I am trying to evaluate an integral of the form \begin{equation}\label{eq:1} \gamma(\tau)=\int_0^\infty\text{d}u\,e^{-\tau u}\Big(1+\frac1u\Big)\frac1{\log(u)^2+\pi^2} \end{equation} and decided to expand the exponential, so that I can write the solution as a power series $$\gamma(\tau)=\sum_{n=0}^\infty\,c_n\tau^n\ ,$$ where $$c_n=\frac{(-1)^n}{n!}\big(\varphi(n+1)+\varphi(n)\big)$$ and \begin{equation}\label{eq2} \varphi(s)=\int_0^\infty\text{d}u\,u^{s-1}\frac1{\log(u)^2+\pi^2}\equiv\big\{\mathcal{M}f\big\}(s) \end{equation} is the Mellin transform this question is about.
Though I have no proof for this, the first integral for $\gamma(\tau)$ seems to converge for all $\tau>0$. By rewriting it as a power series I exchanged summation and integration which is only a valid operation if the Mellin transform $\varphi$ exists for all $s\in\mathbb{N}_0$.
I think it does not exist for any $s$ with $\text{Re}(s)>0$ as the integrand in its definition diverges for large $u$ but WolframAlpha happily gives me the Mellin transform as $$\big\{\mathcal{M}f\big\}(s)=e^{-i\pi s}\ ,$$ implying that all $c_n$ vanish which, as numerical integration suggests, can not be the case.
My questions now is, what is the domain of the Mellin transform of $f$? Does the vanishing of the coefficients $c_n$ come down to the fact that I should not have exchanged summation and integration? Also any help in evaluating $\gamma(\tau)$ is greatly appreciated!