Is there a nice general formula for $\int \frac{dx}{x^n-1}$ and/or $\int \frac{dx}{\Phi_n(x)}$?

Question

Question is in the title, where $\Phi_n$ denotes the $n$-th cyclotomic polynomial.

Motivation: I'm just teaching my calculus students basic integration of rational functions with $\log$ and $\arctan$, so I wondered about this.

Observation and more precise question: Pairing complex conjugates, over $\mathbb R$ these polynomials split into factors of the forms $x-1$, $x+1$, and $$f_{k,n}(x) = x^2-2\cos(2k\pi/n)x +1,$$ where the reciprocal of the last one, if I'm not mistaken, has antiderivative $$\frac{1}{\sin(2k\pi/n)}\arctan\left(\frac{x-\cos(2k\pi/n)}{\sin(2k\pi/n)}\right).$$ Now using partial fractions and then integrating, one gets summands of the following types:

• A) $\log|x-1|$
• B) $\log|x+1|$
• C) $\frac{1}{\sin(2k\pi/n)}\arctan\left(\frac{x-\cos(2k\pi/n)}{\sin(2k\pi/n)}\right)$
• D) $\log (f_{k,n}(x))$.

(The last two, which come from taking apart something of the form $$\frac{\text{linear}}{f_{k,n}(x)},$$ are of course the only ones in $\int \frac{dx}{\Phi_n(x)}$ for $n\ge 3$.)

These bits I recognise in the example formulae below and I'm happy to have these as "building blocks" in a general formula. So I guess all I'm really asking for is a formula which gives the coefficients of each of the terms A)-D) in the integrals. In other words, we have

$$\int \frac{dx}{x^n-1} = \color{red}{A_n} \cdot \log|x-1| + \color{red}{B_n} \cdot \log|x+1| + \sum_{1\le k <\frac{n}{2}} \frac{\color{red}{C_{n,k}}}{\sin(2k\pi/n)}\arctan\left(\frac{x-\cos(2k\pi/n)}{\sin(2k\pi/n)}\right) + \sum_{1\le k <\frac{n}{2}} \color{red}{D_{n,k}} \log(f_{n,k})$$

and for $n \ge 3$,

$$ \int \frac{dx}{\Phi_n(x)} = \sum_{1\le k <\frac{n}{2}\\gcd(k,n)=1} \frac{\color{red}{E_{n,k}}}{\sin(2k\pi/n)}\arctan\left(\frac{x-\cos(2k\pi/n)}{\sin(2k\pi/n)}\right) + \sum_{1\le k <\frac{n}{2}\\gcd(k,n)=1} \color{red}{F_{n,k}} \log(f_{n,k})$$

and I'm looking for closed formulae depending on $n$ and $k$ for the coefficients in red.

Idea: I assume if one goes to $\mathbb C$ one just gets terms of the form $$ \log (x-\zeta)$$ where $\zeta$ runs through the respective roots of unity, and choosing the right branches of $\log$ and grouping complex conjugates together might give a formula for the coefficients. But going through this is beyond me at this point.

Edit:

From Eric Wofsey's answer one gets, by pairing complex conjugates and using the well-known relation (1, 2, 3) between arctan and complex logarithm (leaving out domain restriction and integration constants):

$$\int \frac{dx}{x^n-1} = \color{red}{\frac{1}{n}} \cdot \log|x-1| + \color{blue}{(}(\color{red}{-\frac{1}{n}}) \cdot \log|x+1|\color{blue}{)} + \sum_{1\le k <\frac{n}{2}} \color{red}{\frac{-2\sin(2k\pi/n)}{n}}\arctan\left(\frac{x-\cos(2k\pi/n)}{\sin(2k\pi/n)}\right) + \sum_{1\le k <\frac{n}{2}} \color{red}{\frac{\cos(2k\pi/n)}{n}} \log(f_{n,k})$$

(where the term in $\color{blue}{\text{blue}}$ brackets appears iff $n$ is even), i.e my original normalisation of the arctan-term was bad, would give $C_{n,k} = -2\sin^2(2k\pi/n)/n$),;

and for $n \ge 3$,

$$\int \frac{dx}{\Phi_n(x)} = \sum_{1\le k <\frac{n}{2}\\gcd(k,n)=1} \color{red}{-2\cdot\mathfrak{Im}\left(\frac{1}{\Phi'_n(\zeta_k)}\right)}\arctan\left(\frac{x-\cos(2k\pi/n)}{\sin(2k\pi/n)}\right) + \sum_{1\le k <\frac{n}{2}\\gcd(k,n)=1} \color{red}{\mathfrak{Re}\left(\frac{1}{\Phi'_n(\zeta_k)}\right)} \log(f_{n,k})$$

(where $\zeta_{k,n} = \exp(2k\pi i/n)$).

Remains only the question if there is a nice general formula for the real/imaginary part of $\displaystyle \frac{1}{\Phi'_n(\zeta_k)}$. I have asked that as a new question here.

Answer 1

You can get a pretty simple expression using partial fractions over $\mathbb{C}$. In general, if $f(x)=(x-a_1)\dots(x-a_n)$ is a monic polynomial with distinct roots over $\mathbb{C}$, then we have the partial fractions decomposition $$\frac{1}{f(x)}=\sum_{k=1}^n\frac{1}{f'(a_k)(x-a_k)}.$$ (You can verify that this decomposition is correct by multiplying both sides by $x-a_k$ and taking the limit as $x\to a_k$: on the left side you have the limit of $\frac{x-a_k}{f(x)}$ which is exactly the reciprocal of the difference quotient for $f'(a_k)$ and so converges to $\frac{1}{f'(a_k)}$.)

Thus an antiderivative can be computed as $$\int\frac{1}{f(x)}dx=\sum_{k=1}^n\frac{\log(x-a_k)}{f'(a_k)}+C.$$ You can apply this to either $f(x)=x^n-1$ or $f(x)=\Phi_n(x)$. So, when you write everything over $\mathbb{C}$ and do not attempt to combine terms to get something that is obviously real-valued, the coefficients just come from the reciprocal of the derivative of the polynomial evaluated at each root. In the case of $x^n-1$, at least, you can write this quite explicitly since the derivative is easy to evaluate: $$\int\frac{1}{x^n-1}dx=\frac{1}{n}\sum_{k=1}^n\zeta^k\log(x-\zeta^k)+C$$ where $\zeta$ is a primitive $n$th root of unity (here the $f'(\zeta^k)$ in the denominator became $n(\zeta^k)^{n-1}=n(\zeta^k)^{-1}$).

Answer 2

In terms of a general formula $$ \displaystyle \int \frac{dx}{x^n-1} = -x\;_2F_1\left(1,\frac{1}{n};1+\frac{1}{n},x^n\right) $$ which contains a hypergeometric function. I have played around with the more general cyclotomic case but did not get anywhere in terms of hypergeometric functions.

We have in general from a geometric series $$ \frac{1}{x^n-1} = -1 - x^n -x^{2n} -x^{3n} -\cdots = -\sum_{k=1}^\infty x^{n k} $$ if we integrate term by term then we have $$ \int \frac{1}{x^n-1}\;dx = -x - \frac{x^{n+1}}{n+1} -\frac{x^{2n+1}}{2n+1} -\frac{x^{3n+1}}{3n+1} -\cdots = - \sum_{k=1}^\infty \frac{x^{kn+1}}{kn+1} $$ $$ \int \frac{1}{x^n-1}\;dx = - x\sum_{k=1}^\infty \frac{x^{kn}}{kn+1} $$ If we consider the definition of the hypergeometric function $$ \;_2F_1(a,b;c,x) = \sum_{k=1}^\infty \frac{(a)_k (b)_k x^k}{(c)_k k!} $$ with $(\cdot)_k$ the Pochhammer symbol. For this example we have $$ \;_2F_1\left(1,\frac{1}{n};1+\frac{1}{n},x\right) = \sum_{k=1}^\infty \frac{\left(1\right)_k \left(\frac{1}{n}\right)_k x^k}{\left(1+\frac{1}{n}\right)_k k!} $$ we have that $(1)_k=k!$ so $$ \;_2F_1\left(1,\frac{1}{n};1+\frac{1}{n},x\right) = \sum_{k=1}^\infty \frac{ \left(\frac{1}{n}\right)_k x^k}{\left(1+\frac{1}{n}\right)_k} =\sum_{k=1}^\infty \frac{x^{kn}}{kn+1} $$