions other than $\delta$. Among them 'principal Value of 1/x", notation $P.V. \dfrac{1}{x}$ which is fundamental in Radon and/or Hilbert transform (http://mathworld.wolfram.com/CauchyPrincipalValue.html) or (http://physics.stackexchange.com/q/105729)
    • – Jean Marie Dec 23 '16 at 00:44