On A Relation of the Gamma Function to a Certain Condition

Question

I have researched to find an answer to this question to no avail. Does $$\Gamma \left( s \right) =- \int\limits_0^\infty \frac{t^{s - 1}}{(e^{2t}-e^{t})}\,dt$$ iff $Re(s) = \frac 1 2$? (Where $\Gamma \left( s \right)$ is normally defined as the well known gamma function $\Gamma \left( s \right) = \int\limits_0^\infty {t^{s - 1}e^{-t}\,dt}$). Please prove or find a counterexample.

Answer 1

The answer is negative. $$ \int_{0}^{+\infty}\frac{t^{s-1}e^{-t}}{e^{t}-1}\,dt = \Gamma(s)\left(\zeta(s)-1\right) $$ if $\text{Re}(s)>1$. The integrand function has a non-integrable singularity in a right neighbourhood of the origin if $\text{Re}(s)\leq 1$, hence it doesn't make sense to ask for the value of the LHS for $\text{Re}(s)=\frac{1}{2}$.