The function $I(z)$ is defined by

$$I(z)=\frac{1}{\Gamma(z)} \int_{0}^{\infty} \frac{t^{z-1}}{e^{t}+1} d t$$

For what values of $z$ is $I(z)$ analytic?

By considering $I(z)-\zeta(z)$, where $\zeta(z)$ is the Riemann zeta function which you may assume is given by

$$\zeta(z)=\frac{1}{\Gamma(z)} \int_{0}^{\infty} \frac{t^{z-1}}{e^{t}-1} d t \quad(\operatorname{Re} z>1)$$

show that $I(z)=\left(1-2^{1-z}\right) \zeta(z)$. Deduce from this result that the analytic continuation of $I(z)$ is an entire function. [You may use properties of $\zeta(z)$ without proof.]