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  • alober

    [latex]I = \int_{0}^{m\pi}\ln(1+2r\cos x+r^2)dx = \int_{0}^{m\pi}(\ln(1+re^{ix})+\ln(1+re^{-ix}))dx = \int_{0}^{m\pi}(\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}r^ke^{kix}+\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}r^ke^{-kix})dx = \int_{0}^{m\pi}(\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}r^k2\cos kx)dx = \sum_{k=1}^{\infty}(\frac{(-1)^{k-1}}{k}r^k\int_{0}^{m\pi}\cos kxdx) = 0[/latex]

  • peterflyer

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  • ÇóÖª~óÆÖ¾

    ÒýÓûØÌû:
    2Â¥: Originally posted by alober at 2017-08-20 20:46:08
    I = \int_{0}^{m\pi}\ln(1+2r\cos x+r^2)dx = \int_{0}^{m\pi}(\ln(1+re^{ix})+\ln(1+re^{-ix}))dx = \int_{0}^{m\pi}(\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}r^ke^{kix}+\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k ...

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    ÒýÓûØÌû:
    2Â¥: Originally posted by alober at 2017-08-20 20:46:08
    I = \int_{0}^{m\pi}\ln(1+2r\cos x+r^2)dx = \int_{0}^{m\pi}(\ln(1+re^{ix})+\ln(1+re^{-ix}))dx = \int_{0}^{m\pi}(\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}r^ke^{kix}+\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k ...

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  • alober

    ÒýÓûØÌû:
    6Â¥: Originally posted by ÇóÖª~óÆÖ¾ at 2017-08-23 10:45:57
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  • ÇóÖª~óÆÖ¾

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    7Â¥: Originally posted by alober at 2017-08-23 10:53:54
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