A curious family of integrals that give $\pi$

Question

I have edited the question to include what I think is an interesting property of this integral. I have noticed experimentally that:

$$\int_{0}^{1} \frac{1-x^2}{x^4+2x^3+2x^2-2x+1} dx=\frac{\pi}{4},\tag{1}$$

$$\int_{0}^{1} \frac{x}{x^4+2x^3+2x^2-2x+1} dx=\frac{\pi}{8},\tag{2}$$

$$\int_{0}^{1} \frac{1+x-x^2}{x^4+2x^3+2x^2-2x+1} dx=\frac{3\pi}{8}.\tag{3}$$

So a little variation in the numerator seems to produce always something like $n\pi$. Here is my new question: what is exactly the relation between $n$ and the numerator of the integrand?

The closest thing to my formula is formula $(34)$ in this list of formulas, but it is not exactly the same as mine.

Answer 1

Substitute $x=\tan \frac t2$\begin{align} &\int_{0}^{1} \frac{1-x^2}{x^4+2x^3+2x^2-2x+1} dx\\ =& \int_0^{\pi/2} \frac{\cos t}{2-\sin 2t }dt \overset{x\to \frac\pi2 - x}= \frac12\int_0^{\pi/2} \frac{\cos t+\sin t}{2-\sin 2t }dt\\ =&\ \frac12 \int_0^{\pi/2} \frac{d(\sin t-\cos t)}{1+(\sin t -\cos t)^2} =\frac\pi4 \end{align} and, similarly \begin{align} \int_{0}^{1} \frac{x}{x^4+2x^3+2x^2-2x+1} dx = \frac12\int_0^{\pi/2} \frac{\sin t}{2-\sin 2t }dt =\frac\pi8 \end{align} In general \begin{align} \int_{0}^{1} \frac{ax^2 +b x + c}{x^4+2x^3+2x^2-2x+1} dx =\frac\pi8(c+b-a)+\frac\pi{3\sqrt3}(a+c) \end{align}

Answer 2

Rewrite $$x^4+2x^3+2x^2-2x+1=(x-a)(x-b)(x-c)(x-d)$$ where

$$a=-\frac{\sqrt{3}+1}{2} (1+i)\qquad\qquad b=-\frac{\sqrt{3}+1}{2} (1-i)$$ $$c=\frac{\sqrt{3}-1}{2} (1+i)\qquad\qquad d=\frac{\sqrt{3}-1}{2} (1-i)$$

Using partial fraction decomposition $$\frac{1-x^2}{x^4+2x^3+2x^2-2x+1}=\frac{1-a^2}{(a-b) (a-c) (a-d) (x-a)}+$$ $$\frac{b^2-1}{(a-b) (b-c) (b-d) (x-b)}+\frac{c^2-1}{(a-c) (c-b) (c-d) (x-c)}+$$ $$\frac{d^2-1}{(a-d) (d-b) (d-c) (x-d)}$$

And use, for $k$ being a complex $$\int_0^1\frac {dx}{x-k}=\log \left(\frac{k-1}{k}\right)$$

Replacing $(a,b,c,d)$ by their values and simplifying the complex numbers $$-\frac{\left(\frac{1}{144}-\frac{i}{144}\right) \left(3 i+(1+2 i) \sqrt{3}\right) (\pi +6 i \log (2))}{1+\sqrt{3}}+$$ $$\frac{\left(\frac{1}{4}+\frac{i}{4}\right) \left(\sqrt{3}+(2+i)\right) \log \left(\frac{\sqrt{3}+(2+i)}{1+\sqrt{3}}\right)}{3+\sqrt{3}}+$$ $$\frac{\left(\frac{1}{4}-\frac{i}{4}\right) \left(\sqrt{3}+(-2+i)\right) \log \left(\frac{\sqrt{3}+(-2+i)}{\sqrt{3}-1}\right)}{\sqrt{3}-3 }+$$ $$\frac{\left(\frac{1}{4}+\frac{i}{4}\right) \left(\sqrt{3}+(-2-i)\right) \log \left(\frac{\sqrt{3}+(-2-i)}{\sqrt{3}-1}\right)}{\sqrt{3}-3 }=\color{blue}{\huge\frac \pi 4}$$

Answer 3

More generally, let

$$ a_k = \int_0^1 \frac{x^k}{x^4 + 2 x^3 + 2 x^2 - 2 x + 1} \; dx $$

Of course $$a_{k+4} + 2 a_{k+3} + 2 a_{k+2} - 2 a_{k+1} + a_k = \int_0^1 x^k \; dx = \frac{1}{k+1} $$ so we can get $a_k$ for all natural numbers $k$ expressed as linear combinations over the integers of $a_0$ to $a_3$. Using Claude's methods: $$ \eqalign{ a_0 &= \frac{\pi \sqrt{3}}{9} + \frac{\pi}{8}\cr a_1 &= \frac{\pi}{8}\cr a_2 &= \frac{\pi\sqrt{3}}{9} - \frac{\pi}{8}\cr a_3 &= -\frac{\pi \sqrt{3}}{9}+\frac{\pi}{8}+\frac{\ln \! \left(2\right)}{2}\cr} $$ If you want a polynomial $p(x)$ of degree $\le 3$ with rational coefficients so $\int_0^1 \frac{p(x)}{x^4 + 2 x^3 + 2 x^2 - 2 x + 1}\; dx$ is a rational number times $\pi$, you want the coefficient of $x^3$ to be $0$ (because you can't handle $\ln(2)$), and the coefficients of $x^0$ and $x^2$ must add to $0$ (to get rid of the $\sqrt{3}$).