<Home>Qualifying Exams><Analysis>Spring 2012 - Problem 10

Spring 2012 - Problem 10

Gamma function

Let us define the Gamma function via

Γ(z)=∫0∞tz−1e−t dt,\Gamma\p{z} = \int_0^\infty t^{z-1} e^{-t} \,\diff{t},

at least when the integral is absolutely convergent. Show that this function extends to a meromorphic function in the whole complex plane. You cannot use any particular properties of the Gamma function unless you derive them from this definition.
Solution.

First, notice that if Re⁡z>0\Re{z} > 0, then

∫0∞∣tz−1∣e−t dt≤∫0∞tRe⁡z−1e−t dt<∞.\int_0^\infty \abs{t^{z-1}} e^{-t} \,\diff{t} \leq \int_0^\infty t^{\Re{z} - 1} e^{-t} \,\diff{t} < \infty.

In other words, the integral converges absolutely, and so it converges uniformly on compact sets. Thus, Γ\Gamma is a continuous function. Let R⊆{Re⁡z>0}R \subseteq \set{\Re{z} > 0} be a rectangle. Since the integrand is non-negative, we may apply Fubini-Tonelli to get

∫∂RΓ(z) dz=∫∂R∫0∞tz−1e−t dt dz=∫0∞∫∂Re(z−1)log⁡t dz dt=0.\int_{\partial R} \Gamma\p{z} \,\diff{z} = \int_{\partial R} \int_0^\infty t^{z-1} e^{-t} \,\diff{t} \,\diff{z} = \int_0^\infty \int_{\partial R} e^{\p{z-1}\log{t}} \,\diff{z} \,\diff{t} = 0.

Indeed, e(z−1)log⁡te^{\p{z-1}\log{t}} is holomorphic for any t>0t > 0, and so by Cauchy's theorem, ∫∂Re(z−1)log⁡t dz=0\int_{\partial R} e^{\p{z-1}\log{t}} \,\diff{z} = 0 for almost every tt. Thus, by Morera's theorem, we see that Γ\Gamma is holomorphic in the right half-plane.

For 0<ε<R0 < \epsilon < R, observe that

∫εRtz−1e−t dt=−tz−1e−t∣εR+(z−1)∫εRtz−2e−t dt.\int_\epsilon^R t^{z-1} e^{-t} \,\diff{t} = \Bigl. -t^{z-1}e^{-t} \Bigr\rvert_\epsilon^R + \p{z - 1}\int_\epsilon^R t^{z-2} e^{-t} \,\diff{t}.

Sending ε→0\epsilon \to 0 and R→∞R \to \infty, we get

Γ(z)=(z−1)Γ(z−1).\Gamma\p{z} = \p{z - 1}\Gamma\p{z - 1}.

Thus, for n∈Nn \in \N, this gives

Γ(z+n+1)=(z+n)Γ(z+n)=⋯=(z+n)⋯(z+1)zΓ(z)Γ(z)=Γ(z+n+1)(z+n)⋯(z+1)z.\begin{gathered} \Gamma\p{z + n + 1} = \p{z + n}\Gamma\p{z + n} = \cdots = \p{z + n} \cdots \p{z + 1} z\Gamma\p{z} \\ \Gamma\p{z} = \frac{\Gamma\p{z + n + 1}}{\p{z + n} \cdots \p{z + 1} z}. \end{gathered}

Thus, Γ(z)\Gamma\p{z} extends meromorphically to Re⁡(z+n+1)>0  ⟺  Re⁡z>−n−1\Re\p{z + n + 1} > 0 \iff \Re{z} > -n - 1 for any n∈Nn \in \N, except for poles at {−n,…,−2,−1,0}\set{-n, \ldots, -2, -1, 0}. Hence, Γ\Gamma extends meromorphically to the entire complex plane.