Integral $\int_0^\infty \frac{x^n - 2x + 1}{x^{2n} - 1} \mathrm{d}x=0$

Question

Inspired by some of the greats on this site, I've been trying to improve my residue skills. I've come across the integral

$$\int_0^\infty \frac{x^n - 2x + 1}{x^{2n} - 1} \mathrm{d}x=0$$

where $n$ is a positive integer that is at least $2$. With non-complex methods, I know that the integral is $0$. But I know that it can be done with residue theorem.

The trouble comes in choosing a contour. We're probably going to do some pie-slice contour, perhaps small enough to avoid any of the $2n$th roots of unity, and it's clear that the outer-circle vanishes. But I'm having trouble getting the cancellation for the integral.

Can you help? (Also, do you have a book reference for collections of calculations of integrals with the residue theorem that might have similar examples?)

Answer 1

We want to prove that the integral is $0$ for $n>1$, it is the same thing as $$\int_0^{\infty} \frac{\mathrm{d}x}{x^n+1} = 2\int_0^{\infty} \frac{x-1}{x^{2n}-1} \ \mathrm{d}x.$$ The left hand integral is widely known to be $\frac{\pi}{n} \csc \frac{\pi}{n}$, we want to calculate the right hand integral. let $f(x)=\frac{x-1}{x^{2n}-1}$, and consider the contour $C=C_1\cup C_2\cup C_3$ where $$C_1=[0,r],\ C_2=\left\{z \in \mathbb{C} | |z|=r,\ \arg(z) \in \left[0,\frac{\pi}{2n}\right]\right\},\ \ C_3 =e^{\frac{\pi i}{2n}} C_1. $$ Here's what the contour look like

Notice that $\int_C f(z) \ \mathrm{d}z=0$ (the integral is taken counter clockwise always) since $f$ is holomorphic inside $C$. and $$\left|\int_{C_2} f(x)\ \mathrm{d}x \right| =\mathcal{O}(r^{-1}) \to 0.$$ And \begin{align*} \int_{C_3}f(z) \ \mathrm{d}z &= e^{\frac{\pi i}{2n}}\int_0^r f\left(x e^{\frac{\pi i }{2n}}\right) \ \mathrm{d}x \\ &=e^{\frac{\pi i}{2n}}\int_0^r \frac{e^{\frac{\pi i}{2n}}x -1}{x^{2n}+1} \ \mathrm{d}x \\ &= e^{\frac{\pi i}{n}}\int_0^r \frac{x }{x^{2n}+1} \ \mathrm{d}x-e^{\frac{\pi i}{2n}}\int_0^r \frac{1}{x^{2n}+1} \ \mathrm{d}x. \end{align*} Note that $\int_{0}^{\infty} \frac{x}{x^{2n}+1} \ \mathrm{d}x = \frac{\pi }{2n} \csc \frac{\pi}{n}$, then by taking $r\to \infty$ we get $$\int_0^{\infty} f(x) \ \mathrm{d}x =-e^{\frac{\pi i}{n}}\cdot \frac{\pi }{2n} \csc \frac{\pi}{n} + e^{\frac{\pi i}{2n}} \frac{\pi }{2n} \csc \frac{\pi}{2n} = \frac{\pi}{2n} \csc \frac{\pi}{n}.$$ Which is what we were looking for.

Answer 2

\begin{align*} \int_{0}^{\infty} \frac{x^n-2x+1}{x^{2n}-1} \, dx &= \int_{0}^{1} \frac{x^n-2x+1}{x^{2n}-1} \, dx - \int_{0}^{1} \frac{x^{n-2}-2x^{2n-3}+x^{2n-2}}{x^{2n}-1} \, dx \\ &= \frac{1}{2n} \left( \int_{0}^{1} \frac{x^{\frac{1}{2n}+\frac{1}{2}}-2x^{\frac{1}{n}}+x^{\frac{1}{2n}}}{x-1} \, \frac{dx}{x} - \int_{0}^{1} \frac{x^{\frac{1}{2}-\frac{1}{2n}}-2x^{1-\frac{1}{n}}+x^{1-\frac{1}{2n}}}{x-1} \, \frac{dx}{x} \right). \end{align*} But the duplication formula for the digamma function \begin{equation*} \psi_{0}(z) + \psi_{0}(z+\tfrac{1}{2}) - 2 \psi_{0}(2z) = -\log 4 \end{equation*} shows that \begin{equation*} \int_{0}^{1} \frac{x^{z} + x^{z+\frac{1}{2}} - 2 x^{2z}}{x - 1} \, \frac{dx}{x} = -\log 4. \end{equation*} Plugging this to the formula above, we obtain the desired conclusion.

Answer 3

Just in case someone wonders how it can be done the "normal" way, here it is.

By factorization what we are trying to prove is: $$ \int_0^\infty \frac{1}{x^n-1} dx = 2 \int_0^\infty \frac{x}{x^{2n}-1} dx $$ On the right hand side let $x \rightarrow \sqrt{t}$.

Answer 4

Resolving the integrand into two partial fractions, we have $$\displaystyle I=\int_0^{\infty} \frac{x^n-2 x+1}{\left(1+x^n\right)\left(1-x^n\right)} d x=\int_0^{\infty}\left(\frac{x}{1+x^n}-\frac{1-x}{1-x^n}\right) d x=J-K\tag*{} $$ For the integral $J$, we are going to transforms it into a Beta function by letting $y=\frac{1}{1+x^n}$. $$\displaystyle \begin{aligned}J & =\frac{1}{n} \int_0^1 y^{-\frac{2}{n}}(1-y)^{\frac{2}{n}-1} d y \\& =\frac{1}{n} B\left(-\frac{2}{n}+1, \frac{2}{n}\right) \\& =\frac{\pi}{n} \csc \left(\frac{2 \pi}{n}\right) \quad \textrm{ (By Euler Reflection Formula)}\end{aligned}\tag*{} $$

---

Next, we are going to evaluate the integral $ \displaystyle K=\displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{} $ by the theorem $ \displaystyle \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \textrm{ where } z\notin Z.\tag*{} $ We first split the integral into two integrals $ \displaystyle \displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{1-x}{1-x^{n}} d x+\int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{}$ Transforming the latter integral by the inverse substitution $ x\mapsto \frac{1}{x}$ m, we have $\displaystyle \displaystyle \int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{x^{n-3}-x^{n-2}}{1-x^{n}} d x \tag*{} $ Putting back yields $ \begin{aligned}\displaystyle K&=\int_{0}^{1} \frac{1-x+x^{n-3}-x^{n-2}}{1-x^{n}} d x\\\displaystyle &=\int_{0}^{1}\left[\left(1-x+x^{n-3}-x^{n-2}\right) \sum_{k=0}^{\infty} x^{n k}\right] d x\\ \displaystyle & =\sum_{k=0}^{\infty} \int_{0}^{1}\left[x^{n k}-x^{n k+1}+x^{n(k+1)-3}-x^{n(k+1)-2}\right] d x\\ & =\sum_{k=0}^{\infty}\left(\frac{1}{n k+1}-\frac{1}{n k+2}+\frac{1}{n(k+1)-2}-\frac{1}{n(k+1)-1}\right)\\ & =\sum_{k=0}^{\infty}\left[\frac{1}{n k+1}-\frac{1}{n(k+1)-1}\right]+\sum_{k=0}^{\infty}\left[\frac{1}{n(k+1)-2}-\frac{1}{n k+2}\right] \end{aligned}\tag*{} $ Modifying yields $$\displaystyle \begin{aligned} K&=\frac{1}{n}\left[\sum_{k=0}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-1}^{-\infty} \frac{1}{k+\frac{1}{n}}\right]+\frac{1}{n}\left(\sum_{k=1}^{\infty} \frac{1}{k-\frac{2}{n}}+\sum_{k=0}^{-\infty} \frac{1}{k-\frac{2}{n}}\right)\\& =\frac{1}{n}\left(\sum_{k=-\infty}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-\infty}^{\infty} \frac{1}{k-\frac{2}{n}}\right)\end{aligned} \tag*{} $$ By the Theorem, $$ \displaystyle \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \tag*{} $$ where $ \displaystyle z\notin Z,$ we have $$ \displaystyle \displaystyle K=\frac{1}{n}\left[\pi \cot \left(\frac{\pi}{n}\right)+\pi \cot \left(\frac{-2 \pi}{n}\right)\right]=\frac{\pi}{n}\left[\cot \left(\frac{\pi}{n}\right)-\cot \left(\frac{2 \pi}{n}\right)\right] =\frac{\pi}{n} \csc \frac{2 \pi}{n} =J\tag*{} $$

---

We can now conclude that $\displaystyle \boxed{I=J-K=0 }\tag*{} $