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Thread: integration by parts

  1. #1
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    integration by parts

    I need to prove:

    I(subscript 2k)=integral[-1,1](x^2k)*(cos(pi*x))dx

    is equal to

    I(subscript 2k)=-(4k/pi^2)-((2k)(2k-1)/(pi^2))*I(subscript 2k-2)

    but i keep getting

    I(subscript 2k)=-(4k/pi^2)+((2k)(2k-1)/(pi^2))*I(subscript 2k-2)

    can anyone help?
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \displaystyle\int_{-1}^1x^{2k}\cos\pi xdx=\int_{-1}^1x^{2k}\left(\frac{\sin\pi x}{\pi}\right)'dx=$
    $\displaystyle \displaystyle =\left.\frac{1}{\pi}x^{2k}\sin\pi x\right|_{-1}^1-\frac{2k}{\pi}\int_{-1}^1x^{2k-1}\sin\pi xdx=$
    $\displaystyle \displaystyle =\frac{2k}{\pi}\int_{-1}^1x^{2k-1}\left(\frac{\cos\pi x}{\pi}\right)'dx=$
    $\displaystyle \displaystyle =\left. \frac{2k}{\pi^2}x^{2k-1}\cos\pi x\right|_{-1}^1-\frac{2k(2k-1)}{\pi^2}I_{2k-2}=-\frac{4k}{\pi^2}-\frac{2k(2k-1)}{\pi^2}I_{2k-2}$

    Edited.
    Last edited by red_dog; Aug 24th 2007 at 10:45 AM.
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  3. #3
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    Thanks,

    but this is not the solution that I need to prove,

    I need to prove the first version of the proof,

    but I keep getting the second???
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ubhik View Post
    I need to prove:

    I(subscript 2k)=integral[-1,1](x^2k)*(cos(pi*x))dx

    is equal to

    I(subscript 2k)=-(4k/pi^2)-((2k)(2k-1)/(pi^2))*I(subscript 2k-2)

    but i keep getting

    I(subscript 2k)=-(4k/pi^2)+((2k)(2k-1)/(pi^2))*I(subscript 2k-2)

    can anyone help?
    Here, if you have any questions please say so. I think you just mixed up the signs somewhere when you had to multiply the second integral by $\displaystyle - \frac {2k}{\pi}$


    $\displaystyle I_{2k} = \int_{-1}^{1} x^{2k} \cos ( \pi x)~dx$

    Using integration by parts with $\displaystyle u = x^{2k}$ and $\displaystyle dv = \cos ( \pi x)$, we get:

    $\displaystyle I_{2k} = \left. \frac {x^{2k}}{\pi} \sin ( \pi x) \right|_{-1}^{1} - \frac {2k}{\pi} \int_{-1}^{1} x^{2k - 1} \sin ( \pi x )~dx$

    $\displaystyle = - \frac {2k}{\pi} \int_{-1}^{1} x^{2k - 1} \sin ( \pi x )~dx$

    Again, using by parts with $\displaystyle u = x^{2k - 1}$ and $\displaystyle dv = \sin( \pi x)$, we get:

    $\displaystyle I_{2k} = - \frac {2k}{\pi} \left[ \left. - \frac {x^{2k - 1}}{\pi} \cos (\pi x) \right|_{-1}^{1} + \frac {2k - 1}{\pi} { \color {red} \int_{-1}^{1} x^{2k - 2} \cos (\pi x )~dx} \right]$ .........Note, the red is $\displaystyle I_{2k - 2}$

    $\displaystyle = \left. \frac {2kx^{2k - 1}}{\pi^2} \cos (\pi x) \right|_{-1}^{1} - \frac {2k(2k - 1)}{\pi^2} I_{2k - 2}$

    $\displaystyle = \frac {-4k}{\pi^2} - \frac {2k(2k - 1)}{\pi^2} I_{2k - 2}$

    as desired
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  5. #5
    MHF Contributor red_dog's Avatar
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    I was also wrong!
    I edited. See again my previous post.
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  6. #6
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    Thank you, thats brilliant!
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