Compute the integral $\int \frac{1}{-x + \alpha x^n + \frac{\alpha}{2} n (1-x) x^{n-1}}dx$

Question

Let $n \in \{2,3,\dots\}$ and $\alpha \in (0,1)$ I wish to compute the following integral: $$ \int \frac{1}{-x + \alpha x^n + \frac{\alpha}{2} n (1-x) x^{n-1}}dx. $$ When computing it with Mathematica, I get no result for general $n$ and when I fix $n$ to be some specific value, I get nasty expressions like, for example for $n=5$.

I was wondering if there is some trick to compute the integral for general $n$ as the integral does not look that complex.

This integral comes up in the context of a differential equation:

$x' = -x + \alpha x^n + \frac{\alpha}{2} n(1-x)x^{n-1},$

which I attempt to solve analytically. I have solved it numerically this is not a problem. For the analytical solution I have used WA to obtain an expression for this integral but the expression it gives is not something I can use to find a closed formula for $x$.

Answer 1

Wolfram Alpha works here with the expansion into partial fractions ( https://en.wikipedia.org/wiki/Partial_fraction_decomposition , sections: “Application to symbolic integration” and "Procedure"), which is the standard method for such a problem (= Polynomial in the denominator) .

The solution for $\,n=2\,$ is $\,\displaystyle \frac{\ln x}{\alpha -1} + constant$ .

The solution for $\,n>2\,$ is:

$$ \int\frac{dx}{-x+ \alpha x^n+\frac{1}{2}\alpha n(1-x)x^{n-1}} =$$ $$= \frac{1}{n-2}\sum\limits_{\{\omega: (n-2) \alpha \omega^{n-1}-n\alpha \omega^{n-2}+2=0\}}\frac{-n\ln(x-\omega)+(n-2)\omega\ln(x-\omega)}{(n-1)\omega-n} -\ln x + constant$$

Answer 2

Not an answer to the original question, but a remark about the ODE.

To me this looks like a job for perturbation methods. In case an approximate solution would be useful, and using the fact that $\alpha$ is small, we can formally write:

$$x(t)=x_0+ \alpha x_1+\alpha^2 x_2+\dots$$

And substitute this into the original equation:

$$x_0'+ \alpha x_1'+\alpha^2 x_2'+\dots = \\ = -\left(x_0+ \alpha x_1+\alpha^2 x_2+\dots \right) + \alpha \left(x_0+ \alpha x_1+\alpha^2 x_2+\dots \right)^n + \\ +\alpha \frac{n}{2}\left(1-x_0- \alpha x_1-\alpha^2 x_2-\dots \right)\left(x_0+ \alpha x_1+\alpha^2 x_2+\dots \right)^{n-1}$$

Keeping only certain powers of $\alpha$ at each step, we obtain a sequence of ODEs:

$$x_0=C_1 e^{-t}$$

$$x_1'=-x_1+C_1^n e^{-nt}+C_1^{n-1} \frac{n}{2} \left(1-C_1 e^{-t} \right)e^{-(n-1)t}$$

And so on.

From the form of the equation we can guess that for large $t$ the asymptotic solution should approach $x=x_0=C_1 e^{-t}$.

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For small $t$ one could use a Taylor series approximation instead of perturbations:

$$x(t)=a_0+a_1 t+a_2 t^2+\dots$$