I try to solve the following integral:

$$\int\frac{x^{\alpha}}{1+\psi x}\,\mathrm dx$$

What I try is to use the substitution method. I define $1+\psi x=u$ and this yields $\mathrm du=\psi\, \mathrm dx$

This gives me at final:

$$\int\frac{\left(u-1\right)^{\alpha}}{u\psi^{2}}\,\mathrm du$$

This one did not help me to solve the integral. What can I do to solve it?

  • 2
    $\begingroup$ you will need special functions for this one $\endgroup$ – tired Nov 17 '16 at 14:16

Let $y=-bx$ \begin{align} \int \frac{x^{a}}{1+bx} dx &= \left(\frac{1}{-b}\right)^{a+1} \int y^{a} (1-y)^{-1} dy \\ &= \left(\frac{1}{-b}\right)^{a+1} \mathrm{B}_{y}(a+1,0) \\ &= \left(\frac{1}{-b}\right)^{a+1} \mathrm{B}_{-bx}(a+1,0) \\ &= \left(\frac{1}{-b}\right)^{a+1} \frac{(-bx)^{a+1}}{a+1} {}_{2}\mathrm{F}_{1}(a+1,1;a+2;-bx) \\ &= \frac{x^{a+1}}{a+1} {}_{2}\mathrm{F}_{1}(a+1,1;a+2;-bx) \end{align}

Note: $$\mathrm{B}_{z}(x,y) = \int\limits_{0}^{z} t^{x-1} (1-t)^{y-1} dt = \frac{z^{x}}{x} {}_{2}\mathrm{F}_{1}(x,1-y;x+1;z)$$ the incomplete beta function and Gauss's hypergeometric function respectively.

  • $\begingroup$ Thanks for the incomplete beta function and gauss's hypergeometric function ! $\endgroup$ – optimal control Nov 17 '16 at 16:17
  • $\begingroup$ By the way, is it possible to use this method by using a definite integral (for example, summing between 0 and 1). Or is it necessary to use analytical continuation of gauss's hypergeometric function ? $\endgroup$ – optimal control Nov 18 '16 at 20:46
  • $\begingroup$ do you mean $\int_{0}^{1}$ for the original integral? $\endgroup$ – poweierstrass Nov 18 '16 at 21:56
  • $\begingroup$ yes exactly. I think it is possible to replace $z$ by some value that we should find, no ? as we use substitution, I think we should as well change for the upper bound. or is it just sufficient to replace $z$ by $1$ ? $\endgroup$ – optimal control Nov 18 '16 at 23:28
  • 1
    $\begingroup$ you can plug in the values for x or if you start with $\int_{0}^{1}$ you can get the hypergeometric function directly by using the integral definition. either way, you obtain the same answer. $\endgroup$ – poweierstrass Nov 19 '16 at 14:00

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