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Thread: Proof of derivative of $\Gamma'(1)=-\gamma$

  1. #1
    Senior Member Vinod's Avatar
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    Proof of derivative of $\Gamma'(1)=-\gamma$

    Hello,

    Look at this page https://brilliant.org/discussions/th...amma-function/. On this page, in solution to problem number 2, $\displaystyle\int_0^1 s^n ln(1-s)ds =\frac{-1}{n+1}\displaystyle\int_0^1 \frac{s^{n+1}-1}{s-1}ds$ is computed.

    I have computed $\displaystyle\int_0^1 s^n ln(1-s)ds =\frac{1}{n+1}\displaystyle\int_0^1 \frac{s^{n+1}}{s-1}ds$ where i am wrong? I used integration by parts technique where $u=\ln{(1-s)}$ and $v=\frac{s^{n+1}}{n+1}$ So uv is 0 and the other term is $\frac{1}{n+1}\displaystyle\int_0^1 \frac{s^{n+1}}{s-1}ds$
    Last edited by Vinod; Mar 2nd 2019 at 04:29 AM.
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  2. #2
    Senior Member Vinod's Avatar
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    Re: Proof of derivative of $\Gamma'(1)=-\gamma$

    Quote Originally Posted by Vinod View Post
    Hello,

    Look at this page https://brilliant.org/discussions/th...amma-function/. On this page, in solution to problem number 2, $\displaystyle\int_0^1 s^n ln(1-s)ds =\frac{-1}{n+1}\displaystyle\int_0^1 \frac{s^{n+1}-1}{s-1}ds$ is computed.

    I have computed $\displaystyle\int_0^1 s^n ln(1-s)ds =\frac{1}{n+1}\displaystyle\int_0^1 \frac{s^{n+1}}{s-1}ds$ where i am wrong? I used integration by parts technique where $u=\ln{(1-s)}$ and $v=\frac{s^{n+1}}{n+1}$ So uv is 0 and the other term is $\frac{1}{n+1}\displaystyle\int_0^1 \frac{s^{n+1}}{s-1}ds$
    I solved my query.
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