Integral of $\frac{1}{x^{n}+1}$ for odd $n$

Question

I am trying to find a general formula for $$ \int \frac{1}{x^{n}+1} \, dx $$ where $n$ is an odd integer.

I can't find any answers on this site for this question, and I have already tried the following.

We can write $$ x^{n} + 1 = (x+1)\sum_{k=0}^{n-1} (-1)^{k}x^{k} $$ and so we get $$ \int \frac{1}{x^{n}+1} \, dx = \int \frac{1}{x+1} + \frac{\sum_{k=0}^{n-2} a_{k}x^{k}}{\sum_{k=0}^{n-1} (-1)^{k}x^{k}} \, dx$$ for some numbers $a_{k}$.

When $n=3$, this is easy enough to solve using partial fractions. When $n=5$ it is tricky but doable. For $n \geq 7$ it becomes intractable and even Mathematica fails to evaluate.

Is there a way to continue with my approach? Is there a better way to find a general formula?

Answer 1

We know that $$x^n+1=\prod _{k=1}^{n} x-e^{i{2k+1\over n}\pi}$$since all the roots are of order $1$, we can write $${1\over x^n+1}={1\over\prod _{k=1}^{n} (x-e^{i{2k+1\over n}\pi})}={1\over n}\sum_{k=1}^{n} {e^{-i\pi {n-1\over n}(2k+1)}\over x-e^{i{2k+1\over n}\pi}}$$by integrating we obtain$$\int{1\over x^n+1}dx={1\over n}\sum_{k=1}^{n} {e^{-i\pi {n-1\over n}(2k+1)} \ln | x-e^{i{2k+1\over n}\pi}|}+C$$

Answer 2

It's doable but the details are pretty ugly so sorry for not completing this calculation. But I will give a complete algorithm for it.

The idea is to write $\frac{1}{x^n+1} = \sum_{i=1}^{n}\frac{1}{f'(\psi_i)(x-\psi_i)}$, where $\psi_i$ are the n-th roots of unity. Now you can further write the sum by grouping $\frac{1}{f'(\psi_i)(x-\psi_i)} + \frac{1}{f'(\psi_{n-i})(x-\psi_{n-i})} = \frac{(2cos\frac{2\pi}{n})x-2}{x^2-(2cos\frac{2\pi}{n})x+1}$ so we have actually managed to split our poly into fractions of irreducible polys over R(of degree 1 and 2!), which we know how to integrate.

Answer 3

$\frac 1 {x^{n}+1}=1-x^{n}+(-x^{n})^{2}+\cdots$ and you can integrate this term by term for $|x| <1$. Note that for $n$ odd the integrand has a singularity at $-1$.

Answer 4

Let's find the Taylor series for $x^a$. Assume that $D_b^kf(x)=\frac{\mathrm{d}^k}{\mathrm{d}x^k}f(x)\big|_{x=b}$. $$D^0x^a=x^a$$ $$D^1x^a=ax^{a-1}$$ $$D^2x^a=a(a-1)x^{a-2}$$ $$\cdots$$ $$D^kx^a=x^{a-k}\prod_{i=0}^{k-1}(a-i)$$ We'll center our Taylor series about the point $x=1$. Thus, we know that our Taylor-Series coefficients are given by $$c_k=\frac1{k!}\prod_{i=0}^{k-1}(a-i)$$ And hence $$x^a=\sum_{k\geq0}\frac{(x-1)^k}{k!}\prod_{i=0}^{k-1}(a-i)$$ This series has a radius of convergence of $1$, which is something to keep in mind.

We proceed by noting that $$(1-x)^a=\sum_{k\geq0}(-1)^k\frac{x^k}{k!}\prod_{i=0}^{k-1}(a-i)$$ And plugging in $a=-1$ gives $$\frac1{1-x}=\sum_{k\geq0}(-1)^k\frac{x^k}{k!}\prod_{i=0}^{k-1}(-1-i)$$ $$\frac1{1-x}=\sum_{k\geq0}(-1)^k(-1)^k\frac{x^k}{k!}\prod_{i=0}^{k-1}(i+1)$$ $$\frac1{1-x}=\sum_{k\geq0}x^k$$ $$\frac1{1+x}=\sum_{k\geq0}(-1)^kx^k$$ $$\frac1{1+x^n}=\sum_{k\geq0}(-1)^kx^{nk}$$ So if we assume that $|x|<1$, we have that $$ \begin{align} \int\frac{\mathrm{d}x}{1+x^n}=&\int\sum_{k\geq0}(-1)^kx^{nk}\mathrm{d}x\\ =&\sum_{k\geq0}(-1)^k\int x^{nk}\mathrm{d}x\\ =&\sum_{k\geq0}(-1)^k\frac{x^{nk+1}}{nk+1}\\ \end{align} $$ And there you go.

Answer 5

Another answer in real functions

$$\begin{aligned} \int\frac{1}{x^{2n+1}+1}\mathop{\mathrm{d}x}&=\boxed{\frac{\ln\left(\left|x+1\right|\right)}{2n+1}\\ -\frac{1}{2n+1}\sum_{k=1}^{n}\left(\cos\left(\frac{2k-1}{2n+1}\pi\right)\ln\left(\left|x^{2}-2\cos\left(\frac{2k-1}{2n+1}\pi\right)x+1\right|\right)\\ +2\sin\left(\frac{2k-1}{2n+1}\pi\right)\tan^{-1}\left(\cot\left(\frac{2k-1}{2n+1}\pi\right)-\csc\left(\frac{2k-1}{2n+1}\right)x\right)\right)+C} \end{aligned}$$ where $n\in\mathbb{N}$.

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Steps:

Let

$$I=\int\frac{1}{x^{n}+1}\mathop{\mathrm{d}x}$$

Do partial fraction decomposition

$$\frac{1}{x^{n}+1}=\frac{A_{1}}{x-z_{1}}+\frac{A_{2}}{x-z_{2}}+\dots+\frac{A_{n}}{x-z_{n}}$$ where $z_{1},z_{2},\ldots,z_{n}$ are the $n$ complex roots of $x^{n}+1$.

To find the value of $A_{i}$, multiply both sides of the equation with $x-z_{i}$ and find the limit of both sides as $x$ approaches $z_{i}$.

Only $A_{i}$ remains on the right side. The limit of the left side is

$$\lim_{x\to z_{i}}\frac{x-z_{i}}{x^{n}+1}=\lim_{x\to z_{i}}\frac{\left(x-z_{i}\right)'}{\left(x^{n}+1\right)'}=\lim_{x\to z_{i}}\frac{1}{nx^{n-1}}=\frac{1}{n{z_{i}}^{n-1}}=-\frac{z_{i}}{n}$$

(Note that ${z_{i}}^n=-1$).

Becase $n$ is an odd number, the integral can be rewritten as

$$I=\int\frac{1}{x^{2n+1}+1}\mathop{\mathrm{d}x}$$

where $n\in\mathbb{N}$.

The $2n+1$ complex roots of $x^{2n+1}+1$ are $-1,z_{1},z_{2},\dots,z_{n},\overline{z_{1}},\overline{z_{2}},\dots,\overline{z_{n}}$, where

$$z_{k}=\mathrm{e}^{\mathrm{i}\frac{-\pi+2k\pi}{2n+1}}=\cos\left(\frac{2k-1}{2n+1}\pi\right)+\mathrm{i}\sin\left(\frac{2k-1}{2n+1}\pi\right)$$

where $k=1,2,\dots,n$.

So the decomposition can be rearranged as

$$\begin{aligned} I&=\int\left(-\frac{-1}{2n+1}\cdot\frac{1}{x+1}+\sum_{k=1}^{n}\left(-\frac{z_{k}}{2n+1}\cdot\frac{1}{x-z_{k}}-\frac{\overline{z_{k}}}{2n+1}\cdot\frac{1}{x-\overline{z_{k}}}\right)\right)\mathop{\mathrm{d}x}\\ &=\int\left(\frac{1}{2n+1}\cdot\frac{1}{x+1}-\frac{1}{2n+1}\sum_{k=1}^{n}\left(\frac{z_{k}}{x-z_{k}}+\frac{\overline{z_{k}}}{x-\overline{z_{k}}}\right)\right)\mathop{\mathrm{d}x}\\ &=\frac{1}{2n+1}\int\frac{1}{x+1}\mathop{\mathrm{d}x}-\frac{1}{2n+1}\sum_{k=1}^{n}\int\left(\frac{z_{k}}{x-z_{k}}+\frac{\overline{z_{k}}}{x-\overline{z_{k}}}\right)\mathop{\mathrm{d}x}\\ &=\frac{1}{2n+1}I_{0}-\frac{1}{2n+1}\sum_{k=1}^{n}I_{k} \end{aligned}$$

$I_{0}$ is easy

$$I_{0}=\int\frac{1}{x+1}\mathop{\mathrm{d}x}=\ln{\left|x+1\right|}+C$$

Then $I_{k}$

$$\begin{aligned} I_{k}&=\int\left(\frac{z_{k}}{x-z_{k}}+\frac{\overline{z_{k}}}{x-\overline{z_{k}}}\right)\mathop{\mathrm{d}x}\\ &=\int\frac{\left(z_{k}+\overline{z_{k}}\right)x-2z_{k}\overline{z_{k}}}{x^2-\left(z_{k}+\overline{z_{k}}\right)x+z_{k}\overline{z_{k}}}\mathop{\mathrm{d}x} \end{aligned}$$

Let $z_{k}=a_{k}+b_{k}\mathrm{i}$, then $\overline{z_{k}}=a_{k}-b_{k}\mathrm{i}$. Note that $z_{k}\overline{z_{k}}={a_{k}}^2+{b_{k}}^2=\left|z_{k}\right|^2=1$, so

$$\begin{aligned} I_{k}&=\int\frac{2a_{k}x-{2a_{k}}^2-{2b_{k}}^2}{x^2-2a_{k}x+{a_{k}}^2+{b_{k}}^2}\mathop{\mathrm{d}x}\\ &=\int\left(a_{k}\cdot\frac{2x-2a_{k}}{x^2-2a_{k}x+1}+2b_{k}\cdot\frac{-b_{k}}{{a_{k}}^2-2a_{k}x+x^2+{b_{k}}^2}\right)\mathop{\mathrm{d}x}\\ &=a_{k}\int\frac{2x-2a_{k}}{x^2-2a_{k}x+1}\mathop{dx}+2b_{k}\int\frac{-\frac{1}{b_{k}}}{\frac{{a_{k}}^2}{{b_{k}}^2}-\frac{2a_{k}x}{{b_{k}}^2}+\frac{x^2}{{b_{k}}^2}+1}\mathop{\mathrm{d}x}\\ &=a_{k}\int\frac{\mathop{\mathrm{d}}\left(x^2-2a_{k}x+1\right)}{x^2-2a_{k}x+1}+2b_{k}\int\frac{\mathop{\mathrm{d}}\left(\frac{a_{k}-x}{b_{k}}\right)}{\left(\frac{a_{k}-x}{b_{k}}\right)^2+1}\\ &=a_{k}\ln(\left|x^2-2a_{k}x+1\right|)+2b_{k}\tan^{-1}\left(\frac{a_{k}-x}{b_{k}}\right)+C \end{aligned}$$

Substitute $a_{k}=\cos\left(\frac{2k-1}{2n+1}\pi\right),b_{k}=\sin\left(\frac{2k-1}{2n+1}\pi\right)$ into $I_{k}$ and you can get the final result above.

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This answer can be futher generalized to integrals like $\int\frac{x^m}{x^n\pm 1}\mathop{\mathrm{d}x}$. See my answer to this question.