$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$

Question

How do we compute this integral ? $$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$ I have tried partial fraction but it is quite hard to factorize the denominator. Any help is appreciated.

Answer 1

$$ \begin{align} \int_0^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,\mathrm{d}x &=\int_0^1\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x\tag{1a}\\ &=\int_1^\infty\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x\tag{1b}\\ &=\frac12\int_0^\infty\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x\tag{1c} \end{align} $$ Explanation:
$\mathrm{(1a)}$: multiply by $\frac{1+x^2}{1+x^2}$
$\mathrm{(1b)}$: substitute $x\mapsto1/x$ and simplify
$\mathrm{(1c)}$: average $\mathrm{(1a)}$ and $\mathrm{(1b)}$

Depending on the context of the question, there are a few ways to compute $\mathrm{(1c)}$.

One is to use this answer, which gets $\int_0^\infty\frac{x^n}{1+x^{10}}\,\mathrm{d}x=\frac\pi{10}\csc\!\left(\frac{(n+1)\pi}{10}\right)$ by contour integration: $$ \begin{align} \int_0^\infty\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x &=\int_0^\infty\frac1{1+x^{10}}\,\mathrm{d}x +\int_0^\infty\frac{x^2}{1+x^{10}}\,\mathrm{d}x\\ &+\int_0^\infty\frac{x^6}{1+x^{10}}\,\mathrm{d}x +\int_0^\infty\frac{x^8}{1+x^{10}}\,\mathrm{d}x\\ &=\frac\pi{10}\left[\csc\left(\frac{\pi}{10}\right) +\csc\left(\frac{3\pi}{10}\right) +\csc\left(\frac{7\pi}{10}\right) +\csc\left(\frac{9\pi}{10}\right)\right]\\ &=\frac\pi5\left[\csc\left(\frac{\pi}{10}\right)+\csc\left(\frac{3\pi}{10}\right)\right]\\ &=\frac{2\pi}{\sqrt5}\tag{2} \end{align} $$ Therefore, $$ \int_0^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,\mathrm{d}x=\frac\pi{\sqrt5}\tag{3} $$

Answer 2

Per partial fractions \begin{align} & \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx\\ =& \ \frac1{\sqrt5} \int_{0}^1 \frac{\phi^{-1}(x^2+1)}{x^4-\phi x^2+1}+ \frac{\phi(x^2+1)}{x^4-\phi^{-1} x^2+1} \ dx\\ =& \ \frac1{\sqrt5} \bigg( {\tan^{-1}\frac{x-\frac1x}{\phi^{-1}}} + \tan^{-1}\frac{x-\frac1x}{\phi}\bigg)\bigg|_0^1=\frac\pi{\sqrt5} \end{align}

Answer 3

Let $I$ be the integral given by

$$I= \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$

Now, make the substitution $x=u^{-1}$. Then $dx=-u^{-2}du$ and the limits of integration over $u$ extend from $\infty$ to $1$. Thus, we may write

$$ \begin{align} I & = \int_{\infty}^1 \frac{1+u^{-6}}{1-u^{-2}+u^{-4}-u^{-6}+u^{-8}}\left(-u^{-2}\right)du \\ & = \int_1^{\infty} \frac{1+u^{6}}{1-u^{2}+u^{4}-u^{6}+u^{8}}du \end{align} $$

whereby

$$I=\frac12 \int_0^{\infty} \frac{1+u^{6}}{1-u^{2}+u^{4}-u^{6}+u^{8}}du$$ Next, we exploit the even symmetry of the integrand, which reveals that

$$I=\frac14 \int_{-\infty}^{\infty} \frac{1+u^{6}}{1-u^{2}+u^{4}-u^{6}+u^{8}}du$$

Using the Residue Theorem along with Jordan's Lemma, the integral $I$ is given by

$$I=2\pi i \left(\frac14\right)\sum_{i=1}^4 Res_i \left(\frac{1+z^{6}}{1-z^{2}+z^{4}-z^{6}+z^{8}}\right)$$ where the summation is over the four residues of the integrand in the upper-half plane.

All that remains is to find the four complex roots $z_i$ of the denominator that lie in the upper-half plane. These are $z_1=e^{\frac{\pi}{10}}$, $z_2=e^{\frac{3\pi}{10}}$, $z_3=e^{\frac{7\pi}{10}}$, and $z_4=e^{\frac{9\pi}{10}}$.

To find the residue of $z_i$, take the limit

$$\lim_{z \to z_i} (z-z_i) \left(\frac{1+z^{6}}{1-z^{2}+z^{4}-z^{6}+z^{8}}\right)=\frac{1+z_i^{6}}{-2z_i+4z_i^{3}-6z_i^{5}+8z_i^{7}}$$

Answer 4

This is not a final answer but it is too long for a comment.

Beside the elegant solution given by Dr.MV using the residue theorem, there is something I found interesting in the problem. $$\frac{1+x^6}{1-x^2+x^4-x^6+x^8}=(1+x^2+x^6+x^8)\sum_{k=0}^{\infty}(-1)^k x^{10k}$$ So $$I=\int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,dx=\sum_{k=0}^{\infty}\int_{0}^1 (-1)^k(1+x^2+x^6+x^8)x^{10k}\,dx$$ which leads to $$I=\sum_{k=0}^{\infty}(-1)^k \left(\frac{1}{10 k+1}+\frac{1}{10 k+3}+\frac{1}{10 k+7}+\frac{1}{10 k+9}\right)$$ and $I$ has a very nice and simple expression.

Edit

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{10 k+a}=\frac{1}{20} \left(\psi \left(\frac{a}{20}+\frac{1}{2} \right)-\psi\left(\frac{a}{20}\right)\right)$$

Answer 5

Here is a solution that utilizes the Cauchy–Schlömilch transformation twice to obtain the answer:

Let $I$ denote the integral. By substituting $x \mapsto x^{-1}$, we find that

$$ I = \int_{1}^{\infty} \frac{1+x^6}{1-x^2+x^4-x^6+x^8} \, \mathrm{d}x. $$

Note that the integrand is invariant! Using this symmetry, we obtain:

$$ \begin{align*} I &= \frac{1}{2} \int_{0}^{\infty} \frac{1+x^6}{1-x^2+x^4-x^6+x^8} \, \mathrm{d}x \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{(1 - x^2 + x^4)(1 + x^2)}{1-x^2+x^4-x^6+x^8} \, \mathrm{d}x \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{(x - x^{-1})^2 + 1}{(x - x^{-1})^4 + 3(x - x^{-1})^2 + 1} \, \mathrm{d}(x - x^{-1}) \\ &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{u^2 + 1}{u^4 + 3u^2 + 1} \, \mathrm{d}u \\ &= \int_{0}^{\infty} \frac{1}{(u - u^{-1})^2 + 5} \, \mathrm{d}(u - u^{-1}) \\ &= \int_{-\infty}^{\infty} \frac{1}{v^2 + 5} \, \mathrm{d}v \\ &= \frac{\pi}{\sqrt{5}} \end{align*} $$