Arc length of gamma function curve equals its value - coincidence?

Question

While perusing Wolfram Alpha, I stumbled upon a strange curiosity involving the Gamma function:

The arc length of $\Gamma(x + 1)$ from 1 to $x$ very closely equals the numeric value of $\Gamma(x)$.

Examples for $x = $...

1. (Not evaluated)
2. $\Gamma(2+1) = 2;~~~~~~$ Arc length of $\Gamma(x+1)$ from 1 to 2 = 1.442...
3. $\Gamma(3+1) = 6;~~~~~~$ Arc length of $\Gamma(x+1)$ from 1 to 3 = 5.5846...
4. $\Gamma(4+1) = 24;~~~~$ Arc length of $\Gamma(x+1)$ from 1 to 4 = 23.6185...
5. $\Gamma(5+1) = 120;~~$ Arc length of $\Gamma(x+1)$ from 1 to 5 = 119.625...
6. $\Gamma(6+1) = 720;~~$ Arc length of $\Gamma(x+1)$ from 1 to 6 = 719.626...

Question

Is this an unusual or seldom-discussed-but-fascinating mathematical coincidence?

Or, am I missing some obvious numerical relationship staring me right in my eyes? ;-)

P.S. I am aware of how the gamma function extends factorial to $\mathbb{R}$, but with the $+1$ offset.

Answer 1

Your arc length can be written as $$L=\int_1^x \sqrt{\Gamma'^2(t)+1}\,dt$$ where $\Gamma'^2(t)$ is the derivative of $\Gamma(t)$. When $x$ is large enough, $\Gamma'^2(t)\gg 1$, the integral is dominated by the large values of $t$, then $$L\sim \int_1^x\Gamma'(t)\,dt\sim \Gamma(x)-1\sim \Gamma(x)$$

Answer 2

This is a trivial property.

For any function with a large derivative,

$$\int\sqrt{1+f'^2(x)}\,dx\approx\int f'(x)\,dx=f(x)+C.$$

For example with the exponential $e^x$,

$$\int\sqrt{1+e^{2x}}\,dx=\sqrt{1+e^{2x}}+\text{arcoth}\sqrt{1+e^{2x}}+C$$

which is asymptotic to $e^x$.

Intuitively, this is because the horizontal displacement brings a neglectable contribution when the curve gets close to vertical.