Integration involving the Gauss 2F1 Hypergeometric Function

Question

According to WolframAlpha, the solution to the indefinite integral $\int{\frac{x^n-1}{x-1}}dx$, where n is some constant is

-$\frac{x^{n+1}{_2F_1}(1,n+1;n+2;x)}{n+1}-\log(1-x)+C$, where $\log$ denotes the natural logarithm and ${_2F_1}$ is the Gauss' hypergeometric function. I am still new to hypergeometric functions and that has just lost me completely; I have no idea how they arrived at this solution. Please if anyone could kindly explain to me how they came by this solution, I would be highly indebted to them. Please help. Thank you in advance.

Answer 1

tl;dr: Wolfram is giving the general solution that works even when $n$ isn't an integer.

Long version:

Let's look at the definition of $_2F_1$: $$ _2F_1(a,b;c;x) = \sum_{k=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c)}{\Gamma(a)\Gamma(b)\Gamma(c+k)}\frac{x^k}{\Gamma(k+1)} $$ Plugging in $a = 1$, $b = n+1$, $c = n+2$ gives \begin{align} -\frac{x^{n+1}}{n+1}\,_2F_1(1,n+1;n+2;x) = &-\frac{x^{n+1}}{n+1}\sum_{k=0}^\infty \frac{\Gamma(1+k)\Gamma(n+1+k)\Gamma(n+2)}{\Gamma(1)\Gamma(n+1)\Gamma(n+2+k)}\frac{x^k}{\Gamma(k+1)} \\= &-\frac{x^{n+1}}{n+1}\sum_{k=0}^\infty \frac{n+1}{n+1+k}x^k \\ = &-\sum_{k=0}^\infty \frac{x^{n+1+k}}{n+1+k}. \end{align} Additionally using $\ln(1-x) = -\sum_{j=1}^\infty x^j/j$, Wolfram is saying $$ \int\frac{x^n-1}{x-1}dx = \sum_{j=1}^\infty\frac{x^j}{j} -\sum_{k=0}^\infty \frac{x^{n+1+k}}{n+1+k}. $$ Now at this point you may have noticed that the second sum will cancel out every term of the first sum beyond $j = n$. This is true, but only if $n$ is an integer. Wolfram never assumes any information you don't give it, so it gives the answer in terms of the hypergeometric function. If $n$ is an integer, the cancellation I mentioned happens, and you get $$ \int\frac{x^n-1}{x-1}dx = \sum_{j=1}^n\frac{x^j}{j} $$ as expected.

Answer 2

I remember observing something like this myself. I believe the key will be to expand out your fraction in a similar way as this:

$$ \frac{1}{x-1} = -\frac{1}{1-x} = -\left(1+x+x^2 \cdots\right)$$

As then the deriviative will be an infinite of fractions which correspond to the geometric series. For the case above we have:

$$ \left(\frac{1}{1-x}\right)' = 1+\frac{x}{2}+\frac{2x^2}{3}\cdots$$

With higher degrees of $n$, you just instead have:

$$ \frac{x^n-1}{x-1} = \frac{1-x^n}{1-x} = (1-x^n)\left(1+\frac{x}{2}+\frac{2x^2}{3}\cdots \right) $$

I'll leave the final steps of the derivative to you to figure out. Hopefully this is enough of an intuitional nudge.