$\int\limits_0^\infty {x^4 \over (x^4-x^2+1)^4}\ dx$

Question

I want to calculate $$\int\limits_0^\infty \frac{x^4}{(x^4-x^2+1)^4}dx$$

I have searched with keywords "\frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results

Answer 1

For $b > 0$, define $I_b(a)$ by

$$ I_b(a) = \int_{0}^{\infty} \frac{x^{a}}{((x - x^{-1})^2 + 1)^b} \, \mathrm{d}x = \int_{0}^{\infty} \frac{x^{a+2b}}{(x^4 - x^2 + 1)^b} \, \mathrm{d}x. $$

This integral converges if $|a+1| < 2b$. We can also prove that $I_b(a) = I_b(-a-2)$ holds, by using the substitution $x \mapsto 1/x$. Then

\begin{align*} \int_{0}^{\infty} \frac{x^4}{(x^4 - x^2 + 1)^4} \, \mathrm{d}x &= I_4(-4) = I_4(2) \\ &= I_4(2) - I_4(0) + I_4(-2) \\ &= \int_{0}^{\infty} \frac{1}{((x - x^{-1})^2 + 1)^3} \, \mathrm{d}x \end{align*}

So, by the Glasser's master theorem,

\begin{align*} \int_{0}^{\infty} \frac{x^4}{(x^4 - x^2 + 1)^4} \, \mathrm{d}x &= \int_{0}^{\infty} \frac{1}{(u^2 + 1)^3} \, \mathrm{d}u \\ &= \int_{0}^{\frac{\pi}{2}} \cos^4\theta \, \mathrm{d}\theta \tag{$(u=\tan\theta)$} \\ &= \frac{3\pi}{16} \approx 0.58904862254808623221 \cdots. \end{align*}

Answer 2

Here is another way to get to the same point as what @Sangchul Lee gives.

Let $$I = \int_0^\infty \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx.$$ Then $$I = \int_0^1 \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx + \int_1^\infty \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx.$$ Enforcing a substitution of $x \mapsto 1/x$ in the right most integral leads to \begin{align} I &= \int_0^1 \frac{x^4 (1 + x^6)}{(x^4 - x^2 + 1)^4} \, dx\\ &= \int_0^1 \frac{x^4 (1 + x^2)(x^4 - x^2 + 1)}{(x^4 - x^2 + 1)^4} \, dx\\ &= \int_0^1 \frac{x^4 (1 + x^2)}{(x^4 - x^2 + 1)^3} \, dx\\ &= \int_0^1 \frac{1 + 1/x^2}{\left [\left (x - \frac{1}{x} \right )^2 + 1 \right ]^3} \, dx. \end{align} On setting $-u = x - 1/x$, $-du = (1 + 1/x^2) \, dx$ one has \begin{align} I &= \int_0^\infty \frac{du}{(u^2 + 1)^3}\\ &= \int_0^{\frac{\pi}{2}} \cos^4 \theta \, d\theta\\ &= \int_0^{\frac{\pi}{2}} \left (\frac{1}{2} \cos 2\theta + \frac{1}{8} \cos 4\theta + \frac{3}{8} \right ) \, d\theta\\ &= \frac{3\pi}{16}, \end{align} as expected.

Answer 3

Letting $x\mapsto \frac{1}{x} $ transforms the integral into $$ I= \int_0^ {\infty} \frac{x^4}{\left(x^4-x^2+1\right)^4} d x =\int_0^{\infty} \frac{x^{10}}{\left(x^4-x^2+1\right)^4} d x $$ Averaging them gives \begin{equation} I=\frac{1}{2} \int_0^{\infty} \frac{x^4\left(x^2+1\right)}{\left(x^4-x^2+1\right)^3} d x= \frac{1}{2}\int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x^2+\frac{1}{x^2}-1\right)^3} d x=\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^2+1\right]^3} \end{equation} Putting $\tan \theta=x-\frac{1}{x}$ yields $$ \boxed{\int_0^ {\infty} \frac{x^4}{\left(x^4-x^2+1\right)^4} d x =\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^4 \theta d \theta=\frac{3 \pi}{16} } $$