I'm having a hard time proving the following $$\int_0^{\infty} \frac{x^8 - 4x^6 + 9x^4 - 5x^2 + 1}{x^{12} - 10 x^{10} + 37x^8 - 42x^6 + 26x^4 - 8x^2 + 1} \, dx = \frac{\pi}{2}.$$

Mathematica has no problem evaluating it while I haven't the slightest idea how to approach it. Of course, I would like to prove it without the use of a computer. Is this an explicit form of a special function I fail to recognize?

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    $\begingroup$ Where did you find this integral? $\endgroup$ – Fabian Dec 27 '12 at 23:59
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    $\begingroup$ Our professor gave it to us as a fun problem. $\endgroup$ – user54031 Dec 28 '12 at 0:14
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    $\begingroup$ Maybe try a comparison test instead of trying to evaluate the integral directly. I seriously doubt your prof gave you this so you guys could spend 5 hours trying to decompose the fraction $\endgroup$ – Hawk Dec 28 '12 at 1:16
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    $\begingroup$ @user1551: The substitution $x = \cos t$ wouldn't work since $x \in (0,\infty)$. $\endgroup$ – user54031 Dec 29 '12 at 15:21
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    $\begingroup$ Please see the very nice solution by @achillehui to my question in an attempt to approach this from another angle. I have suggested that he post it as a solution here. $\endgroup$ – hypergeometric Dec 3 '16 at 16:53

The integral over $\mathbb{R}$ of a meromorphic function $f(z)$, $O(|z|^{-2})$ at infinity, non-vanishing over $\mathbb{R}$, is equal to $2\pi i$ times the sum of residues in the poles located in the complex upper-half plane. Since:

$$p(y) = y^6-2y^5-2y^4+4y^3+3y^2-4y+1 = p_{+}(y)\cdot p_{-}(y),$$ $$p_{+}(y)= y^3-(i+1)y^2+(i-2)x+1,\qquad p_{-}(y)=y^3+(i-1)y^2-(2+i)y+1,$$

(I got this through a numerical calculation of the roots of $p(y)$, followed by a separation of the roots with positive and negative imaginary part, say $\zeta_1,\zeta_2,\zeta_3$ and $\bar{\zeta_1},\bar{\zeta_2},\bar{\zeta_3}$ - so $p_{+}(z)$ is just $\prod_{j=1}^3 (z-\zeta_j)$) we have:

$$ I = \int_{\mathbb{R}}\frac{y^2 dy}{p(y)} = 2\pi i\sum_{j=1}^{3}\operatorname{Res}_{z=\zeta_j}\left(\frac{z^2}{p_{+}(z)\cdot p_{-}(z)}\right),$$

but $p_{-}(x)-p_{+}(x)=2i(x^2-x)$, so:

$$ I = \int_{\mathbb{R}}\frac{y^2 dy}{p(y)} = 2\pi i\sum_{j=1}^{3}\operatorname{Res}_{z=\zeta_j}\left(\frac{z^2}{p_{+}^2(z)+2i(z^2-z)p_{+}(z)}\right).$$

By De l'Hopital theorem, and since $\zeta_j$ is a double zero of $p_{+}^2(x)$:



$$ I = \int_{\mathbb{R}}\frac{y^2 dy}{p(y)} = \pi\sum_{j=1}^{3}\frac{\zeta_j}{(\zeta_j-1)p_{+}'(\zeta_j)}.$$

Now we compute the remainder between $(z-1)p_{+}'(z)$ and $p_{+}(z)$, in order to have:

$$ I = \int_{\mathbb{R}}\frac{y^2 dy}{p(y)} = \pi\sum_{j=1}^{3}\frac{\zeta_j}{-(1+i)+6\zeta_j-(2-i)\zeta_j^2}.$$

If now we take $\alpha=\frac{3+\sqrt{6-i}}{2-i}$ and $\beta=\frac{3-\sqrt{6-i}}{2-i}$ we can re-write the last line as:

$$ I = \frac{\pi}{(i-2)(\alpha-\beta)}\left(\sum_{j=1}^{3}\frac{\alpha}{\zeta_j-\alpha}-\sum_{j=1}^{3}\frac{\beta}{\zeta_j-\beta}\right)=-\frac{\pi}{2\sqrt{6-i}}\left(\Sigma_1-\Sigma_2\right).$$

Now $\Sigma_1$ is the sum of the reciprocal of the roots of the polynomial $p_{+}(\alpha(z+1))$, and $\Sigma_2$ is the sum of the reciprocal of the roots of the polynomial $p_{+}(\beta(z+1))$. This quantities can be computed through the coefficients of $p_{+}$, since the sum of the reciprocal of the roots of a polynomial $q(z)$ is just $-\frac{q'(0)}{q(0)}$. This gives:

$$\Sigma_1 = -\alpha\frac{p_{+}'(\alpha)}{p_{+}(\alpha)},\qquad \Sigma_2 = -\beta\frac{p_{+}'(\beta)}{p_{+}(\beta)}.$$

Up to a massive amount of long but straightforward computations, we get:

$$\Sigma_1 = (i-2)-\sqrt{6-i},\qquad \Sigma_2 = (i-2)+\sqrt{6-i}, $$

from which $\color{red}{I=\pi}$ finally follows.

I am really grateful to Jon Haussmann for the proof that

$$\int_0^{\infty} \frac{x^8 - 4x^6 + 9x^4 - 5x^2 + 1}{x^{12} - 10 x^{10} + 37x^8 - 42x^6 + 26x^4 - 8x^2 + 1}dx = \frac{1}{2}\int_{\mathbb{R}}\frac{y^2 dy}{p(y)},$$

where only the second integral is treated here.

IMPORTANT UPDATE: In fact, there is no need to compute the coefficients of $p_{+}(x)$ and $p_{-}(x)$ (we only need the identity $p_{-}(x)-p_{+}(x)=2i(x^2-x)$), or introduce $\alpha$ and $\beta$. Since $p_{+}(x)$ is a third-degree polynomial with roots in the upper half-plane, $$0=\int_{\mathbb{R}}\frac{dz}{p_{+}(z)}=\sum_{j=1}^{3}\frac{1}{p_{+}'(\zeta_j)}.$$ This gives: $$ I = \pi\sum_{j=1}^{3}\frac{\zeta_j}{(\zeta_j-1)p_{+}'(z)} = \pi\sum_{j=1}^{3}\frac{1}{(\zeta_j-1)p_{+}'(\zeta_j)}, $$ but if we decompose $\frac{1}{p_{+}(z)}$ in simple fractions, we get: $$\frac{1}{p_{+}(z)}=\sum_{j=1}^{3}\frac{1}{p_{+}'(\zeta_j)(z-\zeta_j)}, $$ so the magic gives: $$ I = -\frac{\pi}{p_{+}(1)}.$$ Since $p(x)=p_{+}(x)\cdot p_{-}(x)$, $p(1)=1$, $I\in\mathbb{R}^+$ and $p_{-}(1)$ is the conjugate of $p_{+}(1)$, $p_{+}(1)$ can be only $+1$ or $-1$, so $I=\pi$.

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    $\begingroup$ I have the strong feeling that this proof can be significantly shortened, since I do not think that $(\Sigma_1-\Sigma_2)=(i-2)(\alpha-\beta)$ is a mere coincidence. Probably the reduction to elementary functions of the roots of $p_{+}(z)$ is possible directly from: $$I=\pi\sum_{j=1}^{3}\frac{\zeta_j}{(\zeta_j-1)p_{+}'(\zeta_j)}.$$ $\endgroup$ – Jack D'Aurizio Dec 4 '13 at 14:59
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    $\begingroup$ In fact, the feeling was right, the only indespensable ingredient of the proof is the identity: $$p(x)=(x^3-x^2-2x+1)^2+(x^2-x)^2.$$ $\endgroup$ – Jack D'Aurizio Dec 4 '13 at 17:15
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    $\begingroup$ The same technique leads to: $$\frac{\pi}{10}=\int_{\mathbb{R}}\frac{x^2 dx}{x^6+x^4+4x^2+4},\qquad \frac{\pi(\sqrt{2}-1)}{4}=\int_{\mathbb{R}}\frac{x^2 dx}{x^6+x^4+x^2+1}.$$ $\endgroup$ – Jack D'Aurizio Dec 4 '13 at 19:55
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    $\begingroup$ Excellent job! Do you think Wolfram Alpha / Mathematica uses this method to calculate these kind of integrals? $\endgroup$ – user54031 Dec 4 '13 at 22:07
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    $\begingroup$ Honestly, I do not know it for sure. Probably the Sigma algorithm for symbolic integration can handle a large amount of logarithm and dilogarithm identities in order to compute, by the residue theorem or whatsoever, the indefinite integral of rational functions. In fact we only need the resultant between a polynomial and its derivative, a localization algorithm for the roots of such polynomials (above/below the real axis) and proper handling of polynomial symmetric functions of the roots - all these tasks are quite expensive to be achieved by hand computations, but very CPU-like. $\endgroup$ – Jack D'Aurizio Dec 5 '13 at 9:37

Some progess: The integrand actually decomposes as $$\frac{1}{2} \left( \frac{x^2 + 2x + 1}{x^6 + 4x^5 + 3x^4 - 4x^3 - 2x^2 + 2x + 1} + \frac{x^2 - 2x + 1}{x^6 - 4x^5 + 3x^4 + 4x^3 - 2x^2 - 2x + 1} \right).$$ Note that the second term is the same as the first term, except with $-x$ instead of $x$. Thus, with some substitutions, the integral becomes $$\frac{1}{2} \int_{-\infty}^\infty \frac{y^2}{y^6 - 2y^5 - 2y^4 + 4y^3 + 3y^2 - 4y + 1} \; dy.$$

  • $\begingroup$ $$ \int_{-\infty}^\infty \frac{y^2}{y^6 - 2y^5 - 2y^4 + 4y^3 + 3y^2 - 4y + 1} \; dy=\pi, $$ by Wolframalpha, so the result is correct. (I guess if you have a pro version then it is possible to obtain some details.) $\endgroup$ – vesszabo Jan 2 '13 at 20:57


$$I=\int_0^{\infty} \frac{x^8 - 4x^6 + 9x^4 - 5x^2 + 1}{x^{12} - 10 x^{10} + 37x^8 - 42x^6 + 26x^4 - 8x^2 + 1} \, dx$$

As noted by Jon Haussmann,

$$2I=\int_0^{\infty} \frac{(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 - 2x^2 -2x + 1}dx + \int_0^{\infty}\frac{(x+1)^2}{x^6 +4x^5 + 3x^4 - 4x^3 - 2x^2 + 2x + 1}dx$$

Perform the change of variable $x=\dfrac{1}{u-1}$ in the second integral,

$\begin{align}2I&=\int_0^{\infty} \frac{(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 - 2x^2 -2x + 1}dx+\int_1^{\infty} \frac{x^2(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 - 2x^2 -2x + 1}dx\\ &=\left(\int_0^1 \frac{(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 - 2x^2 -2x + 1}dx+\\ \int_1^{\infty} \frac{(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 - 2x^2 -2x + 1}dx\right)+\\&\int_1^{\infty} \frac{x^2(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 - 2x^2 -2x + 1}dx\end{align}$

Perform the change of variable $x=\dfrac{u-1}{u}$ in the first integral of the latter equality,

$\begin{align}2I&=\int_1^{\infty} \frac{x^2}{x^6-4x^5 + 3x^4 +4x^3 - 2x^2 -2x + 1}dx+\int_1^{\infty} \frac{(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 - 2x^2 -2x + 1}dx+\int_1^{\infty} \frac{x^2(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 - 2x^2 -2x + 1}dx\\ &=\int_1^{\infty} \frac{x^2+(x-1)^2+x^2(x-1)^2}{x^6-4x^5 + 3x^4 +4x^3 - 2x^2 -2x + 1}dx\\ &=\Big[\arctan\left(\dfrac{x^3-2x^2-x+1}{x(x-1)}\right)\Big]_0^{+\infty}\\ &=\pi \end{align}$

(Problem found in American Mathematical Monthly, vol. 112, april 2005.

Solution found in vol. 114)

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    $\begingroup$ (+1) So by considering $\arctan(\text{rational function})$ and reverse-engineering this approach, we get a lot of nasty integrals that equal $\pi$ :D $\endgroup$ – Jack D'Aurizio Dec 30 '16 at 19:30

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