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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Christian Ivicevic Jun 26 '16 at 14:17
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Jun 26 '16 at 14:33
  • $\begingroup$ @ChristianIvicevic: that is simple to show through De l'Hopital's theorem since $\cos(\pi n)=(-1)^n$. $\endgroup$ – Jack D'Aurizio Jun 26 '16 at 14:38
  • $\begingroup$ 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? $\endgroup$ – Christian Ivicevic Jun 26 '16 at 14:40
  • $\begingroup$ @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). $\endgroup$ – Jack D'Aurizio Jun 26 '16 at 14:43
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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 $$ \Gamma(z)\Gamma(1-z)\sin(\pi z) = \pi. $$

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