Is there an actual expansion of the Gamma function's integral?

Question

$$\int_0^{\infty} x^{t-1} e^{-x} \, \mathrm{d}x = (t-1)! = \Gamma (t)$$ Is the expression $(t-1)!$ the actual result of integrating the gamma integral? Meaning, if you were to compute the integral through the standard integral computation techniques, would you end up with $(t-1)(t-2)(t-3)...$? If so, how do we integrate the following?: $$\int t! \, \mathrm{d}t = \int \int_0^{\infty} x^{t} e^{-x} \, \mathrm{d}x\, \mathrm{d}t$$

Answer 1

You should be able to obtain the recursion formula $\Gamma(t+1)=t\Gamma(t)$, i.e. $$ \int_0^\infty x^te^{-x}\,\mathrm dx = t\cdot \int_0^\infty x^{t-1}e^{-x}\,\mathrm dx.$$ Then together with an explicit computatoin of $\Gamma(1)$, you obtain the result that $\Gamma(t)0(t-19!$ for $t\in\mathbb N$.

The integral $\int t!\,\mathrm dt$ doesn't make much sense unless you write it as $\int \Gamma(t+1)\,\mathrm dt$ and I am not aware of any nice result / closed formula for that (or a reason for there existing one)