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.

There must be some confusion, since $;x^n-1=(x-1)(x^{n-1}+x^{n-2}+\ldots +x+1);$ , so $$\int\frac{x^n-1}{x-1}dx=\int(x^{n-1}+x^{n-2}+\ldots +x+1),dx=\frac{x^n}n+\frac{x^{n-1}}{n-1}+\ldots+\frac{x^2}2+x+C$$ I can't see any hypergeometric function here. — DonAntonio, Jul 25 '19 at 16:49

score 2 · Accepted Answer · answered Jul 25 '19 at 17:27

2

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.

answered Jul 25 '19 at 17:27

eyeballfrog

22,485

the substitution you made for ln(1-x), does that expression not only work for |x|<1, or it works for all cases of x except x=1? – Poisson Jul 25 '19 at 17:46
@Poisson Right, neither of those sums converge for $|x|>1$, though they can be analytically continued there. – eyeballfrog Jul 25 '19 at 18:02
your clarification is noted. – Poisson Jul 25 '19 at 18:11
One more thing please. I will like to cite the hypergeometric series formula you gave as part the solution. Do you know of any book or paper where this particular formula is found or used so that I could cite it? Thanks – Poisson Aug 05 '19 at 18:33
@Poisson This is just the standard series representation, with the Pochhammer symbols written in terms of the gamma function: $(a)_n = \Gamma(a+n)/\Gamma(a)$. – eyeballfrog Aug 05 '19 at 18:37
Ooh okay. Well noted! – Poisson Aug 05 '19 at 18:50

Zach Hunter · Answer 2 · 2019-07-25T17:10:32.413

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.

Integration involving the Gauss 2F1 Hypergeometric Function

2 Answers2