Find $\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^4}dx$

Question

I was asked to prove that

$$\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^{4}}dx=\pi\frac{14325195794+(2815367209\sqrt{26})}{14623232(9+2\sqrt{26})^\frac{7}{2}}$$

I checked the result numerically and the first digits correct using W|F

$$\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^4}dx\approx 0.19874620328$$

I tried to start with trig substitution but the high power in the integral make it more complicated. Is there any way to evaluate this integral?

That can be tackled through the residue theorem with the same technique shown here: http://math.stackexchange.com/questions/266181/a-nasty-integral-of-a-rational-function — Jack D'Aurizio, Sep 13 '16 at 16:50
@JackD'Aurizio he is my Mathematics professor :). I don't know how he get this horrible integral — Refaat M. Sayed, Sep 13 '16 at 17:00
My bet is on a Chinese mathematical competition. Anyway, the question ultimately boils down to writing $x^8+5x^6+14 x^4+5x^2+1$ as a sum of two squares, see my other answer. The fourth power is not a huge complication. — Jack D'Aurizio, Sep 13 '16 at 17:02
@tired I admit that most of the hardest integrals comes from Russia, but not this one because my professor is Egyptian also I am not sure if this integral is made by him or not. — Refaat M. Sayed, Sep 13 '16 at 17:49
$y^4+5y^3+14y^2+5y+1=(y^2+\frac{1}{2}(5+\sqrt{23}i)y+1)(y^2+\frac{1}{2}(5-\sqrt{23}i)y+1)$ — Rene Schipperus, Sep 13 '16 at 18:04

score 11 · Accepted Answer · answered Sep 13 '16 at 22:52

Hint. A route.

One may recall the following result, which goes back at least to G. Boole (1857).

Proposition. Let $f \in L^1(\mathbb{R})$ and let $f$ be an even function. Then $$ \int_{-\infty}^{+\infty}x^{2n}f\left(x-\frac1x\right) dx=\sum_{k=0}^n \frac{(n+k)!}{(2k)!(n-k)!}\int_{-\infty}^{+\infty} x^{2k}f(x)\: dx. \tag1 $$

Then one may write $$ \begin{align} &\int_0^{+\infty}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^{4}}\:dx \\\\&=\int_0^{+\infty}\frac{x^{-16}}{\left(\left(x^4+\dfrac1{x^4}\right)+5\left(x^2+\dfrac1{x^2}\right)+14\right)^{4}}\:dx \\\\&=\int_0^{+\infty}\frac{x^{-16}}{\left(\left[\left(x-\dfrac1x\right)^2+2\right]^2+5\left(x-\dfrac1x\right)^2+22\right)^{4}}\:dx \\\\&=\int_0^{+\infty}\frac{x^{14}}{\left(\left[\left(x-\dfrac1x\right)^2+2\right]^2+5\left(x-\dfrac1x\right)^2+22\right)^{4}}\:dx \qquad \left(x \to \dfrac1x \right) \\\\&=\frac12\int\limits^{\infty}_{-\infty}\frac{x^{14}}{\left(\left[\left(x-\dfrac1x\right)^2+2\right]^2+5\left(x-\dfrac1x\right)^2+22\right)^{4}}\:dx \\\\&=\frac12\sum_{k=0}^7 \frac{(7+k)!}{(2k)!(7-k)!}\int_{-\infty}^{+\infty} \frac{x^{2k}}{\left(\left(x^2+2\right)^2+5x^2+22\right)^{4}}\:dx \qquad (\text{using}\,\,(1)) \\\\&=\sum_{k=0}^7 \frac{(7+k)!}{(2k)!(7-k)!}\int_0^{+\infty} \frac{x^{2k}}{\left(x^4+6x^2+26\right)^{4}}\:dx \\\\&=\pi\:\frac{14325195794+(2815367209\sqrt{26})}{14623232(9+2\sqrt{26})^\frac{7}{2}} \end{align} $$ where we have concluded by using Theorem $3.1$ (p.$6$) here, in G. Boros and V. Moll's paper.

Awesome solution sir O. Oloa. (+1). Just question. is there a proof for the Proposition? — Refaat M. Sayed, Sep 13 '16 at 23:22
+1. Unexpected and quite fine. I was thinking in the factor $x + x^{-1}$ but I didn't go further. — Felix Marin, Sep 13 '16 at 23:32

Find $\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^4}dx$

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