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## Inverse Geometric Series

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### Inverse Geometric Series

Fri Aug 05, 2016 2:50 pm

Posts: 27
Evaluate $\displaystyle \int_0^1 \frac{1-x}{1-x^n} \, \text{d}x$ where $\displaystyle n \in \mathbb{N}$

Evaluate $\displaystyle \int_0^\infty \frac{1-x}{1-x^n} \, \text{d}x$ where $\displaystyle n \in \mathbb{N} \, \backslash \{ 0 \}$

### Re: Inverse Geometric Series

Sat Aug 06, 2016 9:58 am

Posts: 48

Evaluate $\displaystyle \int_0^1 \frac{1-x}{1-x^n} \, \text{d}x$ where $\displaystyle n \in \mathbb{N}$

Evaluate $\displaystyle \int_0^\infty \frac{1-x}{1-x^n} \, \text{d}x$ where $\displaystyle n \in \mathbb{N} \, \backslash \{ 0 \}$

A somewhat brief solution ...

1.

\begin{align*}
\int_{0}^{1}\frac{1-x}{1-x^n} &=\int_{0}^{1} \frac{1-x}{\left ( 1-x \right ) \left ( 1+x+x^2+\cdots+x^{n-1} \right )} \, {\rm d}x \\
&= \int_{0}^{1}\frac{{\rm d}x}{1+x+x^2+\cdots+x^{n-1}}\\
&=\frac{\psi^{(0)}\left ( \frac{2}{n} \right )- \psi^{(0)}\left ( \frac{1}{n} \right )}{n} , \; n \geq 1
\end{align*}

2.

We need some restrictions first. Appartently for $n=0$ the integral diverges. For $n=1$ the integral converges and is equal to $0$, whereas for $n=2$ the integral diverges again since it is equal to

$$\int_{0}^{\infty} \frac{1-x}{1-x^2} \, {\rm d}x = \int_{0}^{\infty} \frac{1-x}{(1-x)(1+x)} \, {\rm d}x = \int_{0}^{\infty} \frac{{\rm d}x}{1+x} = + \infty$$

Now for $n\geq 3$ we have that:

$$\int_{0}^{\infty} \frac{1-x}{1-x^n} \, {\rm d}x =\int_{0}^{\infty} \frac{{\rm d}x}{1+x+x^2+\cdots+x^{n-1}}$$

but I don't think that a closed form exists. What does the rest of the community think?

### Re: Inverse Geometric Series

Sat Aug 06, 2016 10:18 am

Posts: 27
I was hoping it would have a closed form in terms of the Beta or Gamma Function but that does not seem to be the case.

However, running a few cases for Mathematica returns a closed form in terms of the nth roots of unity, so there may be some hope lying in there somewhere.

### Re: Inverse Geometric Series

Sat Aug 06, 2016 9:41 pm

Posts: 38
There is, in fact, a simple closed form that is quite easy to derive:

$$I(n)=\int_{0}^{\infty}\frac{1-x}{1-x^{n}}\text{d}x=\frac{\pi}{n}\csc\left(\frac{2\pi}{n}\right)$$

To begin, split the integral up into 2 parts
$$I(n)=\underbrace{\int_{0}^{1}\frac{1-x}{1-x^{n}}\text{d}x}_{J(n)}+\underbrace{\int_{1}^{\infty}\frac{1-x}{1-x^{n}}\text{d}x}_{K(n)}$$
Then,
\begin{align*} J(n)&=\int_{0}^{1}\frac{1-x}{1-x^{n}}\text{d}x \\ &=\frac{1}{n}\int_{0}^{1}\frac{x^{\frac{1}{n}-1}}{1-x}-\frac{x^{\frac{2}{n}-1}}{1-x}\text{d}x \\ &=\frac{1}{n}\lim_{a\to 0}\left(B\left(a,\frac{1}{n}\right)-B\left(a,\frac{2}{n}\right)\right) \\ &=\frac{1}{n}\left(\psi\left(\frac{2}{n}\right)-\psi\left(\frac{1}{n}\right)\right) \end{align*}
and,
\begin{align*} K(n)&=\int_{1}^{\infty}\frac{1-x}{1-x^{n}}\text{d}x \\ &=\int_{0}^{1}\frac{1-x}{1-x^{n}}x^{n-3}\text{d}x \\ &=\frac{1}{n}\int_{0}^{1}\frac{x^{-2/n}}{1-x}-\frac{x^{-1/n}}{1-x}\text{d}x \\ &=\frac{1}{n}\lim_{a\to 0}\left(B\left(a,1-\frac{2}{n}\right)-B\left(a,1-\frac{1}{n}\right)\right) \\ &=\frac{1}{n}\left(\psi\left(1-\frac{1}{n}\right)-\psi\left(1-\frac{2}{n}\right)\right) \end{align*}
\begin{align*} I(n)&=\frac{1}{n}\left(-\psi\left(\frac{1}{n}\right)+\psi\left(1-\frac{1}{n}\right)+\psi\left(\frac{2}{n}\right)-\psi\left(1-\frac{2}{n}\right)\right) \\ &=\frac{\pi}{n}\left(\cot\left(\frac{\pi}{n}\right)-\cot\left(\frac{2\pi}{n}\right)\right) \\ &=\frac{\pi}{n}\csc\left(\frac{2\pi}{n}\right) \end{align*}