Generalized integral

**4Math** · September 15th 2010, 12:40 PM

How do I solve this generalized integral??

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

I appriciate any guidance. Thank you

**Defunkt** · September 15th 2010, 12:45 PM

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.

**Soroban** · September 15th 2010, 02:12 PM

Hello, 4Math!

This is a truly ugly problem, but it can be solved . . .

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

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

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

The quartic can also be factored . . .

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

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

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

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

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

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

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

You know what to do, right?

Apply partial fraction decomposition . . .

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

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

I'll wait in the car . . .

**Archie Meade** · September 15th 2010, 03:15 PM

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!

**simplependulum** · September 16th 2010, 12:42 AM

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

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

Consider the integral next to $2$ :

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

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

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

Summation gives :

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

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

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

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

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

If i am right , your generalized integral

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

$\frac{\pi}{n \sin( \frac{\pi}{n} ) }$

**4Math** · September 16th 2010, 03:45 AM

Originally Posted by simplependulum

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

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

Consider the integral next to $2$ :

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

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

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

Summation gives :

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

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

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

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

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

If i am right , your generalized integral

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

$\frac{\pi}{n \sin( \frac{\pi}{n} ) }$

I took Defunkt's advised (See the attached pdf): Complexifying integral.pdf

Originally Posted by Defunkt

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.

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

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