So my book proves the convergence of $\Gamma(z) = \int_0^{\infty}t^{z-1}e^{-t}dt$ in the right half plane $Re(z) > 0$, and then goes on to prove the initial recurrence relation $\Gamma(z+1)=z\Gamma(z)$ by applying integration by parts to $\Gamma(z+1)$:

$$\int_0^{\infty}t^{z}e^{-t}dt = -t^ze^{-t}|_0^{\infty} + z\int_0^{\infty}t^{z-1}e^{-t}dt$$

The book explicitly states this equality to be true only in the right half plane, since otherwise $-t^ze^{-t}|_0^{\infty} = \infty$, instead of equaling zero. With this initial recurrence relation we are 'supposably' able to analytically continue the Gamma function to $Re(z) > -1$ (not including the origin) by writing the relation in the form:

$$\Gamma(z) = \frac{\Gamma(z+1)}{z}$$

What I don't understand is this relation is still only true in the right half plane, since otherwise $-t^ze^{-t}|_0^{\infty}\neq 0$. I don't see what reason we have to believe that, for instance, $\Gamma(-\frac{1}{2}) = \frac{\Gamma(\frac{1}{2})}{-\frac{1}{2}}$.

Furthermore $\int_0^{\infty}t^{z-1}e^{-t}dt$ is clearly not convergent in the left half plane, so I can't even imagine why it would be plausible to think that a recurrence relation directly based on it could possibly lead to a genuine analytic continuation of its domain.

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    $\begingroup$ One lateral application of this analytic cont. formula is that we get $$\lim_{z\to 0}z\Gamma(z)=\lim_{z\to 0}\Gamma(z+1)=\Gamma(1)=1$$which gives the residue of $\,\Gamma(z)\,$ at $\,z=0\,$ . Appying this to all the negatative integers you find quite a curious and nice formula for the function's residues there... $\endgroup$ – DonAntonio Jun 17 '12 at 10:42

As you have said, $\displaystyle \int_0^{\infty}t^{z-1}e^{-t}dt$ does not even make sense on the left half plane.

However, if we define $\Gamma(z) = \dfrac{\Gamma(z+1)}{z}$ on the left half plane, (and this is not equal to the integral above on the left half plane), this is consistent with the fact that $\Gamma(z) = \displaystyle \int_0^{\infty}t^{z-1}e^{-t}dt$ on the right half plane and also provides us a possible extension on the left half plane except at negative integers.

Further, this has to be the only analytic continuation since $\Gamma(z) = \dfrac{\Gamma(z+1)}{z}$ is analytic everywhere except at negative integers and matches with the integral on the right half plane and hence by uniqueness of analytic continuation this is the only possible extension.


To see that the extension, is in fact analytic except at negative integers, you proceed strip by strip. On the right half plane, you have $\Gamma(z)$ defined by the integral, $\displaystyle \int_0^{\infty} t^{z-1} \exp(-t) dt$. You can verify that the integral satisfies the functional equation and is in fact analytic.

We will now extend the function to the strip corresponding to $(-1,0)$ i.e. the strip $\text{Re}(z) \in (-1,0)$ based on the functional equation.

Define $\Gamma(z) = \dfrac{\Gamma(z+1)}{z}$ on the strip $\text{Re}(z) \in (-1,0)$.

First note that $\Gamma(z+1)$ is analytic on the strip $\text{Re}(z) \in (-1,0)$ since $z+1 \in$ right half plane. Also, $\dfrac1z$ is analytic on the strip $\text{Re}(z) \in (-1,0)$. Hence, $\Gamma(z) = \dfrac{\Gamma(z+1)}{z}$ is analytic on the strip $\text{Re}(z) \in (-1,0)$. Hence, now we have extended the definition of the $\Gamma$ function to the region $\text{Re}(z) > -1$.

Now repeat the same argument for the strip $\text{Re}(z) \in (-2,-1)$, since we now have that the $\Gamma$-function is analytic on the region $\text{Re}(z) > -1$. And so on...

  • $\begingroup$ @zyx There is at-most one analytic extension of $f(z) = \int_0^{\infty} t^{z-1} e^{-t} dt$. $\endgroup$ – user17762 Jun 17 '12 at 5:31
  • $\begingroup$ To be slightly more precise, you first define $\Gamma(z) = \Gamma(z+1)/z$ on $\{z: \text{Re}(z) > -1\} \backslash \{0\}$, observing that this agrees with the previous definition on the right half plane. Then you define it in $\{z: \text{Re}(z) > -2\} \backslash \{-1,0\}$ ... $\endgroup$ – Robert Israel Jun 17 '12 at 5:53
  • $\begingroup$ @RobertIsrael Yes. Thanks. I have written that as a comment to Gerry's answer. (In the comment, I need to change the interval from $(-1,0)$ to the corresponding strip.) I will add this to my answer soon. $\endgroup$ – user17762 Jun 17 '12 at 5:56
  • $\begingroup$ Sorry, I thought your third paragraph was asserting unique analytic continuation of the factorial, not of the integral used to define gamma on the right half plane. $\endgroup$ – zyx Jun 17 '12 at 6:53

The recurrence is used to define Gamma in places where the integral doesn't work. What's needed is to prove that a function defined that way is analytic; then it follows that it's the analytic continuation of Gamma.

  • $\begingroup$ To prove that it's analytic for complex numbers with a negative real part.. But how? and I still don't see why anyone would think that it would be, a priori, that is. $\endgroup$ – Thoth Jun 17 '12 at 5:16
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    $\begingroup$ @NollieTré You go interval by interval to see it is analytic except on negative integers. In the interval $(-1,0)$, do you see that $\Gamma(z) = \dfrac{\Gamma(z+1)}{z}$ is analytic? If so, then proceed next to interval $(-2,-1)$. And so on. $\endgroup$ – user17762 Jun 17 '12 at 5:18
  • $\begingroup$ Ohhhh ok I see it now! thanks =] $\endgroup$ – Thoth Jun 17 '12 at 5:31

An excellent introduction to this topic can be found in the book The Gamma Function by James Bonnar. An entire chapter is devoted to analytic continuation of the factorials, as well as why the Gamma function is defined as it is -- Hölder's theorem and the Bohr-Mullerup theorem are discussed.

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    $\begingroup$ Here you can get a free version from 2017 updated: http://vixra.org/abs/1702.0305 - Ameyah $\endgroup$ – suomynonA Mar 25 '17 at 17:48

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