Gamma Function Integral with discontinuity $$\Gamma(z) = \int_0^{\to +\infty}t^{z-1}e^{-t} \, \mathrm dt$$
For $\operatorname{Re}\left({z}\right) > 0$, and analytic continuation elsewhere, except for non positive integers.
But then, for example, 
$$\Gamma\left({\frac 1 2}\right) = \int_0^{\to +\infty}t^{-1/2}e^{-t} \, \mathrm dt$$
This integrand isn't defined at $t = 0$, so what kind of integral is this? And why does the above integral make more sense than:
$$\Gamma\left({0}\right) \large{\stackrel{\text{don't}}{\normalsize=}}\normalsize \int_0^{\to +\infty}t^{-1}e^{-t} \, \mathrm dt$$
which has a similar problem at $t = 0$?
 A: 
The function $f(x)=\frac{1}{x^a}$ is Lebesgue Integrable on $[0,1]$ for all $a<1$ and converges as an Improper Riemann Integral on $[0,1]$ for all $a<1$. 

To evaluate this integral, we can write for $a<1$
$$\begin{align}
\int_0^1 f(x)\,dx&=\lim_{\epsilon \to 0^+}\int_{\epsilon}^1\frac{1}{x^a}\,dx\\\\
&=\lim_{\epsilon \to 0}\left.\left(\frac{1}{(1-a)x^{a-1}}\right)\right|_{\epsilon}^{1}\\\\
&=\frac{1}{1-a}\lim_{\epsilon\to 0}\left(1-\frac{1}{\epsilon^{a-1}}\right)\\\\
&=\frac{1}{1-a}
\end{align}$$
Note that for $a=1$, the integral diverges logarithmically.  
A: There is a difference between the two integrands, we have, as $a \to 0^+$,
$$
\int_a^1t^{-1/2}\: \mathrm dt=\left[2\sqrt{t}\right]_a^1=2-2\sqrt{a}\to 2<\infty
$$ whereas, as $a \to 0^+$,
$$
\int_a^1t^{-1}\: \mathrm dt=\left[\ln |t|\right]_a^1\to \infty.
$$
These are improper integrals.
A: Notice that we can define the first integral as
$$\Gamma(1/2)=\lim_{a\to0}\int_a^\infty t^{-1/2}e^{-t}dt=\sqrt\pi$$
On the other hand,
$$\Gamma(0)=\lim_{a\to0}\int_a^\infty t^{-1}e^{-t}dt\to\infty$$
As you might expect, since we have
$$\Gamma(n)=\frac{\Gamma(n+1)}n$$
And having $n=0$ gives $\Gamma(0)\to\infty$
