# the gamma function being undefined

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?

• Welcome to MSE. First of all try to use MathJax to format your questions properly. Furthermore question of the form "Here is the task. Solve it for me!" are poorly received on this site. So try to provide some context and especially show what you have done so far with an edit :) – mrtaurho Sep 30 '18 at 14:33
• @mrtaurho i have completed part a and b but am unsure of how to even start c, that is why i am asking for some help so i know where to start – beth Sep 30 '18 at 14:38
• That is something you should maybe add within your question. So that other users exactly know how they could help you. For example if you already know how to show the property for $\Gamma(0)$ it would be priceless if someone would add this as an answer since it would not help you at all. Therefore just show what you have done, where you are struggling an so on :) – mrtaurho Sep 30 '18 at 14:41

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.