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I'm having trouble integrating this: $$\int_{0}^{\infty} \frac{1}{(1+x^{c})^{k+1}} \, \mathrm{d}x $$ Where $c,k>0$. Any ideas on how to do it?

Toney Shields
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    Mathematica says, under suitable convergence hypothesis ($\Re(c)>0$ and $\Re(c(k+1))>1$), that your integral equals $$\frac{\Gamma\left( 1+ \frac{1}{c}\right) \Gamma\left(-\frac{1+c(k+1)}{c}\right)}{\Gamma(-(k+1))}.$$

    Or do you need a hand-calculation?

     – Giovanni De Gaetano Aug 25 '16 at 12:42
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    This is a very difficult problem, and has no general approach. At least, are $c,k$ positive integers or they may be any real number? – Crostul Aug 25 '16 at 12:42
  • @GiovanniDeGaetano, I need hand-calculation, but it's better to know the results so I know what I should have. – Toney Shields Aug 25 '16 at 12:44
  • @Crostul, I don't know if they're real or integers, the only thing I know is that they're positive. – Toney Shields Aug 25 '16 at 12:47
  • Well, then I also signal, if you do not already know it, the page http://dlmf.nist.gov/5 . And observe that only positive doesn't make it always convergent. Good luck! – Giovanni De Gaetano Aug 25 '16 at 12:47
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    @Crostul "This is a very difficult problem, and has no general approach." Hmmm... For every $c$ positive (otherwise the integral diverges), the change of variable $$t=\frac1{1+x^c}$$ does the job neatly. – Did Aug 25 '16 at 14:28

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Let $x^{c} = z$, \begin{align} \int\limits_{0}^{\infty} (1+x^{c})^{-k-1} \mathrm{d} x & = \frac{1}{c} \int\limits_{0}^{\infty} z^{\frac{1}{c}-1} (1+z)^{-k-1} \mathrm{d} z \\ & = \frac{1}{c} \mathrm{B}\left(\frac{1}{c}, k+1 - \frac{1}{c}\right) \\ & = \frac{\Gamma\left( \frac{1}{c} \right)\Gamma\left( k+1-\frac{1}{c} \right)}{c\Gamma(k+1)} \end{align}

poweierstrass
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  • Isn't the beta function integrable only on $[0,1]$? – Toney Shields Aug 25 '16 at 12:56
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    No, see http://dlmf.nist.gov/5.12 – poweierstrass Aug 25 '16 at 12:57
  • @ poweierstrass, can you help me find this? http://math.stackexchange.com/questions/1898465/how-to-calculate-int-0-infty-fraceitx1xck1-mathrmd – Toney Shields Aug 25 '16 at 13:14
  • That integral is related to the inverse Fourier transform. My knowledge of Fourier transforms is insufficient to answer the question via this method. Also, as a commentator noted, one could try contour integration but this will be messy. I have added an up vote to the question. – poweierstrass Aug 25 '16 at 16:49