8.1.19 Prove que
|\Gamma(\alpha+i \beta)|=|\Gamma(\alpha)| \prod_{n=0}^{\infty}\left[1+\frac{\beta^{2}}{(\alpha+n)^{2}}\right]^{-1 / 2}
Essa equação tem sido útil em cálculos da teoria da degradação beta.

Passo 1

O enunciado pede para mostrar que, para \alpha>0 e \beta\in\mathbb{R}, vale
\bigl|\Gamma(\alpha+i\beta)\bigr| =\bigl|\Gamma(\alpha)\bigr| \prod_{n=0}^{\infty}\Bigl[1+\frac{\beta^{2}}{(\alpha+n)^{2}}\Bigr]^{-1/2}.
O que vamos precisar é da forma de Weierstrass para a função \Gamma, isto é, o produto infinito
\frac{1}{\Gamma(z)} = z\,e^{\gamma z} \prod_{n=1}^{\infty}\Bigl(1+\frac{z}{n}\Bigr)e^{-z/n},
onde \gamma é a constante de Euler–Mascheroni.

Passo 2

Substituímos z=\alpha+i\beta na representação de Weierstrass e tomamos o módulo em ambos os lados. De um lado,
\Bigl|\tfrac1{\Gamma(\alpha+i\beta)}\Bigr| =|\alpha+i\beta|\, \bigl|e^{\gamma(\alpha+i\beta)}\bigr|\, \prod_{n=1}^\infty \bigl|1+\tfrac{\alpha+i\beta}{n}\bigr|\, \bigl|e^{-(\alpha+i\beta)/n}\bigr|.
Agora use:
  1. \bigl|e^{\gamma(\alpha+i\beta)}\bigr|=e^{\gamma\alpha}.
  2. \bigl|e^{-(\alpha+i\beta)/n}\bigr|=e^{-\alpha/n}.
  3. \bigl|\alpha+i\beta\bigr|=\sqrt{\alpha^2+\beta^2}.
  4. \displaystyle\bigl|1+\tfrac{\alpha+i\beta}{n}\bigr| =\sqrt{\Bigl(1+\tfrac\alpha n\Bigr)^2+\tfrac{\beta^2}{n^2}} =\frac{\sqrt{(\alpha+n)^2+\beta^2}}{\,n\,}.

Passo 3

Com isso,
\Bigl|\tfrac1{\Gamma(\alpha+i\beta)}\Bigr| =\sqrt{\alpha^2+\beta^2}\;e^{\gamma\alpha} \prod_{n=1}^\infty \frac{\sqrt{(\alpha+n)^2+\beta^2}}{n}\;e^{-\alpha/n}.
Reescrevendo um pouco,
\Bigl|\tfrac1{\Gamma(\alpha+i\beta)}\Bigr| =\sqrt{\alpha^2+\beta^2}\;e^{\gamma\alpha} \prod_{n=1}^\infty \Bigl(1+\tfrac\alpha n\Bigr)e^{-\alpha/n} \;\prod_{n=1}^\infty \Bigl[1+\tfrac{\beta^2}{(\alpha+n)^2}\Bigr]^{1/2}.

Passo 4

Quando \beta=0, a mesma product formula dá
\frac1{\Gamma(\alpha)} =\alpha\,e^{\gamma\alpha} \prod_{n=1}^\infty\Bigl(1+\tfrac\alpha n\Bigr)e^{-\alpha/n}.
Logo
\frac{\bigl|\Gamma(\alpha)\bigr|}{\bigl|\Gamma(\alpha+i\beta)\bigr|} =\frac{\alpha\,e^{\gamma\alpha}\prod_{n=1}(1+\frac\alpha n)e^{-\alpha/n}} {\sqrt{\alpha^2+\beta^2}\,e^{\gamma\alpha}\prod_{n=1}(1+\frac\alpha n)e^{-\alpha/n}} \;\Bigl[\prod_{n=1}\bigl[1+\tfrac{\beta^2}{(\alpha+n)^2}\bigr]^{1/2}\Bigr]^{-1}.
Quase tudo cancela, sobrando
\frac{\bigl|\Gamma(\alpha)\bigr|}{\bigl|\Gamma(\alpha+i\beta)\bigr|} =\frac{\alpha}{\sqrt{\alpha^2+\beta^2}} \;\prod_{n=1}^\infty \Bigl[1+\tfrac{\beta^2}{(\alpha+n)^2}\Bigr]^{-1/2}.

Passo 5

Observe que
\displaystyle\frac{\alpha}{\sqrt{\alpha^2+\beta^2}} =\Bigl[1+\frac{\beta^2}{\alpha^2}\Bigr]^{-1/2}
é exatamente o termo n=0 no produto desejado. Portanto podemos escrever
\bigl|\Gamma(\alpha+i\beta)\bigr| =\bigl|\Gamma(\alpha)\bigr| \prod_{n=0}^{\infty} \Bigl[1+\frac{\beta^2}{(\alpha+n)^2}\Bigr]^{-1/2},
que é o que se queria demonstrar.

Resposta

\bigl|\Gamma(\alpha+i\beta)\bigr| =\bigl|\Gamma(\alpha)\bigr| \prod_{n=0}^{\infty}\Bigl[1+\tfrac{\beta^{2}}{(\alpha+n)^{2}}\Bigr]^{-1/2}.

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