# Analytic continuation of Euler's reflection formula with the Gamma function

Let $\widetilde\Gamma$ be an analytic continuation of $\Gamma$ on $\mathbb C\setminus(-\mathbb N_0)$. Show that the function $$\widetilde\Gamma(z)\widetilde\Gamma(1-z)-\frac{\pi}{\sin(\pi z)}$$ can be analytically continued to an entire function.

I do assume that the analytic continuation is the classical $$\widetilde\Gamma(z)=\frac{\Gamma(z+n)}{z(z+1)\cdot\ldots\cdot(z+n-1)}$$ for $z\in\mathbb C\setminus(-\mathbb N_0),\operatorname{Re} z>-n$ with residues $$\operatorname{Res}_{-n}(\widetilde\Gamma)=\frac{(-1)^n}{n!}$$ with $n\in\mathbb{N}_0$ which I had to deduce in the excersice leading to this problem. I stumbled upon explanations on how to compute $\widetilde\Gamma(z)\widetilde\Gamma(1-z)$ using the Beta function I am not familiar with. Now I am curious as to what I have to do exactly with the given function and which methods there are available to do so.

The "standard" analytic continuation leads to to Euler's limit product formula, that leads to the Weierstrass product: $$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n} \tag{1}$$ from which: $$z\,\Gamma(z)\Gamma(-z)=\frac{1}{z}\prod_{n\geq 1}\left(1-\frac{z^{2}}{n^2}\right)^{-1}\tag{2}$$ and you may finish the work by recognizing the (reciprocal) Weierstrass product for a sine function in the RHS.

As an alternative, you may just prove that $\Gamma(z)\Gamma(1-z)-\frac{\pi}{\sin(\pi z)}$ has no singular point, since both terms have simple poles at the same points with the same residues: $$\text{Res}\left(\Gamma(z)\Gamma(1-z),z=-n\right)=(-1)^n,$$ $$\text{Res}\left(\frac{\pi}{\sin(\pi z)},z=-n\right)=\lim_{z\to n}\frac{\pi(z-n)}{\sin(\pi z)}\stackrel{DH}{=}\cos(\pi n)=(-1)^n.\tag{3}$$

Yet another way is to prove the red equality: $$\frac{d^2}{dz^2}\log(\Gamma(z)\Gamma(1-z))=\psi''(z)+\psi''(1-z)=\sum_{n\geq 0}\left(\frac{1}{(z+n)^2}+\frac{1}{(1-z+n)^2}\right)\color{red}{=}\frac{\pi^2}{\sin^2(\pi z)}=\frac{d^2}{dz^2}\log\frac{\pi}{\sin(\pi z)}\tag{4}$$ through Fourier series or other means.

• I'd rather keep it simple and stick to the argument that the function has no singular points - this isn't that obvious to me, though. For $\Gamma$ we would have to multiply $\Gamma(z)$ with every $(z+n)$ for each $\mathbb N_0\ni n< z$, wouldn't we? So how can it be that the expression will in fact be defined where $\sin(\pi z)$ is clearly difficult to handle. – Christian Ivicevic Jun 26 '16 at 14:17
• @ChristianIvicevic: from $\text{Res}(\Gamma(z),z=-n)=\frac{(-1)^n}{n!}$ it follows that $\text{Res}(\Gamma(z)\Gamma(1-z),z=-n)=(-1)^n$, so you just have to check that the residues of the cosecant match. – Jack D'Aurizio Jun 26 '16 at 14:33
• @ChristianIvicevic: that is simple to show through De l'Hopital's theorem since $\cos(\pi n)=(-1)^n$. – Jack D'Aurizio Jun 26 '16 at 14:38
• I never stumbled upon this argumentation - did I understand you correctly that $f(z) + g(z)$ is considered entire with no singularities when their residues match s.t. that they "cancel" out? If so, where does this result come from? – Christian Ivicevic Jun 26 '16 at 14:40
• @ChristianIvicevic: it is slightly different. Here we have two meromorphic functions $f(z)$ and $g(z)$ with simple poles at the integers. If their residues match, then $f(z)-g(z)$ is an entire function (as a meromorphic function without poles). – Jack D'Aurizio Jun 26 '16 at 14:43

The Laplace transform of $t^{-1+\alpha}$ for $\alpha > 0$ is given by the following for $s > 0$: \begin{align} \mathscr{L}\{t^{-1+\alpha}\} & = \int_{0}^{\infty}e^{-st}t^{-1+\alpha}dt \\ & = \int_{0}^{\infty}e^{-st}(st)^{-1+\alpha}d(st)\cdot s^{-\alpha} \\ & = \int_{0}^{\infty}e^{-u}u^{-1+\alpha}du\cdot s^{-\alpha}=\Gamma(\alpha)s^{-\alpha} \end{align} Because of this, \begin{align} \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\mathscr{L}\{ t^{-1+\alpha+\beta} \} & = \Gamma(\alpha)\Gamma(\beta)s^{-\alpha-\beta} \\ & = \Gamma(\alpha)s^{-\alpha}\Gamma(\beta)s^{-\beta} \\ & = \mathscr{L}\{ t^{-1+\alpha}\}\mathscr{L}\{ t^{-1+\beta}\} \\ & = \mathscr{L}\{ t^{-1+\alpha} \star t^{-1+\beta} \} \end{align} By uniqueness of transforms, the convolution of $t^{-1+\alpha}$ and $t^{-1+\beta}$ is given by $$t^{-1+\alpha}\star t^{-1+\beta} = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}t^{-1+\alpha+\beta},\;\;\; \alpha,\beta > 0.$$ If $0 < \alpha < 1$ and $\beta = 1-\alpha$, then $$t^{-1+\alpha}\star t^{-\alpha} = \Gamma(\alpha)\Gamma(1-\alpha)$$ Curiously, the above does not depend on $t > 0$. So, set $t=1$: $$\Gamma(\alpha)\Gamma(1-\alpha)=\int_{0}^{1}u^{-1+\alpha}(1-u)^{-\alpha}du.$$ By turning the integral on the right into a contour integral enclosing $[0,1]$ in its interior, you can show that $F(\alpha)=\sin(\pi\alpha)\Gamma(\alpha)\Gamma(1-\alpha)$ is a contour integral that has an entire extension. That's enough for what you want. In fact, you can then evaluate the contour integral by a single residue at $\infty$ to obtain the full identity $$\Gamma(z)\Gamma(1-z)\sin(\pi z) = \pi.$$