Is it possible to evaluate $\pi!$ using the pi fuction without using any approximations? I'm trying to calculate it using the formula but I keep running through the problem that it gives a circular response:
$$x!=\Pi(x)=\int_0^\infty t^x e^{-t}dt$$ $$\pi!=\int_0^\infty t^\pi e^{-t}dt$$ $$\pi!={-t^\pi e^{-t}}\vert_0^\infty+\pi\int_0^\infty t^{\pi-1} e^{-t}dt$$ $$\pi!={-t^\pi e^{-t}}\vert_0^\infty+\pi\Pi(\pi-1)$$
But by defition, $x\Pi(x-1)=\Pi(x)=x!$, so:
$$\pi!={-t^\pi e^{-t}}\vert_0^\infty+\pi!$$ ${-t^\pi e^{-t}}\vert_0^\infty$ converges to $0$, so:
$$\pi!=\pi!$$
I'm doing something wrong?
You are mistaken when you assert that $\Pi(x)=x\Pi(x-1)$ by definition. This is actually a property of the $\Pi$ function. What you did was to prove that this holds when $x=\pi$.