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I want to show that

$$ \lim_{s \to 1} \left( \zeta(s,a) - \frac{1}{s-1} \right) = - \psi(a)$$

where $\zeta(s,a)$ is the Hurwitz zeta function and $\psi(a)$ is the digamma function.

The only approach I can think of is to use the integral representation

$$\zeta(s,a) = 2 \int_{0}^{\infty} \frac{\sin (s \arctan \frac{t}{a} )}{(a^{2}+t^{2})^{s/2} (e^{2 \pi t}-1)} \, dt + \frac{1}{2a^{s}} + \frac{a^{1-s}}{s-1} $$ which is valid for all complex values of $s$.

$$ \begin{align} \lim_{s \to 1} \left( \zeta(s,a) - \frac{1}{s-1} \right) &= 2 \int_{0}^{\infty} \frac{\sin (\arctan \frac{t}{a} )}{(a^{2}+t^{2})^{1/2} (e^{2 \pi t}-1)} \, dt + \frac{1}{2a} \\ &= 2\int_{0}^{\infty} \frac{t}{(a^{2}+t^{2})(e^{2 \pi t}-1)} \, dt + \frac{1}{2a} \\ &= - 2 \, \frac{d}{d a} \int_{0}^{\infty} \frac{\arctan (\frac{t}{a})}{e^{2 \pi t}-1} \, dt + \frac{1}{2a}.\end{align}$$

Binet's second log gamma formula states

$$ 2 \int_{0}^{\infty} \frac{\arctan \left( \frac{x}{z} \right)}{e^{2 \pi x} -1} \, dx = \ln \Gamma(z) - \left( z- \frac{1}{2} \right) \ln z + z - \frac{\ln (2 \pi)}{2}.$$

So

$$ \begin{align} \lim_{s \to 1} \left( \zeta(s,a) - \frac{1}{s-1} \right) &= - \frac{d}{da} \left( \ln \Gamma(a) - \left( a- \frac{1}{2} \right) \ln (a) + a - \frac{\ln (2 \pi)}{2} \right) + \frac{1}{2a} \\ &= -\psi(a) +\ln (a) +1 -\frac{1}{2a} -1 + \frac{1}{2a} \\ &= - \psi(a) \color{#C00000}{+ \ln (a)} . \end{align} $$

Why do I have $\ln (a)$ in my final answer?

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Everything is correct except in the beginning where you take a limit of the integral representation. You conclude that $\lim_{s\rightarrow 1}\frac{a^{1-s}-1}{s-1}=1$, whereas by l'Hopital's rule, the correct answer is $-\ln(a)$ which should cancel the other log that you get later.

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