Derivative of Riemann zeta function

[TriTertButoxy]

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{t^s-1/Γ(s)} = {(s-1)t^s-2Γ(s) - t^s-1Γ'(s)}/Γ(s)²=f(s,t). The integrand is f(s,t)/(e^t-1)

 

[3.1415926535]

\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!

 

[TriTertButoxy]

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

 

[3.1415926535]

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

 

[TriTertButoxy]

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

 

[3.1415926535]

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

 

[TriTertButoxy]

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.