Tripos Type Integral This is an old calculus question:
Integrate
$$\int x^{-2n-2}(1-x)^n(1-cx)^ndx$$ it comes from Edwards Treatise on Integral Calculus Vol I, pg. 65 and is designated as [Oxford I.P. 1917].
I guess my question is what do they have in mind as a solution to this ? Are they expecting a closed form ?
My solution is to expand the $1-cx$ factor to get $(-1)^{k}\binom{n}{k} \int\frac{x^k(1-x)^n}{x^{2n+2}}dx$ as the coefficient of $c^k$. This in turn can be integrated by the binomial theorem as
$\sum_{l=0}^n(-1)^{k+l}\binom{n}{k}\binom{n}{l}\frac{1}{2n-l-k+1}x^{-2n+k+l-1}$.
This almost seems too simple minded. But maybe it was what they intended. Any opinions ?
 A: Mathematica says:
$$\frac{(c-1) (1-x)^{n+1} x^{-2 n} (1-c x)^{n-1} \left(\frac{(c-1) x}{c x-1}\right)^{2 n}
   \, _2F_1\left(n+1,2 (n+1);n+2;\frac{x-1}{c x-1}\right)}{n+1}$$
I am not sure if this is the sort of thing they would want.
A: This integral can be written as a simple summation. To simplify, we change $y=1/x$ to write
\begin{align}
 I_n=&\int x^{-2n-2}(1-x)^n(1-cx)^n\,dx\\
 &=-\int (y-1)^n(y-c)^n\,dy
\end{align}
and to have a symmetric expression, changing $y=z+(c+1)/2$ gives
\begin{align}
 I_n&=-\int \left( z^2-\left( \frac{c-1}{2} \right)^2 \right)^n\,dz
\end{align}
Now, expanding the polynomial,
\begin{align}
 I_n&=-\sum_{k=0}^n\binom{n}{k}\frac{(-1)^{n-k}}{2k+1}\left( \frac{c-1}{2} \right)^{2n-2k}z^{2k+1}+C\\
 &=(-1)^{n+1}\sum_{k=0}^n\binom{n}{k}\frac{(-1)^{k}}{2k+1}\left( \frac{c-1}{2} \right)^{2n-2k}\left(1-\frac{(c+1)x}{2} \right)^{2k+1}x^{-2k-1}+C
\end{align}
A: Using integration by parts, it can be shown that the integrals $I_n$ have the following property: $$I_n= (1-x)^n(1-cx)^n(\frac{x^{-2n-1}}{-2n-1}) + \frac{nc+n}{-2n-1}\int x^{-2n-1}(1-x)^{n-1}(1-cx)^{n-1}dx - \frac{2nc}{2n-1}I_{n-1} $$
That leftover integral does not exactly match the form of an $I_k$, but it is close. To me, it's concievable that $J_n = \int x^{-2n-1}(1-x)^{n-1}(1-cx)^{n-1}dx$ has a decomposition which will look similar to the above but with one of the $I_n$ appearing in the middle and $J_{n-1}$ at the end.
Maybe it is possible to solve the two recurrances, or perhaps the two relationships themselves would suffice for an answer.
