# I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$

I want to show that

$$\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$$

Expand $(x^4-x+\pi)^2=x^4-2x^3+2x^2-2x\pi+\pi{x^2}+\pi^2$

Let see (substitution of $y=x^2$)

$$\int_{-\infty}^{\infty}{x\over (x^4-x^2+1)^2}dx={1\over 2}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$

Substituion of $y=x^3$

$$\int_{-\infty}^{\infty}{x^3\over (x^4-x^2+1)^2}dx={1\over 4}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$

As for $\int_{-\infty}^{\infty}{x^2\over (x^4-x^2+1)^2}dx$ and $\int_{-\infty}^{\infty}{x^4\over (x^4-x^2+1)^2}dx$ are difficult to find a suitable substitution. This is the point where I am shrugged with to find a suitable substitution To lead me to a particular standard integral. Need some help, thank.

standard integral of the form

$$\int{1\over (ax^2+bx+c)^2}dx={2ax+b\over (4ac-b^2)(ax^2+bx+c)}+{2a\over 4ac-b^2}\int{1\over ax^2+bx+c}dx$$ And

$$\int{1\over ax^2+bx+c}dx={2\over \sqrt{4ac-b^2}}\tan^{-1}{2ax+b\over \sqrt{4ac-b^2}}$$

Evaluating the integral of interest can be reduced to evaluating the integral

$$I(a,b)=\int_0^\infty \frac{1}{x^4+bx^2+a} \,dx\tag 1$$

To see this, we exploit first odd symmetry to write the integral of interest as

\begin{align} \int_{-\infty}^\infty \frac{(x^2-x+\pi)^2}{(x^4-x^2+1)^2}\,dx&=2\int_0^\infty \frac{x^4+(2\pi +1)x^2+\pi^2}{(x^4-x^2+1)^2}\,dx \tag 2 \end{align}

Enforcing the substitution $x\to 1/x$ in $(2)$ reveals \begin{align} \int_0^\infty \frac{x^4}{(x^4-x^2+1)^2}\,dx&=\int_0^\infty \frac{x^2}{(x^4-x^2+1)^2}\,dx\\\\ &=-\left.\frac{\partial I(a,b)}{\partial b}\right|_{(a,b)=(1,-1)} \tag 3 \end{align}

Moreover, we have

$$\int_0^\infty \frac{1}{(x^4-x^2+1)^2}\,dx=-\left.\frac{\partial I(a,b)}{\partial a}\right|_{(a,b)=(1,-1)} \tag 4$$

Using $(3)$ and $(4)$ in $(2)$ yields

$$\int_{-\infty}^\infty \frac{(x^2-x+\pi)^2}{(x^4-x^2+1)^2}\,dx=-4(\pi +1)\left.\frac{\partial I(a,b)}{\partial b}\right|_{(a,b)=(1,-1)}-2\pi^2\left.\frac{\partial I(a,b)}{\partial a}\right|_{(a,b)=(1,-1)}$$

where $I(a,b)$ as given by $(1)$ can be found using partial fraction expansion or contour integration, for examples.

• much more systematic then my approach (+1) – tired Jun 22 '16 at 23:59
• @tired Well, your answer inspired me to pursue. – Mark Viola Jun 23 '16 at 1:16

First note that because of

$$\int_{\mathbb{R}}dx\frac{x^4}{(x^4-x^2+1)^2}=\int_{\mathbb{R}}dx\frac{x^4\overbrace{-x^2+1+(x^2-1)}^{=0}}{(x^4-x^2+1)^2}=\\ \int_{\mathbb{R}}dx\frac{1}{(x^4-x^2+1)}+\int_{\mathbb{R}}dx\frac{x^2-1}{(x^4-x^2+1)^2}=\\ \int_{\mathbb{R}}dx\frac{1}{(x^4-x^2+1)}+\int_{\mathbb{R}}dx\frac{-1}{(x^4-x^2+1)^2}+\color{blue}{\underbrace{\int_{\mathbb{R}}dx\frac{x^2}{(x^4-x^2+1)^2}}_{J}}$$

we only need to calulate $\color{\blue}{J}$ because everthing else is covered by your standard formulas.

Now let's define

$$I(a)=\int_{\mathbb{R}}dx\frac{1}{(x^4-a x^2+1)}$$

from which it follows that

$$\color{blue}{J}=I'(1)$$

so it can also derived using ur standard identities ($'$ denotes a derivative w.r.t. $a$)

Edit:

Note that the integrals with numerators containing odd powers of $x$ vanish due to symmetry!

• Feynman comes to the rescue yet again! +1 – Mark Viola Jun 22 '16 at 23:20
• I found that $$\int_0^\infty \frac{x^4}{(x^4-x^2+1)^2}\,dx=\int_0^\infty \frac{x^2}{(x^4-x^2+1)^2}\,dx$$And thus, the entire problem is reduced to evaluating the integral $$I(a,b)=\int_0^\infty \frac{1}{x^4+bx^2+a}\,dx$$and differentiating with respect to $a$ and $b$. – Mark Viola Jun 22 '16 at 23:56
• exactly, the first two integrals in my block of formulas cancel out! :) – tired Jun 22 '16 at 23:58

Trick: the integral over $\mathbb{R}$ of a non-vanishing meromorphic function, $O\left(\frac{1}{\|z\|^2}\right)$ at infinity, is just $2\pi i$ times the sum of the residues for the poles in the upper half-plane. In our case such poles are located at $z=e^{\pi i/6}$ and $z=e^{5\pi i/6}$ (roots of $z^2-iz-1$) and the sum of the residues is $$-\frac{i}{2}\left(1+\pi+\pi^2\right).$$ After that, it is simple math. A worked example of the same technique in a similar (but apparently much harder) problem can be found here.

Elaborating user @Dr. MV's answer, we have

$$\int_0^\infty\frac{1}{a^2x^4+bx^2+c^2}\ dx=\frac{c\pi}{2a\sqrt{b+2ac}}$$

Putting $a=1$, $b=a$, and $c^2=b$, then

$$I(a,b)=\int_0^\infty\frac{1}{x^4+ax^2+b}\ dx=\frac{\pi}{2}\sqrt{\frac{b}{a+2\sqrt{b}}}$$

Hence

$$\frac{\partial}{\partial a}I(a,b)=\int_0^\infty\frac{x^2}{\left(x^4+ax^2+b\right)^2}\ dx=-\frac{\pi}{2b}\sqrt{\left(\!\frac{b}{a+2\sqrt{b}}\!\right)^3}$$ and $$\frac{\partial}{\partial b}I(a,b)=\int_0^\infty\frac{1}{\left(x^4+ax^2+b\right)^2}\ dx=\frac{\pi(a+\sqrt{b})}{2\sqrt{b\left(a+2\sqrt{b}\right)^3}}$$

I think that this is a time for complex analysis....

$\int \frac {(x^2 - x + \pi)^2}{(x^2 x + 1)^2} dx$

$\int \frac {(x^2 - x + \pi)^2}{(x+a)^2(x-\bar a)^2(x- b)^2(x-\bar b)^2} dx$

where $a,b,\bar a, \bar b$ are the complex roots of $(x^4-x^2 + 1)$

The Cauchy integral theorem says: $\oint \frac {f(a)}{(z-a)^2} dz = 2\pi i f'(a)$