The gamma function is the integral of $x^{z-1}e^{-x}$
If you integrate by parts you get two terms.
The first one is $-x^{z-1}e^{-x}$ and this is bound by infinity and zero.
If you plug in infinity, it is equal to zero, and when you plug in zero, it is equal to zero, and $0-0$ is zero, so we can remove it.
But if you plot the graph, or integrate it, you can see the area between the curve and the $x$ axis is not zero.
Are you supposed to be visualising the "$f(x)g(x)$" part of integration by parts $(f(x)g'(x) = f(x)g(x) - f'(x)g(x))$ as an area or not ?
Is there a geometric explanation of this ?
You can write the integration by parts rule as $$\int \limits_a^b [f(x) g'(x) + f'(x) g(x)] \, \mathrm{d} x = \left[f(x) g(x)\right]_{x=a}^{x=b}$$ to see that the term on the right-hand side can indeed be understood as an area (a signed area, to be precise). However, it is not the area under the graph of $f g'$ ! Instead, as can be seen from the equation, it is the area between the graph of $f g' + f' g = (fg)'$ and the $x$-axis .
For the gamma function integral we have $f(x) = x^{z-1}$ and $g(x) = - \mathrm{e}^{-x}$ . Since the term on the right-hand side vanishes in this case, we conclude that the signed area beneath the graph of the function $$x \mapsto x^{z-1} \mathrm{e}^{-x} - (z-1) x^{z-2} \mathrm{e}^{-x} = [x - (z-1)] x^{z-2} \mathrm{e}^{-x}$$ on the interval $(0,\infty)$ is equal to zero (if $z>1$ of course).
The plot below shows this function for $z=3$ . The red area below and the blue area above the axis have (almost) the same size, so their contributions cancel.