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\begin{document}

\title{Zeta type Integrals}



\author[1]{Benedict Irwin}%
\affil[1]{University of Cambridge}%


\vspace{-1em}



  
  \date{May 07, 2024}


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\sloppy


Consider these integrals which all have a component
$$
\frac{1}{t(e^t-1)}
$$
and all related to zeta, gamma and polygamma type functions. This could be seen as a transform of a kernel function $k(s,t)$.

The classic integral is
$$
\Gamma(s)\zeta(s) = \int_0^\infty \frac{t^s}{(e^t-1)} \; \frac{dt}{t}
$$
a somewhat forced addition is
$$
\Gamma(s) = \int_0^\infty \frac{t^s(1-e^{-t})}{(e^t-1)} \; \frac{dt}{t}
$$
we have
$$
\log\Gamma(s+1) = \int_0^\infty \frac{(e^{-st}-1)+s(1-e^{-t})}{(e^t-1)} \; \frac{dt}{t}
$$
and
$$
\Gamma(n)\zeta(n,s+1) = \int_0^\infty \frac{t^n e^{-st}}{(e^t-1)} \; \frac{dt}{t}
$$
also the related polygamma function
$$
(-1)^{m+1}\psi^{(m)}(s)= \Gamma(m+1)\zeta(m+1,s) =  \int_0^\infty \frac{t^m t e^{-s t + t}}{ (e^t-1)} \; \frac{dt}{t}
$$
we have
$$
\Gamma(s)\psi^{(0)}(s)\zeta(s) + \Gamma(s)\zeta'(s) =  \int_0^\infty \frac{t^s \log(t)}{ (e^t-1)} \; \frac{dt}{t}
$$
$$
\Gamma(s)\psi^{(0)}(s)\zeta(s,n+1) + \Gamma(s)\zeta^{(1,0)}(s,n+1) =  \int_0^\infty \frac{t^s \log(t)e^{-nt}}{ (e^t-1)} \; \frac{dt}{t}
$$
also consider the Bernoulli numbers, with an integral $b(s)$ that takes the values $b(2n) = B_{2n}$ (Wikipedia)
$$
b(s) = \int_0^\infty \frac{st^s 2e^{s i \pi/2}}{1-e^{2\pi t}} \frac{(e^t-1)}{(e^t-1)}\frac{dt}{t}
$$
According to Woon [1], the Bernoulli polynomials should be redefined as 
$$
\frac{z e^{x z}}{e^z-1} = \left(\sum_{k=0}^\infty \frac{(-1)^k}{k!}B_k z^k\right)\left(\sum_{m=0}^\infty \frac{(xz)^m}{m!}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{n!} B_n(x) z^n
$$

The link is obviously very strong with the exponential generating functions for Bernoulli numbers
$$
\frac{t}{e^t-1} = \sum_{m=0}^\infty \frac{B^-_m t^m}{m!}
$$
$$
\frac{t}{1-e^{-t}} = \sum_{m=0}^\infty \frac{B^+_m t^m}{m!}
$$


\section*{References}

[1] https://arxiv.org/pdf/physics/9705021.pdf

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