Asymptotic series of Confluent Hypergeometric function $U(a,1,z) $ as $z \to 0$ Consider the Confluent hypergeometric function $U(a,b,z)$, which is a solution of the Kummer's Equation : $$zw''+(b-z)w'-aw=0$$
it has the following integral representation when $- \pi/2 <  \arg z< \pi/2$ $$U(a,z)= \frac{1}{\Gamma(a)} \int^{\infty}_0 t^{a-1}(1+t)^{b-a-1} e^{-tz} dt$$
I know that there is the following asymptotic series for $U(a,1,z)$ : 
$$\mathop{U}\nolimits\!\left(a,1,z\right)=-\frac{1}{\mathop{\Gamma}\nolimits%
\!\left(a\right)}\left(\mathop{\ln}\nolimits z+\mathop{\psi}\nolimits\!%
\left(a\right)+2\gamma\right)+\mathop{O}\nolimits\!\left(z\mathop{\ln}%
\nolimits z\right) \ \ \ \ (*) \ \ \ \ z \to 0$$
http://dlmf.nist.gov/13.2.E19
Where $\gamma$ is the Euler constant and $\psi(z) = \displaystyle \frac{\Gamma'(z)}{\Gamma(z)}$ is the digamma function.
I would be very thankful if one can explain how should I start deriving asymptotic series $(*)$ 
 A: It might be easier to start with one of the other integral representations, but here's one way.  We'll begin with some manipulations to strip away the lower order terms:
$$
\begin{align}
&\int_0^\infty t^{a-1} (1+t)^{-a} e^{-zt}\,dt \\
&\qquad = \left(\int_0^1 + \int_1^\infty\right) t^{a-1} (1+t)^{-a} e^{-zt}\,dt \\
&\qquad = \int_0^1 t^{a-1} (1+t)^{-a}\,dt + O(z) + \int_1^\infty t^{a-1} (1+t)^{-a} e^{-zt}\,dt \\
&\qquad = \int_0^1 t^{a-1} (1+t)^{-a}\,dt + O(z)  + \int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1} + at^{-2}\right]e^{-zt}\,dt \\
&\qquad \qquad - a\int_1^\infty t^{-2}e^{-zt}\,dt + \int_1^\infty t^{-1} e^{-zt}\,dt. \tag{1}
\end{align}
$$
The last integral in the last line diverges logarithmically as $z \to 0^+$ while the other integrals converge.  Thus the leading-order behavior comes from the last integral.  To extract this behavior let's first make the change of variables $zt=s$, then integrate by parts:
$$
\begin{align}
\int_1^\infty t^{-1} e^{-zt}\,dt &= \int_z^\infty s^{-1} e^{-s}\,ds \\
&= e^{-s}\log s \Bigr|_z^\infty + \int_z^\infty e^{-s}\log s\,ds \\
&= -e^{-z}\log z + \int_0^\infty e^{-s}\log s\,ds - \int_0^z e^{-s}\log s\,ds.
\end{align}
$$
To get the last line we split the integral like $\int_z^\infty = \int_0^\infty - \int_0^z$.  Note that we could repeatedly integrate the last integral by parts to obtain further terms of the asymptotic expansion if we desire.  We only need the terms up to $O(z\log z)$, so it's enough to notice that


*

*$e^{-z}\log z = \log z + O(z\log z)$,

*$\int_0^\infty e^{-s}\log s\,ds = -\gamma$, and

*$\int_0^z e^{-s}\log s\,ds = O(z\log z)$ by L'Hopital's rule.
Thus
$$
\int_1^\infty t^{-1} e^{-zt}\,dt = -\log z - \gamma + O(z\log z). \tag{2}
$$
It remains to estimate the remaining integrals in $(1)$, namely
$$
\int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1} + at^{-2}\right]e^{-zt}\,dt \qquad \text{and} \qquad \int_1^\infty t^{-2}e^{-zt}\,dt.
$$
The integrand in the first of these is $O(t^{-3})e^{-zt}$, so we can apply the mean value theorem (i.e. $e^{-zt} = 1 -zte^{-zc(t)}$ with $0 < c(t) < t$) to conclude that
$$
\int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1} + at^{-2}\right]e^{-zt}\,dt = \int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1} + at^{-2}\right]dt + O(z).$$
$$
\tag{3}
$$
For the second integral we proceed as we did with the singular integral in $(2)$:
$$
\begin{align}
\int_1^\infty t^{-2}e^{-zt}\,dt &= \int_1^\infty t^{-2}\,dt + \int_1^\infty t^{-2}(e^{-zt}-1)\,dt \\
&= \int_1^\infty t^{-2}\,dt + z\int_z^\infty s^{-2}(e^{-s}-1)\,ds \\
&= \int_1^\infty t^{-2}\,dt + z\int_1^\infty s^{-2}(e^{-s}-1)\,ds - z\int_1^z s^{-2} (e^{-s}-1)\,ds.
\end{align}
$$
Now
$$
\int_1^z s^{-2} (e^{-s}-1)\,ds = O(\log z)
$$
by L'Hopital's rule, so
$$
\int_1^\infty t^{-2}e^{-zt}\,dt = \int_1^\infty t^{-2}\,dt + O(z\log z). \tag{4}
$$
Combining $(3)$ and $(4)$ we get
$$
\begin{align}
&\int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1} + at^{-2}\right]e^{-zt}\,dt - a\int_1^\infty t^{-2}e^{-zt}\,dt \\
&\qquad = \int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1}\right]dt + O(z\log z).
\end{align}
$$
Substituting this and $(2)$ into the first equation we got, we conclude that

$$
\int_0^\infty t^{a-1} (1+t)^{-a} e^{-zt}\,dt = -\log z - \gamma + f(a) + O(z\log z),
$$
  where
  $$
f(a) = \int_0^1 t^{a-1} (1+t)^{-a}\,dt + \int_1^\infty \left[t^{a-1} (1+t)^{-a} - t^{-1}\right]dt.
$$

It shouldn't be hard to show that
$$
f(a) = \int_0^1 \frac{t^{a-1}-1}{1-t}\,dt = -\psi(a)-\gamma,
$$
from which the result would follow.
