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Math Help - Integration

  1. #1
    Member roshanhero's Avatar
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    Integration

    \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...

    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...

    \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...

    \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} =

    = \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

    \chi \sigma
    Last edited by chisigma; March 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

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

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

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

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

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

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

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

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

    Quote Originally Posted by roshanhero View Post
    \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  x = \tan(\theta) \implies dx = \sec^2(\theta)~d\theta

    The integral becomes :

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

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

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

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

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

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

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

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

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

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

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

    The integral becomes :

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

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

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

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

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

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

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

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

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

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

     = 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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    <br />
\int_{-\infty}^{\infty} \frac{1 + x^2 }{1+x^4}~dx<br />
    How did you get this from the above one.I am confused..
    <br />
2 \int_0^{\infty} \frac{1 + x^2}{1+x^4}~dx<br />
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  9. #9
    Member roshanhero's Avatar
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    I got this answer,tell me where I am wrong
    <br />
\int_{\infty }^{\infty }\frac{1+x^2}{1+x^4}dx<br />
\int_{\infty }^{\infty }\frac{x^2(1+1/x^2)}{x^2(x^2+1/x^2)}dx<br />
    <br />
\int_{\infty }^{\infty }\frac{(1+1/x^2)}{(x-1/x)^2+2}<br />
Put (x-1/x)=u<br />
    <br />
\int_{\infty }^{\infty }\frac{du}{u^2+2}<br />
\int_{\infty }^{\infty }\frac{du}{u^2+(\sqrt2)^2}<br />
[\frac{1}{\sqrt2} \tan^{-1}(u/\sqrt{2})]_{-\infty}^{\infty}<br />
\frac{\pi}{\sqrt2}<br />
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  10. #10
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    Quote Originally Posted by roshanhero View Post
    <br />
\int_{-\infty}^{\infty} \frac{1 + x^2 }{1+x^4}~dx<br />
    How did you get this from the above one.I am confused..
    <br />
2 \int_0^{\infty} \frac{1 + x^2}{1+x^4}~dx<br />
    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
    <br />
\int_{\infty }^{\infty }\frac{1+x^2}{1+x^4}dx<br />
\int_{\infty }^{\infty }\frac{x^2(1+1/x^2)}{x^2(x^2+1/x^2)}dx<br />
    <br />
\int_{\infty }^{\infty }\frac{(1+1/x^2)}{(x-1/x)^2+2}<br />
Put (x-1/x)=u<br />
    <br />
\int_{\infty }^{\infty }\frac{du}{u^2+2}<br />
\int_{\infty }^{\infty }\frac{du}{u^2+(\sqrt2)^2}<br />
[\frac{1}{\sqrt2} \tan^{-1}(u/\sqrt{2})]_{-\infty}^{\infty}<br />
\frac{\pi}{\sqrt2}<br />

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

    The explanation is ,

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

    there is a singularity at  x = 0

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

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

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

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

    The substitution gives

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

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

    But if we do the substitution first , we will obtain  \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 a>0 and f being an even integrable function on [-a,a] then \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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