# Math Help - Tricky integral of a rational function

1. ## Tricky integral of a rational function

1. The problem statement, all variables and given/known data
Verify that $\int_0^{\infty}\frac{x^8-4x^6+9x^4-5x^2+1}{x^{12}-10x^{10}+37x^8-42x^6+26x^4-8x^2+1}dx=\frac{\pi}{2}$
2. Relevant equations
Newton Leibniz formula.
3. The attempt at a solution
This is a particulary interesting integral in that one cannot directly use the partial fractions decomposition nor the methods of residues due to the irreducibility of both numerator and denominator. Advanced methods such as Feynman and Schwinger parametrizations also fail. On the other hand, Mathematica has no difficulty in evaluating this integral so I must be missing on something.
Any help would be greatly appreciated.
Thanks!

2. Have you tried some trigonometric substitution? (x=sin(a) for instance...)

3. Originally Posted by Also sprach Zarathustra
Have you tried some trigonometric substitution? (x=sin(a) for instance...)
Haven't tried it. Then a would have to be complex, right?

4. x=tan(t) doesn't help, either.

5. Still haven't solved it Is there, perhaps, some class of special functions that are rational? Maybe this is a special case of that?

6. I only know of using the residue theorem. The only thing I can think of is that by stating that this function is even so this integral can be re-written as

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

You can turn this into a complex line integral

$\frac{1}{2}\oint_{C}\frac{z^8-4z^6+9z^4-5z^2+1}{z^{12}-10z^{10}+37z^8-42z^6+26z^4-8z^2+1}dz$

Where C is the closed semicircle contour large enough to enclose all the poles in the upper half of the imaginary axis. You then use residue theorem, but to get the poles, the only way I can think of is by going to Mathematica, Maple, MATLAB etc. You would get a ton of roots, and it's very tedious to do this by hand, but at the end, your answer should be $\pi$ which means that the final answer is $\frac{\pi}{2}$

7. A solution is given in page 258 of this book. Apparently, the problem appeared the American Mathematical
Monthly in April 2005. The solution separates the integral into two then uses contour integration. There is
a little lemma about 'positively-oriented simple closed curves' that finishes off the problem (I'm completely
uninitiated when it comes to complex analysis, so I didn't really make much of it -- but perhaps you can).