Insane Integral

Calculus Level 5

If it is given

∫ − ∞ ∞ x 2 x 6 − 2 x 5 − 2 x 4 + 4 x 3 + 3 x 2 − 4 x + 1 d x = π , \int_{-\infty}^\infty\frac{x^2}{x^6 - 2x^5 - 2x^4 + 4x^3 + 3x^2 - 4x + 1} \; dx=\pi,

then the value of

∫ 0 ∞ x 8 − 4 x 6 + 9 x 4 − 5 x 2 + 1 x 12 − 10 x 10 + 37 x 8 − 42 x 6 + 26 x 4 − 8 x 2 + 1 d x \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

can be expressed as p π q \,\dfrac{p\pi}{q} . Find the value of p 2 + q 2 \,p^2+q^2 .


The answer is 5.

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Tunk-Fey Ariawan by Tunk-Fey Ariawan , 35, Indonesia

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1 solution

Gopinath No
Apr 7, 2014

Let

g ( x ) = x 2 x 6 − 2 x 5 − 2 x 4 + 4 x 3 + 3 x 2 − 4 x + 1 f ( x ) = x 8 − 4 x 6 + 9 x 4 − 5 x 2 + 1 x 12 − 10 x 10 + 37 x 8 − 42 x 6 + 26 x 4 − 8 x 2 + 1 \displaystyle \begin{aligned} g(x) &= \dfrac{x^{2}}{x^{6} - 2 \, x^{5} - 2 \, x^{4} + 4 \, x^{3} + 3 \, x^{2} - 4 \, x + 1}\\\\ f(x) &= \dfrac{x^{8} - 4 \, x^{6} + 9 \, x^{4} - 5 \, x^{2} + 1}{x^{12} - 10 \, x^{10} + 37 \, x^{8} - 42 \, x^{6} + 26 \, x^{4} - 8 \, x^{2} + 1} \end{aligned}

They are related as

f ( x ) = g ( 1 + x ) + g ( 1 − x ) 2 f(x)=\dfrac{g(1+x)+g(1-x)}{2}

∴ ∫ 0 ∞ f ( x ) d x = 1 2 ( ∫ 0 ∞ g ( x ) d x + ∫ 0 ∞ g ( − x ) d x ) = 1 2 ( ∫ 0 ∞ g ( x ) d x + ∫ − ∞ 0 g ( x ) d x ) = 1 2 ∫ − ∞ ∞ g ( x ) = π 2 \displaystyle \begin{aligned} \therefore \int_0^\infty f(x) \, dx &= \dfrac{1}{2}\left(\int_0^\infty g(x) \, dx+\int_0^\infty g(-x) \, dx\right)\\ &=\dfrac{1}{2}\left(\int_0^\infty g(x) \, dx+\int_{-\infty}^0 g(x) \, dx\right)\\ &=\dfrac{1}{2}\, \int_{-\infty}^\infty g(x)\\ &=\dfrac{\pi}{2} \end{aligned}

Hence, enter 5.

14 Helpful 0 Interesting 2 Brilliant 1 Confused

It appeared first in AMM. See the solution using complex analysis by this mathematician: the integral

gopinath no - 7 years, 2 months ago

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