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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?

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You can refer to this posting. –  sos440 Mar 31 '12 at 23:27
    
Thanks for the link! –  Argon Mar 31 '12 at 23:31
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3 Answers

up vote 5 down vote accepted

Taken from this answer:

By the recursive relation $\Gamma(x+1)=x\Gamma(x)$, we get $$ \small{\log(\Gamma(x))=\log(\Gamma(n+x))-\log(x)-\log(x+1)-\log(x+2)-\dots-\log(x+n-1)}\tag{1} $$ Differentiating $(1)$ with respect to $x$, evaluating at $x=1$, and letting $n\to\infty$ yields $$ \begin{align} \frac{\Gamma'(1)}{\Gamma(1)}&=\log(n)+O\left(\frac1n\right)-\frac11-\frac12-\frac13-\dots-\frac1n\\ &\to-\gamma\tag{2} \end{align} $$ A discussion of why $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))=\log(x)+O\left(\frac1x\right)$, and a proof of the log-convexity of $\Gamma(x)$, is given in the answer cited above. It is easy to accept since $\log(\Gamma(n+1))=\log(\Gamma(n))+\log(n)$.

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Thanks for the answer! –  Argon Apr 4 '12 at 1:46
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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

$$\frac{\Gamma '(1)}{\Gamma(1)}=\Gamma'(1)=-\gamma-1+1=-\gamma $$

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Hint:

  1. Differentiation under the integral sign $$\frac{\rm d}{\rm{d}x} \int f(x,t)\, dt = \int \frac{\partial}{\partial{x}} f(x,t)\, dt$$

  2. Derivatives of exponential and logarithmic functions $$\frac{\partial}{\partial{x}} t^x = t^{x} \log t$$

  3. You should be able to work out: $$\frac{\rm d}{\rm{d}x} \int_0^\infty e^{-t} t^{x-1}\, dt = \int_0^\infty e^{-t} t^{x-1}\log t\, dt$$

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