I have a question about the following continuation of the Zeta function:
Consider
$$ \intop_{0}^{\infty}e^{-nt}t^{z-1}dt=\frac{1}{n^{z}}\intop_{0}^{\infty}e^{-t}t^{z-1}dt=\frac{\varGamma\left(z\right)}{n^{z}} $$
So that $$ \varGamma\left(z\right)\sum_{n=1}^{\infty}\frac{1}{n^{z}}=\intop_{0}^{\infty}t^{z-1}\left(\sum_{n=1}^{\infty}e^{-nt}\right)dt=\intop_{0}^{\infty}t^{z-1}\frac{1}{e^{t}-1}dt $$
And thus $$ \zeta\left(z\right)=\frac{1}{\varGamma\left(z\right)}\intop_{0}^{\infty}\frac{t^{z-1}}{e^{t}-1}dt $$
Or $$ \zeta\left(z\right)=\frac{1}{\varGamma\left(z\right)}\left[\intop_{0}^{1}\frac{t^{z-1}}{e^{t}-1}dt+\intop_{1}^{\infty}\frac{t^{z-1}}{e^{t}-1}dt\right] $$
Now, the zeta function suppose to have singularity at $ z=1 $ (and that supposed to be the only singularity)
But I dont understand how can we see that $$ \intop_{0}^{1}\frac{t^{z-1}}{e^{t}-1}\begin{cases} converge & z\neq1\\ diverge & z=1 \end{cases} $$
Seems like we'll always have a problem at $ t=0 $
I would really appreciate clarification here. Thanks in advance.
Assume that $\Re z>1$. Using the Maclaurin series of ${\frac{t}{{e^t - 1}}}$ and integrating term-by-term, we find $$ \frac{1}{{\Gamma (z)}}\int_0^1 {t^{z - 2} \frac{t}{{e^t - 1}}dt} = \frac{1}{{\Gamma (z)}}\frac{1}{{z - 1}} - \frac{1}{2}\frac{1}{{\Gamma (z + 1)}} + \frac{z}{{12\Gamma (z + 2)}} - \cdots . $$ Now the RHS can be extended analytically to the whole of $\mathbb C$ except $z=1$. The original integral, however, makes sense only for $\Re z>1$.