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

1. ## 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 $\frac{1}{n+1}\displaystyle\int_0^1 \frac{s^{n+1}}{s-1}ds$

2. ## Re: Proof of derivative of $\Gamma'(1)=-\gamma$

Originally Posted by Vinod
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.