# Thread: Gamma Function logarithmic derivative

1. ## Gamma Function logarithmic derivative

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

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

2. Originally Posted by EinStone
Show that $\displaystyle -\frac{\Gamma'}{\Gamma}(z)=\gamma + \frac{1}{z} + \sum_{j=1}^\infty(\frac{1}{z+j}-\frac{1}{j})$

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

Now take $\displaystyle \log$'s of both sides.

$\displaystyle -\log\Gamma(s) = \log s +\gamma s+\sum_{n=1}^\infty \left(\log\left(1+\frac{s}{n}\right)-\frac{s}{n}\right)$

Now differentiate both sides.

$\displaystyle -\frac{\Gamma'(s)}{\Gamma(s)}=\frac{1}{s} + \gamma + \sum_{n=1}^\infty(\frac{1}{s+n}-\frac{1}{n})$

This process is called logarithmic differentiation.

3. Oh nice it was easy, somehow I havent heard that term before. Here yet another problem:

Show that, for real z > 0, we have $\displaystyle \frac{\Gamma'}{\Gamma}(z) = O(log|z|)$ as $\displaystyle z \to \infty$

4. Originally Posted by EinStone
Oh nice it was easy, somehow I havent heard that term before. Here yet another problem:

Show that, for real z > 0, we have $\displaystyle \frac{\Gamma'}{\Gamma}(z) = O(log|z|)$ as $\displaystyle z \to \infty$
$\displaystyle \frac{\Gamma'(z)}{\Gamma(z)} = O(1) + \sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{z+n}\right) = O\left(\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{\lfloor z\rfloor+n}\right)\right)$

$\displaystyle = O\left(\sum_{n=1}^{\lfloor z \rfloor} \frac{1}{n}\right)$ since the previous sum is telescoping.

Now it's a well known fact that for $\displaystyle x\in\mathbb{R}$ and $\displaystyle x>0$, $\displaystyle \log(x) = \sum_{n=1}^{\lfloor x \rfloor} \frac{1}{n}+O(1)$. (Ask if you'd like to see why.)

Thus $\displaystyle \frac{\Gamma'(z)}{\Gamma(z)} = O(\log(z))$ as $\displaystyle z\to \infty$. (I omitted the absolute value since $\displaystyle z > 0$.)

5. Ok thanks, could you say something about the log x thing? A reference is enough or just give the general idea, I dont need details.

6. Originally Posted by EinStone
Ok thanks, could you say something about the log x thing? A reference is enough or just give the general idea, I dont need details.
Perform LRAM and RRAM with step size $\displaystyle 1$ to approximate $\displaystyle \int_{1}^x \frac{1}{t}dt$, where $\displaystyle x\in\mathbb{N}$.
We know that $\displaystyle \int_{1}^x \frac{1}{t}dt = \log(x)$, but since $\displaystyle \frac{1}{t}$ is monotonically decreasing, RRAM $\displaystyle <\log(x)<$ LRAM.

Now RRAM $\displaystyle = \sum_{n=2}^x \frac{1}{n}$
LRAM $\displaystyle = \sum_{n=1}^{x-1} \frac{1}{n}$

Another way to prove this is by applying the Euler Maclaurin formula to $\displaystyle \sum_{n\leq x} \frac{1}{n}$, if you are familiar.