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So I was playing around with Euler's Reflection Formula ($\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)}$), trying to prove it with calculus, and was able to reduce $$ \Gamma(s)\Gamma(1-s)=2\int_0^{\pi/2}\tan^{2s-1}(x)\,\text dx\;\;(0<s<1) $$

But I have no idea how to show $$ 2\int_0^{\pi/2}\tan^{2s-1}(x)\,\text dx=\frac{\pi}{\sin(\pi s)} $$

I have been able to substitute to find that it is the same integral as $$ 2\int_1^\infty\frac{1}{x(x^2-1)^{1-s}}\text dx=\int_1^\infty\frac{1}{x(x-1)^{1-s}}\text dx $$ But can't get any further.

Any ideas?

Tim Ratigan
  • 7,365

1 Answers1

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ With $\ds{\sin\pars{x} \equiv t}$: \begin{align}&\color{#c00000}{% \int_{0}^{\pi/2}\tan^{2s - 1}\pars{x}\,\dd x} =\int_{0}^{1}{t^{2s - 1} \over \pars{1 - t^{2}}^{s - 1/2}} \,{\dd t \over \pars{1 - t^{2}}^{1/2}} =\int_{0}^{1}{t^{s - 1/2} \over \pars{1 - t}^{s}}\,\half\,t^{-1/2}\,\dd t \\[3mm]&=\half\int_{0}^{1}t^{s - 1}\pars{1 - t}^{-s}\,\dd t =\half\,{\Gamma\pars{s}\Gamma\pars{-s + 1} \over \Gamma\pars{1}} =\half\,{\pi \over \sin\pars{\pi s}} \end{align}

$$ \color{#44f}{\large 2\int_{0}^{\pi/2}\tan^{2s - 1}\pars{x}\,\dd x ={\pi \over \sin\pars{\pi s}}}\,,\qquad\qquad 0 < \Re\pars{s} < 1 $$

See this link.

Felix Marin
  • 94,079