Could someone help me with part C of this question, relating to the Gamma Function. I am aware it is something to do with convergence but I am not sure how to show this
(a) Use integration by parts to show that $Γ(x + 1) = xΓ(x)$.
(b) Hence prove that $Γ(x + 1) = x!$ when $x$ is a positive integer.
(c) The property found in (a) can be used to extend the definition of $Γ(x)$ to some negative values of $x$. Show that a value for $Γ(0)$ cannot be defined in this way. Are there other values of $x$ for which $Γ(x)$ cannot be defined in this way?
We know the recurrence relation of the Gamma Function $\Gamma(x+1)=x\Gamma(x)$. This is equivalent to $\frac{\Gamma(x+1)}{x}=\Gamma(x)$. Plugging in $x=0$ to get a value for $\Gamma(0)$ leads to $\frac{\Gamma(1)}{0}=\Gamma(0)\Leftrightarrow 1=0$ which is clearly a contradiction.
Applying the recurrence equation to a $x$ of $\mathbb{Z}^-$ for example $-2$ yields to
$$\Gamma(-2)=(-2)\Gamma(-1)=(-2)(-1)\Gamma(0)$$
and so again we have to face the problematic $\Gamma(0)$. For any number $x\ne\mathbb{Z}^-$ one can apply the recurrence relation without any doubts.
Another way to think about the last assumption can be done by using the integral representation of the Gamma Function
$$\Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t}dt$$
which becomes
$$\Gamma(-x)=\int_0^{\infty}\frac{e^{-t}}{t^{x+1}}dt$$
for any $x\in\mathbb{Z}^-$ and since the integrand has a pole at $t=0$ the integral does not converges at all.