4
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ This doesn't have any roots over $\Bbb R$ so a partial fraction decomposition is going to be pretty challenging. Wolfram Alpha gives the solution in terms of a product of gamma functions as well. $\endgroup$ – Cameron Williams May 16 '15 at 1:41
  • 1
    $\begingroup$ Using half-angle identities one can compute the complex roots $e^{ik\pi/12} = \cos(k\pi/12) + i\sin(k\pi/12)$ to get the roots over $\Bbb C$, then combine the conjugate terms to get the factorization of the denominator over $\Bbb R$. This is ugly and I don't see any nicer way off the top of my head. $\endgroup$ – Rolf Hoyer May 16 '15 at 1:45
  • 1
    $\begingroup$ Parentheses, please. What you wrote is the integral of $1+x^{12}$, but I am sure you meant $1/(1+x^{12})$ If you know where the roots of $-1$ are, you can group them in conjugate pairs to get a factorization of $1+x^{12}$ into real quadratics $\endgroup$ – Ross Millikan May 16 '15 at 1:45
  • 1
    $\begingroup$ the factorization is $(x^2\pm\sqrt{2}x+1)(x^2\pm\frac{\sqrt{3}-1}{\sqrt{2}}x+1)(x^2\pm\frac{\sqrt{3}+1}{\sqrt{2}}x+1)$ ($\pm$ are for short, every parenthesis counts 2 times, for $+$ and for $-$) @Sam $\endgroup$ – Alexey Burdin May 16 '15 at 1:49
  • $\begingroup$ Someone edit this question $\endgroup$ – Ilaya Raja S May 16 '15 at 1:57
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.