${\eta(s)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{e^x+1}dx}$ eşitliğini ispatlayın

Question

${\eta(s)}$ dirichlet eta fonksiyonu ve ${\Gamma(s)}$ gama fonksiyonu olmak üzere :

$${\large\eta(s)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{e^x+1}dx}$$Eşitliğini ispatlayın.

Answer 1

$\frac{1}{e^x + 1} = \frac{e^{-x}}{1 + e^{-x}} = \sum_{n=1}^{\infty} (-1)^{n-1} e^{-nx}$

$\int_0^\infty \frac{x^{s-1}}{e^x + 1} dx = \sum_{n=1}^{\infty} (-1)^{n-1} \int_0^\infty x^{s-1} e^{-nx} dx$

$\int_0^\infty x^{s-1} e^{-nx} dx = \frac{1}{n^s} \int_0^\infty t^{s-1} e^{-t} dt = \frac{\Gamma(s)}{n^s}$

$\int_0^\infty \frac{x^{s-1}}{e^x + 1} dx = \Gamma(s) \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}$

$\eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}$

$\int_0^\infty \frac{x^{s-1}}{e^x + 1} dx = \Gamma(s) \cdot \eta(s)$

$\boxed{\eta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x + 1} dx}$