# Derivative of Riemann zeta function

I'm trying to evaluate the derivative of the Riemann zeta function at the origin, $\zeta'(0)$, starting from its integral representation
$$\zeta(s)=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}\frac{1}{e^t-1}.$$
I don't want to use a symbolic algebra system like Mathematica or Maple.

I am able to continue to $s=0$ and evaluate the zeta function there $\zeta(0)=-1/2$. I'm just stuck on how to evaluate the derivative.

Can somebody show me how to do this starting from the integral representation? Thanks.

mathman
d/ds{ts-1/Γ(s)} = {(s-1)ts-2Γ(s) - ts-1Γ'(s)}/Γ(s)2=f(s,t). The integrand is f(s,t)/(et-1)

\begin{array}\\\zeta (s)=\frac{1}{\Gamma (s)}\int_{0}^{\infty}\frac{t^{s-1}}{e^{t-1}}dt\\
\frac{d}{ds}\zeta (s)=\frac{d}{ds}\frac{1}{\Gamma (s)}\int_{0}^{\infty}\frac{t^{s-1}}{e^{t-1}}dt+\frac{1}{\Gamma (s)}\frac{d}{ds}\int_{0}^{\infty}\frac{t^{s-1}}{e^{t-1}}dt=\frac{-1}{\Gamma^2 (s)}\frac{d}{ds}\Gamma (s)\int_{0}^{\infty}\frac{t^{s-1}}{e^{t-1}}dt+\frac{1}{\Gamma (s)}\int_{0}^{\infty}\frac{d}{ds}\frac{t^{s-1}}{e^{t-1}}dt\\

\zeta' (s)=\frac{-1}{\Gamma^2 (s)}\int_{0}^{\infty}t^{s-1}e^{-t}\ln{t}dt\int_{0}^{\infty}\frac{t^{s-1}}{e^{t-1}}dt+\frac{1}{\Gamma (s)}\int_{0}^{\infty}\frac{t^{s-1}}{e^{t-1}}\ln{t}dt\end{array}

Now put s=0 and integrate!

Hi thanks, but the -1 is not in the exponential.
Its $\frac{1}{e^t-1}$ not $\frac{1}{e^{t-1}}$.

Hi thanks, but the -1 is not in the exponential.
Its $\frac{1}{e^t-1}$ not $\frac{1}{e^{t-1}}$.
Oh yes you are right! My mistake

\begin{array}\\\zeta (s)=\frac{1}{\Gamma (s)}\int_{0}^{\infty}\frac{t^{s-1}}{e^t-1}dt\\
\frac{d}{ds}\zeta (s)=\frac{d}{ds}\frac{1}{\Gamma (s)}\int_{0}^{\infty}\frac{t^{s-1}}{e^t-1}dt+\frac{1}{\Gamma (s)}\frac{d}{ds}\int_{0}^{\infty}\frac{t^{s-1}}{e^t-1}dt=\frac{-1}{\Gamma^2 (s)}\frac{d}{ds}\Gamma (s)\int_{0}^{\infty}\frac{t^{s-1}}{e^t-1}dt+\frac{1}{\Gamma (s)}\int_{0}^{\infty}\frac{d}{ds}\frac{t^{s-1}}{e^t-1}dt\\

\zeta' (s)=\frac{-1}{\Gamma^2 (s)}\int_{0}^{\infty}t^{s-1}e^{-t}\ln{t}dt\int_{0}^{\infty}\frac{t^{s-1}}{e^t-1}dt+\frac{1}{\Gamma (s)}\int_{0}^{\infty}\frac{t^{s-1}}{e^t-1}\ln{t}dt\end{array}

Now it is correct

Ok, that sounds good; but now what? How do I do those integrals? If I set s=0, the second integral doesn't converge...

Ok, that sounds good; but now what? How do I do those integrals? If I set s=0, the second integral doesn't converge...
If you set s=0 (almost) all integrals(including Γ(s) and Γ'(s)) will diverge but a combination of them could actually converge.
My advice: calculate the integrals(you may need power series for some) and replace them with limits. Then, simplify everything and hope that the final limit won't be infinite

thanks. I've tried taking the power series. But in the end, the final sum doesn't converge. I'm afraid I'm going to need help till the bitter end on this one.