I have an existential doubt regarding the good definition of the Gamma function $\Gamma(x)=\displaystyle\int_0^{+\infty}t^{x-1}e^{-t}dt$, this refers to the convergence of the following integral, I have thought of using the limit comparison criterion with $g(t)=\dfrac{1}{t^2}$ that is $$ \lim_{t\to +\infty} \dfrac{t^{x-1}e^{-t}}{\frac{1}{t^2}}=\lim_{t\to +\infty} \dfrac{t^{x+1}}{e^t}=0. $$ This limit does not depend on the sign that $x$ takes or does it? That way I check that $\Gamma(x)$ with $x\in\mathbb R$ is well defined. I know that it is not defined for negative integers but I don't know where the error is, even so after finding my error how can I prove that $\Gamma(x)$ is not defined for negative integers?
Ok, let me write a slightly more detailed answer. First of all, as noted by @user58697, you correctly deduce that the integral does not diverge at infinity. This criterion tells you nothing of divergence for other reasons (poles at finite points). To see that the integral has poles at all negative integers, divide it into two parts: $$\int_{0}^{\infty}t^{x-1}e^{-t}dt=\left(\int_{0}^{1}+\int_{1}^{\infty}\right) t^{x-1}e^{-t}dt$$ The second integral converges for any value of $x$, and defines, in fact, an entire function. Expand the exponent in the first integral into Taylor series: $$\int_{0}^{1}t^{x-1} \sum_{k=0}^{\infty} \frac{(-1)^kt^k}{k!} dt+\int_{1}^{\infty}t^{x-1}e^{ -t }dt=\sum_{k=0}^{\infty} \frac{(-1)^k}{k!}\int_{0}^{1} t^{x-1+k} dt +\int_{1}^{\infty}t^{x-1}e^{ -t }dt=$$ $$=\sum_{k=0}^{\infty} \frac{(-1)^k }{k! (x+k)}+\int_{1}^{\infty}t^{x-1}e^{ -t }dt$$
You've obtained an expression that has a pole at every negative integer.