The complex gamma function

Question

Show that $$\Gamma (z+1)=z\Gamma (z)$$ $\forall z\in \Bbb C$ except for $z=-n$ where $n\in \Bbb N$.

---

I know that the gamma function is defined as $\Gamma (z)=\int_{0}^{\infty}e^{-t}t^{z-1}dt$ And I need to extend $\Gamma(z)$ to all of $\Bbb C-\{singular points\} $

Answer 1

You already have a good definition for the gamma function for $\Re z > 0$. The gamma function has a fairly natural extension by transforming your integral definition into one over a contour in the complex plane. To do this, define $h(w)=w^{z-1}$ to be the complex function with a branch cut along the positive real axis. This can be written as $$ h(w) = e^{\log(w)(z-1)} $$ where $\log$ its branch cut along the positive real axis. Then $h(w)$ has a jump across the real axis given by $$ h(x+2\pi i)-h(x) = (e^{2\pi i z}-1)h(x),\;\;\; 0 \le x < \infty. $$ Letting $C$ be the complex contour starting at $\infty + 2\pi i$, circling clockwise around the origin and ending at $\infty + i0$, the Gamma function becomes $$ \begin{align} \Gamma(z) & = \frac{1}{e^{2\pi i z}-1}\int_{C}e^{-wz}w^{-1+z}dw \\ & = \frac{e^{-i\pi z}}{2i\sin(\pi z)}\int_{C}e^{-wz}w^{-1+z}dw. \end{align} $$ The contour $C$ can be distorted to avoid the origin, without altering the value of the gamma function for $\Re z > 0$. Once this is done, the above integral extends naturally to $\Re z \le 0$ because nothing keeps the integral from converging. One sees, however, that the $\sin$ function prevents the extension of $\Gamma$ to $z=0,-1,-2,-3,\ldots$. For positive integers $\sin \pi z$ vanishes, but so does the contour integral, which is why the definition remains valid there after removing the singularity.

Now it is apparent that $\Gamma$ has a holomorphic extension to $\mathbb{C}\setminus\{0,-1,-2,\ldots\}$. The identity $z\Gamma(z)=\Gamma(z+1)$ must hold everywhere the identity makes sense, because of the Identity Theorem of Complex Analysis. Or you can verify the extended identity directly from the new integral definition.

At first glance, it is a strange fact that $$ \Gamma(z)\Gamma(1-z)\sin\pi z $$ extends to an entire function of $z$. In fact, this function is a constant: $$ \Gamma(z)\Gamma(1-z)\sin \pi z = \pi. $$ So, the sin function factors into products of two reciprocal gamma functions: $$ \sin \pi z = \pi\frac{1}{\Gamma(z)}\frac{1}{\Gamma(1-z)}. $$ Using this identity, one can obtain Hankel's classical contour integral for the reciprocal gamma function: $$ \frac{1}{\Gamma(z)} = \frac{1}{2\pi i}\int_{C} e^{-wz} (-w)^{-z}\,dw. $$ Hankel's formula for the reciprocal gamma function is great because $1/\Gamma(z)$ extends to an entire function of $z$, and has a concise complex contour integral definition. You do have to juggle the contour around a bit to get the definition to hold everywhere, but it's not that tricky.

Answer 2

Integration by parts yields $$ \Gamma(z+1)=\int_0^\infty \mathrm{e}^{-t}t^{z}dt=\left.-\mathrm{e}^{-t}t^{z}\right|_0^\infty+ \int_0^\infty \mathrm{e}^{-t}z\,t^{z-1}dt=z\Gamma(z). $$ We used that $\dfrac{d}{dt} t^z=z\,t^{z-1}$, for $t>0$ and $z\in\mathbb C$, which follows from the fact that $t^z=\exp(z\ln t)$. All the above not work if $z=r\le -1$, as $\mathrm{e}^{-t}/t^r$ is not integrable, for such $r$. Still, $\Gamma(z)$ extends analytically in $\mathbb C$ minus the negative integers.

Answer 3

Hints:

For Re$\,(z)>0\;$ and integrating by parts:

$$u:=t^z\;\;,\;\;u'=zt^{z-1}\\v'=e^{-t}\;\;v=-e^{-t}$$

$$\Gamma(z+1)=\left.-t^ze^{-t}\right|_0^\infty+z\Gamma(z)=z\Gamma(z)$$

Now, for $\;-1<\,$Re$\,(z)<0\;$, take

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

Note that $\;\Gamma(z+1)\;$ is defined in the above and now you see what the problem with $\;z=0\;$, and later with the negative inters, is going to be...fill up.

Answer 4

The Gamma function is defined by $\Gamma(z) = \int_0^{\infty} t^{z-1}e^{-t} \,dt \text{ when } \Re{z} > 0$. The first step is to see that it is analytic when $\Re{z}>0$ because the integral converges uniformly on each compact subset of the right hand half plane. (Why?)

Now when $n$ is equal to 0 or a positive integer then $n! = \Gamma(n+1)$. We need to show that this is true for all complex numbers with positive real part. Consider the integral

$ \int_0^R t^{z-1}e^{-t} \,dt$.

If we integrate by parts, differentiating $e^{-t}$ to get,

$ e^{-t}.\dfrac{t^z}{z} \bigg |_0^R + \dfrac{1}{z} \int_0^R t^ze^{-t}\,dt$.

When $R \rightarrow \infty$, the first term drops out and the second term becomes $\dfrac{1}{z}\Gamma(z+1)$. Thus we have proved for the right half plane.

We can now define the gamma function in the strip where $-1< \Re{z} < 0$ by the equation $\Gamma(z) = z^{-1}\Gamma(z+1)$. We can repeat this process to define in successive strips where $-n < \Re{z} < -n+1$.