calculating contour integral It would like to find an expression for 
$ \dfrac{1}{2 \pi i}\int \limits_{c-i \infty}^{c+i \infty} n^s (s-1)^{-e^{iu}}/s^2 ds$
 where $0<|u| \leq \pi/6$, so $\Re(e^{iu}) \geq 1/2$ and $\log(s-1) \in \mathbb{R}$ for $s>1$ and $c>1$, I
 have already proved (using Hankel contour)
$ \dfrac{1}{2 \pi i}\int \limits_{c-i \infty}^{c+i \infty} n^s (s-1)^{-e^{iu}}= n \log(n)^{1-e^{iu}}/\Gamma(e^{iu})ds$
where $\Gamma$ is the usual gamma Function. Some ideas? I'm not sure there is an expression for it.
 A: 
There was a typographical error in the OP.  Therefore, we will begin by developing the solution for the integral $\frac{1}{2\pi i}\int_{c-i \infty}^{c+i\infty} n^{z}(z-1)^{-a}\,dz$.  To that end, we proceed.


PART $1$:
First, let $g(n)$ be the function with integral representation
$$g(n)=\frac{1}{2\pi i}\int_{c-i \infty}^{c+i\infty} n^{z}(z-1)^{-a}\,dz \tag 1$$
where $n\in \mathbb{R}$, $n>0$, $c\in \mathbb{R}$, $c>1$, $a\in \mathbb{C}$, $|a|=1$ and $0<\arg(a)\le \pi/6$.  
We choose the branch cut that extends from $z=1$ to $z=-\infty$ along the real axis.  Then, we can deform the contour and write the integral in $(1)$ as
$$\begin{align}
g(n)&=\frac{n}{2\pi i} \left( \int_{0}^{-\infty} n^x e^{-a(\log(|x|)+i\pi)}\,dx-\int_{0}^{-\infty}n^x e^{-a(\log(|x|)-i\pi)}\,dx\right)\\\\
&=\frac{n\sin(a\pi)}{\pi}\int_0^\infty n^{-x}x^{-a}\,dx\\\\
&=\frac{n\sin(a\pi)}{\pi}\int_0^\infty e^{-x\log(n)}x^{-a}\,dx\\\\
&=\frac{n\sin(a\pi)}{\pi}\log^{a-1}(n)\int_0^\infty x^{-a}e^{-x}\,dx\\\\
&=\frac{n\sin(a\pi)}{\pi}\log^{a-1}(n)\Gamma(1-a)\\\\
&=\frac{n\log^{a-1}(n)}{\Gamma(a)}
\end{align}$$ 
Therefore, we find that
$$\bbox[5px,border:2px solid #C0A000]{\frac{1}{2\pi i}\int_{c-i \infty}^{c+i\infty} n^{z}(z-1)^{-a}\,dz=\frac{n\log^{a-1}(n)}{\Gamma(a)}}$$

PART $2$:
Next, let $f(n)$ be the function with integral representation
$$f(n)=\int_{c-i\infty}^{c+i\infty}\frac{n^z(z-1)^{-a}}{z^2}\,dz $$
Then, note that 
$$\begin{align}
n\frac{d}{dn}\left(n\frac{d}{dn}\right)f(n)&=n\frac{d}{dn}\left(n\frac{d}{dn}\right)\int_{c-i\infty}^{c+i\infty}\frac{n^z(z-1)^{-a}}{z^2}\,dz\\\\
&=\int_{c-i\infty}^{c+i\infty}n^z(z-1)^{-a}\,dz\\\\
&=g(n)\\\\
&=\frac{n\log^{a-1}(n)}{\Gamma(a)}
\end{align}$$
Now, solving the ODE
$$n\frac{d}{dn}\left(n\frac{d}{dn}\right)f(n)=\frac{n\log(n)^{a-1}}{\Gamma(a)}$$
we find
$$\bbox[5px,border:2px solid #C0A000]{\int_{c-i\infty}^{c+i\infty}\frac{n^z(z-1)^{-a}}{z^2}\,dz=\frac{1}{\Gamma(a)}\int_0^n \log^{a-1}(n')\log(n/n')\,dn'} \tag 2$$

NOTE:

We could enforce the substitution $n'=e^{-t}$ in $(2)$ and use the definition of the Incomplete Gamma Function to express the integral of interest in terms of the Incomplete Gamma Function.  We will leave that exercise to the interested reader.

