Gamma Function logarithmic derivative

**EinStone** · March 23rd 2010, 03:03 PM

Show that $-\frac{\Gamma'}{\Gamma}(z)=\gamma + \frac{1}{z} + \sum_{j=1}^\infty(\frac{1}{z+j}-\frac{1}{j})$

Hint: use Weierstrass product formula $\frac{1}{\Gamma (s)}=se^{\gamma s} \prod_{n=1}^{\infty}(1 + \frac{s}{n})e^{-s/n}$

**chiph588@** · March 23rd 2010, 03:25 PM

Originally Posted by EinStone

Show that $-\frac{\Gamma'}{\Gamma}(z)=\gamma + \frac{1}{z} + \sum_{j=1}^\infty(\frac{1}{z+j}-\frac{1}{j})$

Hint: use Weierstrass product formula $\frac{1}{\Gamma (s)}=se^{\gamma s} \prod_{n=1}^{\infty}(1 + \frac{s}{n})e^{-s/n}$

Double post... tisk tisk tisk

**EinStone** · March 23rd 2010, 03:29 PM

I first thought posting didnt work in the other thread so I decided to change forum, but now its double and I didnt see how to delete, sry.

**chiph588@** · March 23rd 2010, 04:03 PM

Originally Posted by EinStone

I first thought posting didnt work in the other thread so I decided to change forum, but now its double and I didnt see how to delete, sry.

Haha, it's ok

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