Proof that $Γ'(1) = -γ$?

Question

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?

Answer 1

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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Another Approach $$ \begin{align} \int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t &=\lim_{n\to\infty}\int_0^n\log(t)\,\left(1-\frac{t}{n}\right)^n\,\mathrm{d}t\tag{3a}\\ &=\lim_{n\to\infty}n\int_0^1(\log(t)+\log(n))\,(1-t)^n\,\mathrm{d}t\tag{3b}\\ &=\lim_{n\to\infty}\left(\frac{n}{n+1}\log(n)+n\int_0^1\log(1-t)\,t^n\,\mathrm{d}t\right)\tag{3c}\\ &=\lim_{n\to\infty}\left(\frac{n}{n+1}\log(n)-\frac{n}{n+1}H_{n+1}\right)\tag{3d}\\[6pt] &=-\gamma\tag{3e} \end{align} $$ Explanation:
$\text{(3a)}$: Monotone Convergence and $e^{-t}=\lim\limits_{n\to\infty}\left(1-\frac tn\right)^n$
$\text{(3b)}$: substitute $t\mapsto nt$
$\text{(3c)}$: substitute $t\mapsto1-t$
$\text{(3d)}$: integrate by parts $\int_0^1\log(1-t)\,t^n\,\mathrm{d}t=-\frac1{n+1}\int_0^1\frac{1-t^{n+1}}{1-t}\,\mathrm{d}t$
$\text{(3e)}$: $\gamma=\lim\limits_{n\to\infty}H_n-\log(n)$

Answer 2

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: $$\frac{\rm d}{\rm{d}x} \int_0^\infty e^{-t} t^{x-1}\, dt = \int_0^\infty e^{-t} t^{x-1}\log t\, dt$$

Answer 3

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

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