Calculating this $\int_{|z|=1}\frac{z^a(1+z)^m}{(c-\ln z)^b}dz$ complex integral Solving a problem related with series, it appear this integral in the complex plane
$$\int_{|z|=1}\frac{z^a(1+z)^m}{(c-\ln z)^b}dz$$
where $a\in\mathbb{R}$ ,$b,c>0$ and $m=0,1,2,...$ which I'm not able to solve.
I try to use the standar parametrization of $|z|=1$ or even expand $(1+z)^m$ but it seems not right.
Any help is welcomed.
 A: I think I have reached an interesting form for this integral in terms of a finite sum of special functions at least for $a>-1$. I do hope it doesn't revert back to the series you have been considering directly somehow!
I will prove that for $a,b >-1$

$$\begin{align}I&=\int_{|z|=1}\frac{z^a(1+z)^m}{(c-\ln z)^b}dz\\&\\&=\sum_{n=0}^m(-1)^n {m\choose n}\frac{e^{c(a+n+1)}}{(a+n+1)^{1-b}}\Big[\Gamma(1-b,(c+i\pi)(a+n+1))-\Gamma(1-b,(c-i\pi)(a+n+1))\Big]\end{align}\tag{1}$$

where $\Gamma(z,x)$ is the upper incomplete gamma function.
It is possible for this integral to be evaluated using it's branch cuts in the complex plane, but I think it's way more facile to consider the elementary identity
$$\frac{1}{(c-\ln z)^b}=
\frac{1}{\Gamma(b)}\int_{0}^{\infty}dy ~y^{b-1}e^{-(c-\ln z)y}$$
to rewrite the integral as
$$I=\int_{0}^{\infty}\frac{dy}{\Gamma(b)}y^{b-1}e^{-cy}\int_{|z|=1}z^{a+y}(1+z)^mdz$$
The interchange of integration and summation seems to be justified by Fubini's theorem, since the absolute value of this integral is definitely finite. For $a+y\in \mathbb{R}\setminus\mathbb{N}$ one can use a keyhole contour to show that
$$\int_{|z|=1}dz~z^{a+y}(1+z)^m+\lim_{\epsilon\to 0}\Bigg[2i\sin(\pi(a+y))\int_{\epsilon}^1dr~r^{a+y}(1-r)^m-i\int_{-\pi}^{\pi}d\theta~\epsilon ^{a+y+1}e^{i\theta(a+y+1)}(1+\epsilon e^{i\theta})^m\Bigg]=0$$
Now here's the tricky part: We are integrating over the variable $y$, which is positive. If $a+y>a>-1$ then the two summands enclosed in the limit below converge separately and the second one obviously converges to zero. We thus conclude that

$$\int_{|z|=1}dz~z^{a+y}(1+z)^m=-2i\sin(\pi(a+y))B(a+y+1,m+1)\tag{2}$$

However we note that $B(a+y+1,m+1)$ is a rational function of $y$ because
$$B(a+y+1,m+1)=\frac{\Gamma(m+1)\Gamma(a+y+1)}{\Gamma(m+a+y+2)}=\frac{m!}{(a+y+1)(a+y+2)...(a+y+m)(a+y+m+1)}$$
Keeping in mind that we can perform a partial fraction decomposition on this rational function, we seek to perform the integral
$$J(c,x)=\int_{0}^{\infty}dy\frac{y^{b-1}e^{-cy}}{y+x}$$
Taking a derivative with respect to $c$ we can easily show that
$$\frac{\partial J}{\partial c}=-\frac{\Gamma(b)}{c^b}+xJ\iff J(c,x)=\Gamma(b)\frac{e^{cx}}{x^{1-b}}\Gamma(1-b,cx)$$
where we used the fact that $\lim_{c\to\infty}J(c,x)=0$ which can be easily shown by an appropriate bound.
Now performing the partial fraction decomposition
$$\frac{m!}{(a+y+1)...(a+y+m+1)}=\sum_{n=0}^{m}\frac{A_n}{y+a+n+1}~~,~~ A_n=(-1)^n{m\choose n}$$
and substituting the value of the integral term by term and handling the sine appropriately we obtain the result.
EDIT:
Despite the limitations of the above proof it can actually be shown that the formula is general enough for all $a\in\mathbb{R}$. It's of key importance to note in order to generalize that eq. $(2)$ holds for all $a,y\in\mathbb{R}$ since it is an entire function in both:

$$\int_{|z|=1}dz~z^{a+y}(1+z)^m=\frac{2\pi i}{(m+1)B(a+y+m+2, -a-y)}\tag{2a}$$

Form $(2a)$ is not good for integration, however. To perform the integration it suffices to notice that the resulting poles coming from the partial fraction decomposition are annihilated by $\sin(\pi(a+y))$ because it has zeros precisely at $a+y=n\in \mathbb{Z}$. Then establish the following result
$$\begin{align}J(\lambda;b,c,x)&=\int_{0}^{\infty}dy\frac{y^{b-1}e^{-cy}\sin(\lambda(y+x))}{y+x}\\&=\frac{1}{2i}\frac{e^{cx}}{x^{1-b}}\Gamma(b)\Big[\Gamma(1-b,(c+i\lambda)x))-\Gamma(1-b,(c-i\lambda)x))\Big]\end{align}$$
Done!
