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Math Help - Tougher Integral

  1. #1
    Member RedBarchetta's Avatar
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    Tougher Integral

    I found this one on a Harvard graduate level exam:


    Any clues?
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  2. #2
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    Quote Originally Posted by RedBarchetta View Post
    I found this one on a Harvard graduate level exam:


    Any clues?
    Yes,  \frac{x^2}{x^4+5x^2+4} decomposes into

     \frac{4}{3} \frac{1}{x^2+4} - \frac{1}{3} \frac{1}{x^2+1}

    then integrate each separately.
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  3. #3
    Member pberardi's Avatar
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    Could someone kindly do this improper integral? It has been a while since I have done it, I am very interested in seeing the solution, and would appreciate a proper refresher. Pun intended.
    Last edited by pberardi; December 28th 2008 at 02:39 AM.
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  4. #4
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    Quote Originally Posted by pberardi View Post
    Could someone kindly do this improper integral? It has been a while since I have done it, I am very interested in seeing the solution, and would appreciate a proper refresher. Pun intended.
    Do partial fraction decomposition as danny points out,

    Quote Originally Posted by danny arrigo View Post
    Yes,  \frac{x^2}{x^4+5x^2+4} decomposes into

     \frac{4}{3} \frac{1}{x^2+4} - \frac{1}{3} \frac{1}{x^2+1}

    then integrate each separately.
     \int_0^{\infty} \frac{4}{3} \frac{1}{x^2+4} \, dx - \int_0^{\infty}\frac{1}{3} \frac{1}{x^2+1}\, dx  = \left. \frac{2}{3} \tan^{-1}\frac{x}{2} - \frac{1}{3}\tan^{-1} x \right|_0^{\infty} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}
    Last edited by mr fantastic; December 28th 2008 at 04:17 AM. Reason: Fixed the latex.
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
    Yes,  \frac{x^2}{x^4+5x^2+4} decomposes into

     \frac{4}{3} \frac{1}{x^2+4} - \frac{1}{3} \frac{1}{x^2+1}

    then integrate each separately.
    I hope this is right...

    \frac{4}{3}\int_0^\infty{\frac{1}{x^2 + 4}\,dx} - \frac{1}{3}\int_0^\infty{\frac{1}{x^2 + 1}\,dx} = \frac{4}{3}\left[\frac{1}{2}\arctan{\frac{x}{2}}\right]_0^\varepsilon - \frac{1}{3}\left[\arctan{x}\right]_0^\varepsilon

     = \frac{2}{3}[\lim_{\varepsilon \to \infty}(\arctan{\frac{\varepsilon}{2}}) - \arctan{0}] - \frac{1}{3}[\lim_{\varepsilon \to \infty}(\arctan{\varepsilon}) - \arctan{0}]

     = \frac{2}{3}\left[\frac{\pi}{2} - 0 \right] - \frac{1}{3}\left[\frac{\pi}{2} - 0 \right]

     = \frac{\pi}{3} - \frac{\pi}{6}

     = \frac{\pi}{6}.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    I hope this is right...

    \frac{4}{3}\int_0^\infty{\frac{1}{x^2 + 4}\,dx} - \frac{1}{3}\int_0^\infty{\frac{1}{x^2 + 1}\,dx} = \frac{4}{3}\left[\frac{1}{2}\arctan{\frac{x}{2}}\right]_0^\varepsilon - \frac{1}{3}\left[\arctan{x}\right]_0^\varepsilon

     = \frac{2}{3}[\lim_{\varepsilon \to \infty}(\arctan{\frac{\varepsilon}{2}}) - \arctan{0}] - \frac{1}{3}[\lim_{\varepsilon \to \infty}(\arctan{\varepsilon}) - \arctan{0}]

     = \frac{2}{3}\left[\frac{\pi}{2} - 0 \right] - \frac{1}{3}\left[\frac{\pi}{2} - 0 \right]

     = \frac{\pi}{3} - \frac{\pi}{6}

     = \frac{\pi}{6}.
    Looks good to me but on a personal note, I would use something other than  \varepsilon . For me,  \varepsilon is something very small not large
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  7. #7
    Member pberardi's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    I found this one on a Harvard graduate level exam:


    Any clues?
    Mr. Red,
    Would you mind revealing in what class you saw this problem? I am just curious which classes at the graduate level test students on their calculus abilities. Also, was this already set up or was it part of a word problem? Any more information on this would be very interesting.

    Thanks.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    I found this one on a Harvard graduate level exam:


    Any clues?
    Couldnt you use Countour integration on this? . I am not sure how but I am sure TPH or someone could.
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    Couldnt you use Countour integration on this? . I am not sure how but I am sure TPH or someone could.
    Yes you can but I think the first approach is the best one.

    Notice that, \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{x^4+5x^2+1} dx = \int_0^{\infty} \frac{x^2}{x^4+5x^2+1}dx

    Now define f(z) = \frac{z^2}{z^4+5z^2+1} and find its residues in the upper half plane.
    Read my tutorial on it, it should go smoothly for thee.
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  10. #10
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    Quote Originally Posted by Isomorphism View Post
    Do partial fraction decomposition as danny points out,



     \int_0^{\infty} \frac{4}{3} \frac{1}{x^2+4} \, dx - \int_0^{\infty}\frac{1}{3} \frac{1}{x^2+1}\, dx  = \left. \frac{2}{3} \tan^{-1}\frac{x}{2} - \frac{1}{3}\tan^{-1} x \right|_0^{\infty} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}
    Sorry, I don't know how to write it in proper format.
    Here is what I did:
    The function should have the following form:

    A/(x^2+1) + B/(X^2+4)

    then let A = 1 => solve for B:
    => B = -4/(x^2+1)
    substitute B back into the function
    => we have something like:

    1/(x^2+1) - 4/[(x^2+4)*(x^2+1)]

    The second integral need to split up again . My question is how do you know a proper way to decompose it (as you did)?

    Thank you.
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  11. #11
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    Note that \left( x^{2}+4 \right)-\left( x^{2}+1 \right)=3, hence, no partial fractions method involved.
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  12. #12
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    Quote Originally Posted by duydaniel View Post
    Sorry, I don't know how to write it in proper format.
    Here is what I did:
    The function should have the following form:

    A/(x^2+1) + B/(X^2+4)

    then let A = 1 => solve for B:
    => B = -4/(x^2+1)
    substitute B back into the function
    => we have something like:

    1/(x^2+1) - 4/[(x^2+4)*(x^2+1)]

    The second integral need to split up again . My question is how do you know a proper way to decompose it (as you did)?

    Thank you.
    That's "partial fractions" and you typically learn it in "Calculus II", a Freshman or Sophomore class (unless you learned calculus in secondary school). I see now that Krizalid did not need partial fractions because he was able to do the fractions easily- he's sharper than I am!
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  13. #13
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    Quote Originally Posted by duydaniel View Post
    Sorry, I don't know how to write it in proper format.
    Here is what I did:
    The function should have the following form:

    A/(x^2+1) + B/(X^2+4)

    then let A = 1 => solve for B:
    => B = -4/(x^2+1)
    substitute B back into the function
    => we have something like:

    1/(x^2+1) - 4/[(x^2+4)*(x^2+1)]

    The second integral need to split up again . My question is how do you know a proper way to decompose it (as you did)?

    Thank you.
     A \;\text{and}\; B are numbers, not variables! To decompose, say

     \frac{x^2}{x^4+5x^2+4}

    notice that

     \frac{x^2}{(x^2+1)(x^2+4)}

    so we seek numbers A, B, C and D such that

     \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+4} = \frac{x^2}{(x^2+1)(x^2+4)}

    cross multipying gives

     (Ax+B)(x^2+4) + (Cx+D)(x^2+1) = x^2.

    Expanding and re-grouping gives

    (A+C)x^3 + (B+D)x^2 +(4A+C)x + 4B+D = x^2

    comparing gives the equations

    A+C=0,\;\;\;B+D = 1,\;\;\;4A+C=0,\;\;\;4B+D = 0

    from which we solve giving

    A = 0,\;\;\;B = - \frac{1}{3},\;\;\;C=0,\;\;\;D=\frac{4}{3}

    Thus,

    \frac{- \frac{1}{3}}{x^2+1} + \frac{\frac{4}{3}}{x^2+4} = \frac{x^2}{(x^2+1)(x^2+4)}
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  14. #14
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    Quote Originally Posted by danny arrigo View Post
     A \;\text{and}\; B are numbers, not variables! To decompose, say

     \frac{x^2}{x^4+5x^2+4}

    notice that

     \frac{x^2}{(x^2+1)(x^2+4)}

    so we seek numbers A, B, C and D such that

     \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+4} = \frac{x^2}{(x^2+1)(x^2+4)}

    cross multipying gives

     (Ax+B)(x^2+4) + (Cx+D)(x^2+1) = x^2.

    Expanding and re-grouping gives

    (A+C)x^3 + (B+D)x^2 +(4A+C)x + 4B+D = x^2

    comparing gives the equations

    A+C=0,\;\;\;B+D = 1,\;\;\;4A+C=0,\;\;\;4B+D = 0

    from which we solve giving

    A = 0,\;\;\;B = - \frac{1}{3},\;\;\;C=0,\;\;\;D=\frac{4}{3}

    Thus,

    \frac{- \frac{1}{3}}{x^2+1} + \frac{\frac{4}{3}}{x^2+4} = \frac{x^2}{(x^2+1)(x^2+4)}
    That's great!!!
    I thought that the cross multiplying would ended up x^3 as in (Ax+B)(x^2+4) + (Cx+D)(x^2+1) that I try to find A and B instead... but it was probably a long way to go
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