4

$$\Gamma(z) = \int_0^{\to +\infty}t^{z-1}e^{-t} \, \mathrm dt$$

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 $t = 0$?

GFauxPas
  • 5,047

3 Answers3

4

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.

Mark Viola
  • 179,405
4

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.

Olivier Oloa
  • 120,989
2

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 $\Gamma(0)\to\infty$

Simply Beautiful Art
  • 74,685