Does someone has an idea how to calculate: $$\int_0^\zeta\frac{z^b}{(z-w)^2}dz$$ where $\zeta,w \in \mathbb C$? I need to compute this integral because I need to solve a differential equation on the complex plane which is: $$\frac{d}{dz}\left(\frac{z^b}{(z-w)} f(z)\right)=-\frac{z^b}{(z-w)^2} $$ Edit : $z$ is in the unit disk, $0<|w|<1$ and $b=c|w|^2$, $c>1$ is a real constant and I had to remove a half line passing through $0$ so that $z^b$ will be holomorphic.
2026-04-09 03:49:32.1775706572
Integral on complex plane
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$$\int\frac{z^b}{(z-w)^2}dz=\frac{x^{b+1}}{(b+1)w^2}\:_2\text{F}_1\left(2\:,\:b+1\:;\:b+2\:;\: \frac{z}{w}\right)+\text{constant}$$ $\:_2\text{F}_1(A,B;C;X)$ is the Gauss hypergeometric function. https://en.wikipedia.org/wiki/Hypergeometric_function $$\int_0^\zeta\frac{z^b}{(z-w)^2}dz=\frac{\zeta^{b+1}}{(b+1)w^2}\:_2\text{F}_1\left(2\:,\:b+1\:;\:b+2\:;\: \frac{\zeta}{w}\right)\qquad \text{if}\quad b>-1$$