I was looking around the internet for a simple(r) proof of the Gamma Reflection Formula. I found this: Detailed explanation of the Γ reflection formula understandable by an AP Calculus student, and did not understand the last integration: $$\displaystyle\int\limits_0^\infty\frac{v^{z-1}}{v+1} dv$$

Can anyone help me? Explanations understandable by an AP Calculus student would be great!


Splitting the integral as $\int_0^\infty \frac{v^{z-1}}{1+v}\,dv=\int_0^1\frac{v^{z-1}}{1+v}\,dv+\int_1^\infty \frac{v^{z-1}}{1+v}\,dv$, and enforcing the substation $v\to 1/v$ in the integral that extends from $1$ to $\infty$, expanding $\frac1{1+v}$ as $\sum_{n=0}^\infty (-1)^nv^n$, and interchanging the order of the series and the integral, we can write

$$\begin{align} \int_0^\infty \frac{v^{z-1}}{1+v}\,dv&=\int_0^1\frac{v^{z-1}}{1+v}\,dv+\int_1^\infty \frac{v^{z-1}}{1+v}\,dv\\\\ &=\int_0^1\frac{v^{z-1}+v^{-z}}{1+v}\,dv\\\\ &=\sum_{n=0}^\infty (-1)^n \int_0^1 (v^{n+z-1}+v^{n-z})\,dv\\\\ &=\sum_{n=0}^\infty (-1)^n\left(\frac{1}{n+z}+\frac{1}{n+1-z}\right) \tag 1\\\\ &=\frac{\pi}{\sin(\pi z)} \end{align}$$

where I showed in the appendix of THIS ANSWER using real analysis methods only that $(1)$ is the partial fraction expansion of $\pi \csc(\pi z)$.

  • $\begingroup$ Can you elaborate on your steps? I still don't understand how you got from the 0 to 1 + 1 to infinity integral to the 0 to 1 integral. $\endgroup$ – D.R. Feb 27 '17 at 22:20
  • $\begingroup$ @D.R. MV uses the substitution $v\to\frac{1}{v}$ in the second integral. In other words, make the substitution $v=\frac{1}{u}$. Then $$\int_1^\infty \frac{v^{z-1}}{1+v}\,dv = \int_1^0\frac{u^{1-z}}{1+1/u}\,\left(\frac{-du}{u^2}\right) = \int_0^1\frac{u^{-z}}{u+1}\,du.$$ Now rename $u\to v$. $\endgroup$ – zahbaz Feb 27 '17 at 22:32
  • $\begingroup$ @D.R. More importantly, combining the two integrals with the limits of $v\in(0,1)$ permits the next line: a geometric series $\frac{1}{1+v} = \sum_{n=0}^\infty (-v)^n$. Cool stuff. $\endgroup$ – zahbaz Feb 27 '17 at 22:42
  • $\begingroup$ @D.R. Enforcing the substitution $v\to 1/v$ so that $dv\to -\frac1{v^2}\,dv$ and the limits transform from $0$ to $1$ to $\infty$ to $1$. The negative sign on the transformed differential can be absorbed by flipping the new integration limits so they go from $0$ to $1$. $\endgroup$ – Mark Viola Feb 27 '17 at 23:24
  • 1
    $\begingroup$ @D.R. Note that $\frac{1}{1+v}=\sum_{n=0}^\infty (-1)^n v^n$ for $|v|<1$. We need to ensure that the interchange of operations (i.e., the series and the integral) is justified. We can do so by using the Dominated Convergence Theorem. Then, we can write $$\int_0^1 (v^{z-1}+v^{-z})\sum_{n=0}^\infty (-1)^nv^n\,dv=\sum_{n=0}^\infty (-1)^n \int_0^1 (v^{z-1}+v^{-z})v^n\,dv$$ $\endgroup$ – Mark Viola Feb 28 '17 at 4:10

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\pars{~\Re\pars{z - 1} > - 1\ \mbox{and}\ \Re\pars{z - 1} < 0~} \implies \bbx{\ds{0 < \Re\pars{z} < 1}}}$:

\begin{align} \int_{0}^{\infty}{v^{z - 1} \over v + 1}\,\dd v & \,\,\,\stackrel{t\ =\ 1/\pars{v + 1}}{=}\,\,\, \int_{1}^{0}t\,\pars{{1 \over t} - 1}^{z - 1}\pars{-\,{1 \over t^{2}}}\dd t = \int_{0}^{1}t^{-z}\,\pars{1 - t}^{z - 1}\,\dd t \\[5mm] & = {\Gamma\pars{-z + 1}\Gamma\pars{z} \over \Gamma\pars{1}} = \bbx{\ds{\pi \over \sin\pars{\pi z}}} \end{align}


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