TITLE: Proof that $Γ'(1) = -γ$?
QUESTION [6 upvotes]: I know that $Γ'(1) = -γ$, but how does one prove this?
Starting from the basics, we have that:
$$Γ(x) = \int_0^\infty  e^{-t} t^{x-1} dt$$
How do we differentiate this?  How do we then find that
$$Γ'(1) = \int_0^\infty  e^{-t} \log(t)  dt$$
and how would one solve this integral?

REPLY [4 votes]: I think you can also use the following:
$\Gamma$ can also be expressed as
$$\Gamma(z) = \frac{\exp{(-\gamma z)}}{z}\prod\limits_{n=1}^\infty\frac{\exp \left({\frac z n}\right)}{1+\dfrac z n }$$
So that
$$\log \Gamma(z)=-\gamma z-\log z+\sum\limits_{n=1}^\infty\frac{z}{n}-
\sum\limits_{n=1}^\infty\log \left(1+\frac z n \right)$$
Now I'm not going to address convergence now (which you can look up in Landau's Calculus, on the chapter dedicated to the Gamma function), but differentiating gives
$$\frac{\Gamma '(z)}{\Gamma(z)}=-\gamma-\frac 1 z+\sum\limits_{n=1}^\infty \left(\frac 1 n-\frac{1}{n+z}\right) $$
$$\frac{\Gamma '(z)}{\Gamma(z)}=-\gamma-\frac 1 z+\sum\limits_{n=1}^\infty\frac{z}{n(n+z)} $$ 
Letting $z=1$ gives:
$$\frac{\Gamma '(1)}{\Gamma(1)}=\Gamma'(1)=-\gamma-1 +\sum\limits_{n=1}^\infty\frac{1}{n(n+1)} $$ 
But we know that
$$\sum\limits_{n=1}^\infty\frac{1}{n(n+1)}=1$$
so that
$$\frac{\Gamma '(1)}{\Gamma(1)}=\Gamma'(1)=-\gamma-1+1=-\gamma $$