asymptotics of this sum $ x \to 0 $ 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
 A: 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.
