I am facing some complicated integral, which part of it is

$$\frac{z^{M-1}}{(1+(\eta z)^n)^p}$$

I think if I find the taylor series of this part the integral might be solved. So, can someone help me please to find the taylor series of this?

Note: $${\eta, n, p}$$ are real positive numbers, while $${M}$$ is a positive integer.

Sample plots for different values of these parameters is attached.

enter image description here

  • $\begingroup$ in what point? zero? $\endgroup$ – Max Mar 18 '14 at 8:56
  • $\begingroup$ yes ... at zero $\endgroup$ – kazekage Mar 18 '14 at 9:10
  • $\begingroup$ If M isn't a positive integer the function isn't analytic. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 18 '14 at 9:15
  • $\begingroup$ Assume M to be a positive integer $\endgroup$ – kazekage Mar 18 '14 at 9:40


Just focuse on the denominator and define $y={(\eta z})^n$. So $$\frac{1}{(1+y)^p} \simeq 1-p y+\frac{1}{2} \left(p^2+p\right) y^2+\frac{1}{6} \left(-p^3-3 p^2-2 p\right) y^3+\frac{1}{24} \left(p^4+6 p^3+11 p^2+6 p\right) y^4+\frac{1}{120} \left(-p^5-10 p^4-35 p^3-50 p^2-24 p\right) y^5+O\left(y^6\right) $$ Replace now $y$ by its definition and multiply by the numerator.

I am sure that you can take from here.

  • $\begingroup$ I don't but I have a feeling that this expression (ignoring the numerator) is somehow related to the upper incomplete gamma function. see the plots added. $\endgroup$ – kazekage Mar 18 '14 at 15:05
  • $\begingroup$ Didn't you ask for a Taylor expansion ? $\endgroup$ – Claude Leibovici Mar 19 '14 at 5:15

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