How do I solve this generalized integral??
$\displaystyle \displaystyle\int_{-\infty}^{\infty}{\frac{1}{1+{x^6^}}} dx$
I appriciate any guidance. Thank you
Hello, 4Math!
This is a truly ugly problem, but it can be solved . . .
$\displaystyle \displaystyle \int_{\text{-}\infty}^{\infty}{\frac{dx}{1+x^6}$
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 . . .
$\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} ) } $
I took Defunkt's advised (See the attached pdf): Complexifying integral.pdf
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