Estoy tratando de probar la identidad $$\Gamma(z) e^{i \pi z/2} = \int_0^\infty t^{z-1} e^{it}\, dt$$ for $0 < \Re(z) < 1$, starting from the integral definition of the gamma function $$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\, dt.$% $# %z de #%, o incluso discutiendo por serie de energía, aunque no sé cómo manejaría las dificultades de convergencia. Se agradecería cualquier punteros útiles.
Respuesta
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Que $D=\left\{z=re^{i\theta}\in\mathbb C|-\frac{3\pi}{4}\leq\theta\leq\frac{3\pi}{4}\right\}$ y considerar la función $f(w)=w^{z-1}e^{-w}$ $D$, donde elegimos la rama de $\mathrm{exp}((z-1)(\log|w|+i\mathrm{Arg} w))$ $w^{z-1}$. Que $\varepsilon,M>0$ $\varepsilon <M$ y considerar el contorno siguiente: $$\gamma_1(t)=t,\,\,t\in[\varepsilon, M]\\\gamma_2(t)=Me^{it},\,\,t\in\left[0,-\frac{\pi}{2}\right]\\\gamma_3(t)=ti,\,\,t\in[-M,-\varepsilon]\\\gamma_4(t)=\varepsilon e^{it},\,\,t\in\left[-\frac{\pi}{2},0\right].$$ $ f$ is holomorphic in $D$, therefore $$\int_{\gamma_1}f+\int_{\gamma_2}f=\int_{-\gamma_3}f+\int_{-\gamma_4}f.$$ Then, prove that the second and fourth integral converge to $0$ (for the second you will need that the real part of $z$ is less than $1$, for the fourth that it's greater than $0$), the first integral converges to $\Gamma(z) $, while the third converges to $% $ $e^{-\pi iz/2}\int_0^{\infty}t^{z-1}e^{it}\,dt.$
