given the sum $$ \sum_{n=0}^\infty \frac{\exp(-nx)}{n+a} =f(x) $$
what would be the asymtptic of this series ??
for $a=1$ i believe this series goes as $ f(x) \sim \frac{1}{x}+ \gamma $
for every $a >0$ then $$ f(x)\sim \frac{1}{x}-\Psi (x) $$ digamma function
We will show that, for $a>0$ and $x$ in the neighborhood of $0^+$, we have $$f(x)=-\gamma-\psi(a)-\ln(x)+o(x)$$ where $\gamma$ is the Euler-Mascheroni constant, and $\psi=\Gamma'/\Gamma$ is the Digamma function.
Indeed, let us write $f$ as follows: $$\eqalign{ f(x)&=\frac{1}{a}+ \sum_{n=1}^\infty\frac{e^{-nx}}{n}+ \sum_{n=1}^\infty\left(\frac{1}{n+a}-\frac{1}{n}\right)e^{-nx}\cr &=\frac{1}{a}-\ln(1-e^{-x})+ \sum_{n=1}^\infty\left(\frac{1}{n+a}-\frac{1}{n}\right)+ \sum_{n=1}^\infty\left(\frac{1}{n+a}-\frac{1}{n}\right)(e^{-nx}-1)\cr &=-\gamma-\psi(a)-\ln(1-e^{-x})+a \sum_{n=1}^\infty\frac{1-e^{-nx} }{n(n+a)} } $$ This implies that $$ \vert f(x)+\gamma+\psi(a)+\ln(x)\vert\leq \ln\frac{x}{1-e^{-x}}+a g(a,x)\tag{1} $$ ($1-e^{-x}\leq x$,) with $$ g(a,x)= \sum_{n=1}^\infty\frac{1-e^{-nx} }{n(n+a)}\tag{2} $$ Now, for a positive integer $m$ we have $$\eqalign{ 0<g(a,x)&=\sum_{n=1}^m\frac{1-e^{-nx} }{n(n+a)}+ \sum_{n=m+1}^\infty\frac{1-e^{-nx} }{n(n+a)}\cr &\leq(1-e^{-mx})\sum_{n=1}^m\frac{1}{n(n+a)}+\sum_{n=m+1}^\infty\frac{1}{n(n+a)}\cr &\leq mx\sum_{n=1}^m\frac{1}{n^2}+\sum_{n=m+1}^\infty\frac{1}{n^2}\cr &\leq \frac{mx\pi^2}{6}+\int_{m}^\infty\frac{dt}{t^2}= \frac{mx\pi^2}{6}+\frac{1}{m} } $$ Now, for $0<x<1$, choosing $m=\lfloor 1/\sqrt{x}\rfloor$, we get $$ 0<g(a,x)\leq \left(\frac{\pi^2}{6}+\frac{1}{1-\sqrt{x}}\right)\sqrt{x} $$ This shows that $g(a,x)={\cal O}(\sqrt{x})$ in the neighborhood of $0^+$. Since clearly we have $\ln\frac{x}{1-e^{-x}}{\cal O}( x)$ in the neighborhood of $0^+$, we conclude from $(1)$ that $$ f(x)+\gamma+\psi(a)+\ln(x)={\cal O}(\sqrt{x}) $$ and the announced conclusion follows.