Mathematics Asked on November 30, 2020

How do I integrate the improper integral $$int_{-infty}^{infty} frac{dx}{1+x^{12}}$$ using partial fraction decomposition?

I am restricted to use only the principles taught in Calculus 2, which entails partial fraction decomposition. I have tried factoring the denominator but was quickly stumped at the complicated roots. I am unsure of how to proceed. I even thought of looking at it as a series but I don’t think I can do that.

Thank you very much for your help. I greatly appreciate it.

Decompose the integrand

$$ frac{6}{1+x^{12}} =frac2{x^4+1}-frac{sqrt3x^2-2}{x^4-sqrt3x^2+1} + frac{sqrt3x^2+2}{x^4+sqrt3x^2+1} $$

Then

$$int_{-infty}^{infty} frac{dx}{1+x^{12}} =frac23 J(0) +frac{2+sqrt3}3 J(sqrt3)+ frac{2-sqrt3}3 J(-sqrt3)tag1$$

where

begin{align} J(a) &= int_{0}^{infty} frac{dx}{x^4+ax^2+1} overset{xto frac1x} =int_{0}^{infty} frac{x^2dx}{x^4+ax^2+1} \ & =frac12 int_{0}^{infty} frac{(x^2+1)dx}{x^4+ax^2+1} =frac12 int_{0}^{infty} frac{d(x-frac1x)}{(x-frac1x)^2+2+a} =fracpi{2sqrt{2+a}}\ end{align}

Let $a=0, pm sqrt3$ and plug the results into (1) to obtain

$$int_{-infty}^{infty} frac{dx}{1+x^{12}} = frac{sqrt3+1}{3sqrt2}$$

Answered by Quanto on November 30, 2020

To help you get started: $u^4 + 1 = u^4 + 2u^2 + 1 - 2u^2 $. Then write as a difference of squares and factor.

Answered by Frank Newman on November 30, 2020

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