How do I solve this generalized integral??

$\displaystyle \displaystyle\int_{-\infty}^{\infty}{\frac{1}{1+{x^6^}}} dx$

I appriciate any guidance. Thank you

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- Sep 15th 2010, 12:40 PM4MathGeneralized integral
How do I solve this generalized integral??

$\displaystyle \displaystyle\int_{-\infty}^{\infty}{\frac{1}{1+{x^6^}}} dx$

I appriciate any guidance. Thank you - Sep 15th 2010, 12:45 PMDefunkt
Have you learned how to solve complex integrals? 'Complexifying' the integral and then using the residue theorem can solve this quite easily, but it won't help if you haven't studied those tools.

- Sep 15th 2010, 02:12 PMSoroban
Hello, 4Math!

This is a trulyproblem, but it can be solved . . .*ugly*

Quote:

$\displaystyle \displaystyle \int_{\text{-}\infty}^{\infty}{\frac{dx}{1+x^6}$

The denominator is the sum of cubes: .$\displaystyle x^6 + 1 \;=\;(x^2)^3 + (1)^3$

. . which factors: .$\displaystyle (x^2 + 1)(x^4 - x^2 + 1)$

The quartic can also be factored . . .

Add and subtract $\displaystyle 3x^2\!:$

. . $\displaystyle x^4 - x^2 + 1 + 3x^2 - 3x^2$

. . $\displaystyle \;=\;x^4 + 2x^2 + 1 - 3x^2$

. . $\displaystyle =\;(x^2 + 1)^2 - (\sqrt{3}\;\!x)^2 $ . (difference of squares)

. . $\displaystyle =\;(x^2+1 - \sqrt{3}\;\!x)(x^2+1 + \sqrt{3}\;\1x)$

. . $\displaystyle =\;(x^2 - \sqrt{3}\;\!x + 1)(x^2 + \sqrt{3}\;\!x + 1)$

The denominator is: .$\displaystyle x^6+1 \:=\:(x^2 + 1)(x^2 - \sqrt{3}\;\!x + 1)(x^2 + \sqrt{3}\;\!x + 1)$

You know what to do, right?

Apply partial fraction decomposition . . .

$\displaystyle \displaystyle \frac{1}{(x^2+1)(x^2-\sqrt{3}\;\!x + 1)(x^2 + \sqrt{3}\;\!x + 1)} $

. . . . . . . . . $\displaystyle \displaystyle =\;\frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2-\sqrt{3}\;\!x+1} + \frac{Ex + F}{x^2 + \sqrt{3}\;\!x + 1} $

I'll wait in the car . . . - Sep 15th 2010, 03:15 PMArchie Meade
I'll bring the burgers and chips.

Soroban, I get this picture of square roots, squares and cubes strewn all around inside the boot of your car!

So educational! - Sep 16th 2010, 12:42 AMsimplependulum
$\displaystyle \int_{-\infty}^{\infty} \frac{dx}{1 + x^6} $

$\displaystyle = 2 \int_0^{\infty} \frac{dx}{1 + x^6}$

Consider the integral next to $\displaystyle 2 $ :

$\displaystyle I = \int_0^{\infty} \frac{dx}{1 + x^6} $

Sub. $\displaystyle x \mapsto 1/x $ ( maybe i have been using this substitution much more times than trigo. sub. since I realized its power and magic ! )

$\displaystyle I = \int_0^{\infty} \frac{x^4 dx}{1 + x^6} $

Summation gives :

$\displaystyle 2I = \int_0^{\infty} \frac{1+x^4 }{1 + x^6} ~dx $

$\displaystyle = \int_0^{\infty} \frac{(1-x^2 + x^4 ) + x^2}{1 + x^6} ~dx $

$\displaystyle = \int_0^{\infty} \frac{dx}{1 + x^2} + \frac{1}{3} \int_0^{\infty} \frac{d(x^3)}{1 + (x^3)^2 }$

$\displaystyle = \frac{\pi}{2} + \frac{1}{3} \frac{\pi}{2} = \frac{2\pi}{3} $

so we find that the answer is $\displaystyle \frac{2\pi}{3} $

If i am right , your generalized integral

$\displaystyle \int_0^{\infty} \frac{dx}{1 + x^n } $ should be

$\displaystyle \frac{\pi}{n \sin( \frac{\pi}{n} ) } $ - Sep 16th 2010, 03:45 AM4Math

I took Defunkt's advised (See the attached pdf): Attachment 18948

and complexified the integral. It was not easy. After tedious calculations I finally solved it but didn't know if it was correct. But we both get the same result, so I sure hope its correct. Thank you very much for your response.

Kind regards