Riemann Zeta Function?

Question

The Riemann Zeta Function has, for $Re(z)>1$, the following integral

$\zeta(z)=\frac{1}{\Gamma(z)}\int_0^\infty dt\frac{t^{z-1}}{e^t-1}$

where $\Gamma(z)=\int_0^\infty dt t^{z-1}e^{-t}$ is the Euler Gamma Function.

Show that for $Re(z)>0$ the Riemann Zeta Function can be expressed as

$\zeta(z)=\frac{1}{(1-2^{1-z})\Gamma(z)}\int_0^\infty dt\frac{t^{z-1}}{e^t+1}$

I'm totally lost here. I don't know how I can derive that alternate expression for the Zeta Function just from $Re(z)>0$.

Any help would be appreciated.

Answer 1

Hint. One may observe that, for $t>0$, $$ \frac{t^{z-1}}{e^t+1}=\frac2{2^{z-1}}\frac{(2t)^{z-1}}{e^{2t}-1}-\frac{t^{z-1}}{e^t-1} $$ then one may integrate and use the above integral representation of the Riemann zeta function.

Answer 2

Noting that $$ \frac{t^{z-1}}{e^t+1}+\frac{t^{z-1}}{e^t-1}=\frac{2 e^t t^{z-1}}{e^{2 t}-1}, $$ then $$ \int_0^{\infty} \frac{t^{z-1}}{e^t+1} d t+\int \frac{t^{z-1}}{e^t-1} d t=\int_0^{\infty} \frac{2 e^t t^{z-1}}{e^{2 t}-1} d t $$ $$ \begin{aligned} \int_0^{\infty} \frac{2 e^t t^{z-1}}{e^{2 t}-1} d t = & 2 \int_0^{\infty} \frac{e^{-t} t^{z-1}}{1-e^{-2 t}} d z \\ = & 2 \sum_{n=0}^{\infty} \int_0^{\infty} t^{z-1} e^{-(2 n+1) t} d t \\ =&2 \sum_{n=0}^{\infty} \frac{\Gamma(z)}{(2 n+1)^z} \\ =&2 \Gamma(z)\left[\zeta(z)-\frac{1}{z^z} \zeta(z)\right] \\ =&\left(2-\frac{1}{2^{z-1}}\right) \Gamma(z) \zeta(z) \end{aligned} $$ Hence $$ \begin{aligned} \int_0^{\infty} \frac{t^{z-1}}{e^t+1} d t & =\left(2-\frac{1}{2^z}\right) \Gamma(z) \zeta(z)-\Gamma(z) \zeta(z )\\ & =\left(1-\frac{1}{2^z}\right) \Gamma(z) \zeta(z) \end{aligned} $$