Why are the limits of integration of $\Gamma (z)$ the way they are?

Question

$$\Gamma (z)=\int_0^\infty t^{z-1}e^{-t} dt$$ To make things easier, let's call this $G(z)$. If we take the derivative with respect to $t$ of both sides, we get $\frac {dG(z)}{dt}=\frac {d}{dt} [\int_0^\infty t^{z-1}e^{-t} dt]=0$ . The reason why I think this equals zero is because if we are just looking at $\int_0^\infty t^{z-1}e^{-t} dt$ , $z$ is fixed and this definite integral is equal to a constant (if it's convergent). If $\frac {dG(z)}{dt}=0$ , this implies that $G(z)=C\cdot t^0$ , so it looks like $G(z)$ is absolutely independent of the value of $t$ . Why then does $G$ have those specific limits of integration? Those limits, only affect $t$ , which $G$ is independent of. And I know that this is illogical because by my argument the limits of integration could even be equal to each other, which would obviously lead to a false solution. Thanks.

My background just Calculus $1$ so please don't use infinite series and such.

If you are interested in my motivation, the original motivation for this question is that I wanted to see if it is possible to have an elementary function $f(x)$ that that grows faster than $x!$ (please don't tell me the answer to this or give any hints, I wanna see if I can do this by myself). The motivation of THAT question was to see if it is even possible to have an elementary function of a finite number of terms that models $x!$ (again, please don't tell me the answer to that either.)

Answer 1

$\frac{d}{dt} G(z)=0$ because, as a function of $t$, it is a constant. If $F(z)=z$, then $\frac{d}{dt}F(z)=0$, too. There is no free variable $t$ in the definition of $\Gamma$.

It is sometimes confusing notation to the uninitiated, but the $t$ inside the integral is really a placeholder, indicating what is changing - it only exists inside the integral.

Now, the step you are doing is a bit confusing, because you can't bring $d/dt$ from outside the interval, where $t$ is "unbound" and move it inside the integral, where $t$ is "bound." It simply can't be done, and there is no reason to think you can.

If you don't believe me, try it with other functions. It simply won't work. For example, $F(z)=\int_0^1 t^z dt = \frac{1}{1+z}$. Then, by your reason, $\frac{d}{dt} F = \int_0^1 zt^{z-1} dt = 1$. That doesn't make much sense, does it?

Answer 2

I think your real question got lost in here with the $d/dt$ conundrum. Integrate by parts and see what happens. And note that $\Gamma(1)$ has a very nice value because of the $0$ and $\infty$ limits of integration. Another reasonable question you might have asked is what $x!$ should mean when $x$ isn't a positive integer.