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Thread: Integration

  1. #1
    Member roshanhero's Avatar
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    Integration

    $\displaystyle \int_{-\oe }^{\oe}\frac{1+x^2}{1+x^4}dx$
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  2. #2
    MHF Contributor chisigma's Avatar
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    If we don't want to use complex analysis we can proceed in 'elementary' way first remembering that...

    $\displaystyle 1+x^{4} = (x^{2} - \sqrt{2}\cdot x +1)\cdot (x^{2} + \sqrt{2}\cdot x +1)$ (1)

    ... and that conducts to the identity...

    $\displaystyle \frac{1 + x^{2}}{1+ x^{4}} = \frac{1}{2}\cdot \frac{1}{x^{2} - \sqrt{2}\cdot x +1} + \frac{1}{2}\cdot \frac{1}{x^{2} + \sqrt{2}\cdot x +1} $ (2)

    ... so that is...

    $\displaystyle \int_{- \infty}^{\infty} \frac{1+x^{2}}{1+x^{4}}\cdot dx =\int_{0}^{\infty} \frac{dx}{x^{2} - \sqrt{2}\cdot x +1} + \int_{0}^{\infty} \frac{dx}{x^{2} + \sqrt{2}\cdot x +1} = $

    $\displaystyle = \sqrt{2} \cdot |\tan^{-1} (\sqrt{2}\cdot x + 1) + \tan^{-1} (\sqrt{2}\cdot x - 1)|_{0}^{\infty} = \pi \cdot \sqrt{2}$ (3)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Mar 25th 2010 at 07:52 AM. Reason: Non decisive error corrected...
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  3. #3
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    Oh , a classic integral

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

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

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

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

    Sub $\displaystyle x-1/x = t \implies (1+ 1/x^2)dx=dt$

    $\displaystyle I = 2 \int_{-\infty}^{\infty} \frac{dt}{t^2+2} $

    $\displaystyle = \sqrt{2} [ \tan^{-1}(t/\sqrt{2})]_{-\infty}^{\infty} $

    $\displaystyle = \sqrt{2} \pi$
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  4. #4
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    possible alternative?

    Quote Originally Posted by roshanhero View Post
    $\displaystyle \int_{-\oe }^{\oe}\frac{1+x^2}{1+x^4}dx$
    how about x=tan(theta)??
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  5. #5
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    Quote Originally Posted by Pulock2009 View Post
    how about x=tan(theta)??

    Of course , anything you like could be a perfect solution !



    Sub $\displaystyle x = \tan(\theta) \implies dx = \sec^2(\theta)~d\theta $

    The integral becomes :

    $\displaystyle I = 2 \int_0^{\frac{\pi}{2} } \frac{ \sec^4(\theta)}{1 + \tan^4(\theta)} ~d\theta $

    $\displaystyle = 2 \int_0^{\frac{\pi}{2} } \frac{ d\theta }{ \sin^4(\theta ) + \cos^4(\theta )} $

    $\displaystyle = 4 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ \sin^4(\theta ) + \cos^4(\theta )} $

    $\displaystyle = 4 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ (\sin^2(\theta ) + \cos^2(\theta ))^2 - 2\sin^2(\theta)\cos^2(\theta ) } $

    $\displaystyle = 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ 2 - \sin^2(2\theta)} $

    $\displaystyle = 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ 1 + \cos^2(2\theta)}$

    $\displaystyle = 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ \sin^2(2\theta) + 2\cos^2(2\theta)}$

    $\displaystyle = 8 \int_0^{\frac{\pi}{4} } \frac{\sec^2(2\theta) d\theta }{ \tan^2(2\theta) + 2 } $

    Sub $\displaystyle \tan(2\theta) = t \implies 2\sec^2(2\theta) d\theta = dt $

    $\displaystyle I = 4 \int_0^{\infty} \frac{dt}{t^2+2} $

    $\displaystyle = 2\sqrt{2} \cdot \frac{\pi}{2} = \sqrt{2} \pi$
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  6. #6
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    wonderful!!

    Quote Originally Posted by simplependulum View Post
    Of course , anything you like could be a perfect solution !



    Sub $\displaystyle x = \tan(\theta) \implies dx = \sec^2(\theta)~d\theta $

    The integral becomes :

    $\displaystyle I = 2 \int_0^{\frac{\pi}{2} } \frac{ \sec^4(\theta)}{1 + \tan^4(\theta)} ~d\theta $

    $\displaystyle = 2 \int_0^{\frac{\pi}{2} } \frac{ d\theta }{ \sin^4(\theta ) + \cos^4(\theta )} $

    $\displaystyle = 4 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ \sin^4(\theta ) + \cos^4(\theta )} $

    $\displaystyle = 4 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ (\sin^2(\theta ) + \cos^2(\theta ))^2 - 2\sin^2(\theta)\cos^2(\theta ) } $

    $\displaystyle = 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ 2 - \sin^2(2\theta)} $

    $\displaystyle = 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ 1 + \cos^2(2\theta)}$

    $\displaystyle = 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ \sin^2(2\theta) + 2\cos^2(2\theta)}$

    $\displaystyle = 8 \int_0^{\frac{\pi}{4} } \frac{\sec^2(2\theta) d\theta }{ \tan^2(2\theta) + 2 } $

    Sub $\displaystyle \tan(2\theta) = t \implies 2\sec^2(2\theta) d\theta = dt $

    $\displaystyle I = 4 \int_0^{\infty} \frac{dt}{t^2+2} $

    $\displaystyle = 2\sqrt{2} \cdot \frac{\pi}{2} = \sqrt{2} \pi$
    one more question:how do u typeout the latex so fast???
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  7. #7
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    Quote Originally Posted by Pulock2009 View Post
    one more question:how do u typeout the latex so fast???

    Well , most of them are copied from the previous lines , haha .
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  8. #8
    Member roshanhero's Avatar
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    $\displaystyle
    \int_{-\infty}^{\infty} \frac{1 + x^2 }{1+x^4}~dx
    $
    How did you get this from the above one.I am confused..
    $\displaystyle
    2 \int_0^{\infty} \frac{1 + x^2}{1+x^4}~dx
    $
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  9. #9
    Member roshanhero's Avatar
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    I got this answer,tell me where I am wrong
    $\displaystyle
    \int_{\infty }^{\infty }\frac{1+x^2}{1+x^4}dx
    \int_{\infty }^{\infty }\frac{x^2(1+1/x^2)}{x^2(x^2+1/x^2)}dx
    $
    $\displaystyle
    \int_{\infty }^{\infty }\frac{(1+1/x^2)}{(x-1/x)^2+2}
    Put (x-1/x)=u
    $
    $\displaystyle
    \int_{\infty }^{\infty }\frac{du}{u^2+2}
    \int_{\infty }^{\infty }\frac{du}{u^2+(\sqrt2)^2}
    [\frac{1}{\sqrt2} \tan^{-1}(u/\sqrt{2})]_{-\infty}^{\infty}
    \frac{\pi}{\sqrt2}
    $
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  10. #10
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    Quote Originally Posted by roshanhero View Post
    $\displaystyle
    \int_{-\infty}^{\infty} \frac{1 + x^2 }{1+x^4}~dx
    $
    How did you get this from the above one.I am confused..
    $\displaystyle
    2 \int_0^{\infty} \frac{1 + x^2}{1+x^4}~dx
    $
    Well, that's easy- since the function being integrated is an even function, the integral from -a to a is just twice the integral from 0 to a.
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  11. #11
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    Quote Originally Posted by roshanhero View Post
    I got this answer,tell me where I am wrong
    $\displaystyle
    \int_{\infty }^{\infty }\frac{1+x^2}{1+x^4}dx
    \int_{\infty }^{\infty }\frac{x^2(1+1/x^2)}{x^2(x^2+1/x^2)}dx
    $
    $\displaystyle
    \int_{\infty }^{\infty }\frac{(1+1/x^2)}{(x-1/x)^2+2}
    Put (x-1/x)=u
    $
    $\displaystyle
    \int_{\infty }^{\infty }\frac{du}{u^2+2}
    \int_{\infty }^{\infty }\frac{du}{u^2+(\sqrt2)^2}
    [\frac{1}{\sqrt2} \tan^{-1}(u/\sqrt{2})]_{-\infty}^{\infty}
    \frac{\pi}{\sqrt2}
    $

    When i was doing the integral , i faced the same problem :

    The explanation is ,

    the mapping we have taken $\displaystyle x - 1/x = t $ ,

    there is a singularity at $\displaystyle x = 0 $

    $\displaystyle \lim_{x\to 0^-} (x-1/x) = + \infty $ while

    $\displaystyle \lim_{x\to 0^+} (x-1/x) = - \infty $

    So we must put $\displaystyle \int_{- \infty }^{+ \infty }$

    $\displaystyle = \lim_{a\to 0^+}\left( \int_{- \infty }^{0-a} + \int_{0+a}^{+\infty} \right ) $

    The substitution gives

    $\displaystyle = \int_{ -\infty}^{+\infty} + \int_{-\infty}^{+\infty}$

    $\displaystyle = 2\int_{ -\infty}^{+\infty}$

    But if we do the substitution first , we will obtain $\displaystyle \int_{ -\infty}^{+\infty} $ which is half of the correct one .
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  12. #12
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    this just comes from the fact that given $\displaystyle a>0$ and $\displaystyle f$ being an even integrable function on $\displaystyle [-a,a]$ then $\displaystyle \int_{-a}^a f=2\int_0^a f,$ so there's no much here, the justification comes from dealing with improper integrals.
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